# structures

id: 0
type: category
name: empty category
notation: $\0$
nlab_link: https://ncatlab.org/nlab/show/empty+category
description:
This is the category with no objects and no morphisms. It is the initial object in the category of small categories.


id: 1
type: category
name: trivial category
notation: $\1$
nlab_link: https://ncatlab.org/nlab/show/terminal+category
description:
This is the simplest category, consisting of a single object $0$ and its identity morphism $\id_0 : 0 \to 0$. A concrete representation is the full subcategory of $\Set$ consisting of the empty set. It is the terminal object in the category of small categories.


id: 2
type: category
name: discrete category on two objects
notation: $\2$
description:
A concrete representation is the full subcategory of <a href="/category/CRing">$\CRing$</a> consisting of the two fields $\IF_2$ and $\IF_3$.


id: Alg(R)
type: category
name: category of algebras
notation: $\Alg(R)$
nlab_link: https://ncatlab.org/nlab/show/Alg
description:
This is a generalization of the category of rings, which we get for $R = \IZ$. We assume our rings (and algebras) to be unital. For $R = 0$ we would get the trivial category, which is why we exclude this here.


id: B
type: category
name: category of finite sets and bijections
notation: $\IB$
nlab_link: https://ncatlab.org/nlab/show/permutation+groupoid
description:
This category is also known as the permutation groupoid. It appears in the definition of a combinatorial species.


id: BG_c
type: category
name: delooping of an infinite countable group
notation: $BG$
nlab_link: https://ncatlab.org/nlab/show/delooping
description:
Every group $G$ yields a groupoid $BG$ with a single object $*$, morphisms given by the elements of $G$, and composition given by the group operation. In this example, we consider the case of an infinite countable group $G$ (such as $G = \IZ$).


id: BG_f
type: category
name: delooping of a non-trivial finite group
notation: $BG$
nlab_link: https://ncatlab.org/nlab/show/delooping
description:
Every group $G$ yields a groupoid $BG$ with a single object $*$, morphisms given by the elements of $G$, and composition given by the group operation. In this example, we consider the case of a non-trivial finite group $G$ (such as $G = C_2$).


id: BG_u
type: category
name: delooping of an infinite uncountable group
notation: $BG$
nlab_link: https://ncatlab.org/nlab/show/delooping
description:
Every group $G$ yields a groupoid $BG$ with a single object $*$, morphisms given by the elements of $G$, and composition given by the group operation. In this example, we consider the case of an uncountable group $G$ (such as $G = \IR$).


id: BN
type: category
name: delooping of the additive monoid of natural numbers
notation: $B\IN$
description:
Every monoid $M$ induces a category $BM$ with a single object $*$, morphisms given by the elements of $M$, and composition given by the monoid operation. Some of the properties of this category depend on the specific monoid. In this example, we take the commutative monoid $M = (\IN,+,0)$, so composition is $n \circ m = n + m$.


id: BOn
type: category
name: delooping of the additive monoid of ordinal numbers
notation: $B\On$
description:
Every monoid $M$ induces a category $BM$ with a single object $*$. This also works when $M$ is large, in which case $BM$ is not locally small. In this example, we apply this construction to the large monoid of ordinal numbers with respect to addition, so composition is $\alpha \circ \beta = \alpha + \beta$.


id: CAlg(R)
type: category
name: category of commutative algebras
notation: $\CAlg(R)$
nlab_link: https://ncatlab.org/nlab/show/CommAlg
description:
This is a generalization of the category of commutative rings, which we get for $R = \IZ$. In general, $\CAlg(R) \cong R \,/\, \CRing$. We assume our rings (and algebras) to be unital. For $R = 0$ we would get the trivial category, which is why we exclude this here.


id: Cat
type: category
name: category of small categories
notation: $\Cat$
nlab_link: https://ncatlab.org/nlab/show/Cat
description:
This is the category of small categories and functors between them. It is the prototype of a 2-category, but here we only treat it as a 1-category.


id: CompHaus
type: category
name: category of compact Hausdorff spaces
notation: $\CompHaus$
nlab_link: https://ncatlab.org/nlab/show/compact+Hausdorff+space
description:
This is the full subcategory of $\Top$ consisting of those spaces that are <a href="https://en.wikipedia.org/wiki/Compact_space" target="_blank">compact</a> and <a href="https://en.wikipedia.org/wiki/Hausdorff_space" target="_blank">Hausdorff</a>.


id: Delta
type: category
name: simplex category
notation: $\Delta$
nlab_link: https://ncatlab.org/nlab/show/simplex+category
description:
The simplex category is a skeleton of $\FinOrd \setminus \{\varnothing\}$. It plays an important role in topology and is used to define the <a href="/category/sSet">category of simplicial sets</a>.


id: FS
type: category
name: category of finite sets and surjections
notation: $\FS$
description:
This category is badly-behaved in itself, but it appears in representation theory. It has two connected components, consisting of the empty set and the non-empty finite sets.


id: FiltVect
type: category
name: category of filtered vector spaces
notation: $\FiltVect_K$
nlab_link: https://ncatlab.org/nlab/show/filtered+vector+space
description:
The definition of a <a href="https://stacks.math.columbia.edu/tag/0121" target="_blank">filtered object</a> is available in any abelian category; in this entry, the ambient category is $\Vect_K$. We do <i>not</i> require the filtration to be exhaustive (i.e. $V = \bigcup_n F^n(V)$) or separated (i.e. $\bigcap_n F^n(V) = 0$). In the proofs below, we will often write $F$ for the filtration regardless of the vector space on which it is defined.


id: FinOrd
type: category
name: category of finite ordered sets
notation: $\FinOrd$
nlab_link: https://ncatlab.org/nlab/show/total+order
description:
The finite ordered sets of the form $\{1 < \dotsc < n\}$ for $n \in \IN$ provide a skeleton (including the empty set for $n = 0$), the augmented simplex category.


id: FinVect_c
type: category
name: category of finite-dimensional vector spaces [countable field]
notation: $\FinVect_K$
nlab_link: https://ncatlab.org/nlab/show/FinDimVect
description:
This is the full subcategory of $\Vect_K$ consisting of finite-dimensional vector spaces. Every object is isomorphic to $K^n$ for a unique $n \in \IN$. In this entry, we assume that $K$ is countably infinite (e.g. $K = \IQ$) in order to determine all properties of this category.


id: FinVect_f
type: category
name: category of finite-dimensional vector spaces [finite field]
notation: $\FinVect_K$
nlab_link: https://ncatlab.org/nlab/show/FinDimVect
description:
This is the full subcategory of $\Vect_K$ consisting of finite-dimensional vector spaces. Every object is isomorphic to $K^n$ for a unique $n \in \IN$. In this entry, we assume that $K$ is finite (e.g. $K = \IF_2$) in order to determine all properties of this category.


id: FinVect_u
type: category
name: category of finite-dimensional vector spaces [uncountable field]
notation: $\FinVect_K$
nlab_link: https://ncatlab.org/nlab/show/FinDimVect
description:
This is the full subcategory of $\Vect_K$ consisting of finite-dimensional vector spaces. Every object is isomorphic to $K^n$ for a unique $n \in \IN$. In this entry, we assume that $K$ is uncountable (e.g. $K = \IR$) in order to determine all properties of this category.


id: Grp_c
type: category
name: category of countable groups
notation: $\Grp_\c$
description:
A group is called countable if its underlying set is countable. In particular, every finite group is countable, but also every finitely generated group is countable.


id: Haus
type: category
name: category of Hausdorff spaces
notation: $\Haus$
nlab_link: https://ncatlab.org/nlab/show/Hausdorff+space
description:
This is the full subcategory of $\Top$ consisting of those spaces that are <a href="https://en.wikipedia.org/wiki/Hausdorff_space" target="_blank">Hausdorff</a>.


id: LRS_R
type: category
name: category of locally ringed spaces
notation: $\LRS_R$
nlab_link: https://ncatlab.org/nlab/show/locally+ringed+topological+space
description:
Here, $R$ is a fixed non-trivial commutative ring. The general properties of this category will not depend on the choice of $R$. For $R = \IZ$ we obtain the category $\LRS$ of locally ringed spaces.


id: M-Set
type: category
name: category of M-sets
notation: $M{-}\Set$
nlab_link: https://ncatlab.org/nlab/show/MSet
description:
Here, $M$ can be any monoid. But the most important special case is that of a group. To settle (future) non-properties, we assume that $M$ is non-trivial, since otherwise we just get the category <a href="/category/Set">$\Set$</a>.


id: Man
type: category
name: category of smooth manifolds
notation: $\Man$
nlab_link: https://ncatlab.org/nlab/show/Diff
description:
Here, a smooth manifold is defined as a second-countable Hausdorff space with a smooth atlas. The dimension is locally constant, not necessarily constant.


id: Meas
type: category
name: category of measurable spaces
notation: $\Meas$
nlab_link: https://ncatlab.org/nlab/show/Meas
description:
This is very similar to the category of topological spaces. Accordingly, limits and colimits can be constructed in the same way.


id: Met
type: category
name: category of metric spaces with non-expansive maps
notation: $\Met$
nlab_link: https://ncatlab.org/nlab/show/Met
description:
In contrast to continuous maps, which only refer to the induced topology, non-expansive maps are closer related to the metrics themselves. This category is badly-behaved, though, especially when compared with $\Met_{\infty}$.


id: Met_c
type: category
name: category of metric spaces with continuous maps
notation: $\Met_c$
nlab_link: https://ncatlab.org/nlab/show/metrisable+topological+space
description:
This category is equivalent to the subcategory of $\Top$ (or $\Haus$) that consists of metrizable topological spaces. Hence, the metrics only play a secondary role here.


id: Met_oo
type: category
name: category of metric spaces with ∞ allowed
notation: $\Met_{\infty}$
nlab_link: https://ncatlab.org/nlab/show/Met
description:
The fact that we allow $\infty$ means that universal constructions work much better when compared to $\Met$.


id: Mono
type: category
name: category of sets with a distinguished subset
notation: $\Mono$
nlab_link: https://ncatlab.org/nlab/show/M-category#def
description:
This is equivalent to the full subcategory of objects $(X, Y, f)$ of <a href="/category/Set_arrow">$\Set^{\rightarrow}$</a> where $f : X \to Y$ is an injective function, i.e. a monomorphism. This explains our notation.


id: N
type: category
name: poset of natural numbers
notation: $(\IN,\leq)$
description:
This can also be seen as the path category of the infinite linear graph $\bullet \to \bullet \to \bullet \to \cdots$.


id: Pos
type: category
name: category of posets
notation: $\Pos$
nlab_link: https://ncatlab.org/nlab/show/Pos
description:
Even though there are many similarities with $\Prost$, the main difference is that the forgetful functor $\Pos \to \Set$ has no right adjoint.


id: Prost
type: category
name: category of prosets
notation: $\Prost$
nlab_link: https://ncatlab.org/nlab/show/Prost
description:
Even though there are many similarities with $\Pos$, the main difference is that the forgetful functor $\Prost \to \Set$ has a right adjoint, mapping $X$ to $(X , X \times X)$ (chaotic preorder).


id: R-Mod
type: category
name: category of left modules over a ring
notation: $R{-}\Mod$
nlab_link: https://ncatlab.org/nlab/show/module
description:
This is the prototype of an abelian category. The category of right modules is the same with the opposite ring $R^{\op}$, hence not listed here.
To settle the unsatisfied properties, we make the assumption that $R$ is <i>not</i> semisimple: If $R$ is semisimple, then by the Artin-Wedderburn theorem, the category is equivalent to a finite direct product of categories $D{-}\Mod$ for division rings $D$, and the case of division rings is in a separate entry. In particular, $R \neq 0$ and $R$ is not a field.


id: R-Mod_div
type: category
name: category of left modules over a division ring
notation: $R{-}\Mod$
nlab_link: https://ncatlab.org/nlab/show/module
description:
Here, we assume that $R$ is a non-commutative division ring, i.e. a skew-field which is not a field. The category of modules behaves mostly the same as in the commutative case.


id: Rel
type: category
name: category of sets and relations
notation: $\Rel$
nlab_link: https://ncatlab.org/nlab/show/Rel
description:
This category is self-dual as it can be: There is an isomorphism $\Rel \cong \Rel^{\op}$ that is the identity on objects and maps a relation to its opposite relation. It is the prototype of a dagger-category.


id: Sch_R
type: category
name: category of schemes
notation: $\Sch_R$
nlab_link: https://ncatlab.org/nlab/show/scheme
description:
Here, $R$ is a fixed non-trivial commutative ring. The general properties of this category will not depend on the choice of $R$. For $R = \IZ$ we obtain the category $\Sch$ of schemes.


id: SemiGrp
type: category
name: category of semigroups
notation: $\SemiGrp$
nlab_link: https://ncatlab.org/nlab/show/semigroup
description:
In contrast to monoids, semigroups do not need to have a neutral element, and in fact, they can be empty. This small difference has a huge impact on the categorical properties. For example, we do not have a zero object anymore.


id: Set
type: category
name: category of sets
notation: $\Set$
nlab_link: https://ncatlab.org/nlab/show/Set
dual_structure_id: Set_op
description:
The category of sets plays a fundamental role in category theory. Due to the Yoneda embedding, many results about general categories can be reduced to the category of sets. It is also usually the first example of a category that one encounters.


id: Set_arrow
type: category
name: category of set functions and commutative squares
notation: $\Set^{\rightarrow}$
nlab_link: https://ncatlab.org/nlab/show/Sierpinski+topos
description:
This category is an example of the <a href="https://ncatlab.org/nlab/show/arrow+category" target="_blank">arrow category</a> $\Arr(\C)$, where $\C$ is the category of sets. It is also known as the Sierpinski topos, since it is equivalent to the category of sheaves on the Sierpinski space.


id: Set_c
type: category
name: category of countable sets
notation: $\Set_\c$
description:
A set is countable if it admits a surjection from $\IN$. In particular, every finite set is countable.


id: Set_f
type: category
name: category of sets with finite-to-one maps
notation: $\Set_\f$
description:
In this variant of $\Set$ we only consider maps with finite fibers, which are commonly called <i>finite-to-one</i>. Equivalently, every preimage of a finite set is again finite, and this description makes it obvious that composition is well-defined.


id: Set_op
type: category
name: dual of the category of sets
notation: $\Set^{\op}$
nlab_link: https://ncatlab.org/nlab/show/opposite+category
dual_structure_id: Set
description:
By definition, this category is the dual (or opposite) of the category <a href="/category/Set">$\Set$</a>.


id: Set_*
type: category
name: category of pointed sets
notation: $\Set_*$
nlab_link: https://ncatlab.org/nlab/show/pointed+set
description:
This is the category of sets with a distinguished element, often called the base point. A map is called pointed when it preserves the base point.


id: Setne
type: category
name: category of non-empty sets
notation: $\Setne$
nlab_link: https://ncatlab.org/nlab/show/inhabited+set
description:
This entry demonstrates that removing an object (the empty set) can drastically change the properties of a category. In particular, this category is neither complete nor cocomplete.


id: SetxSet
type: category
name: category of pairs of sets
notation: $\Set \times \Set$
description:
This is an example of the <a href="https://ncatlab.org/nlab/show/product+category" target="_blank">product of categories</a>. It inherits most (but not all) properties from $\Set$. It can also be seen as the category $\Sh(1+1)$ of sheaves on a discrete space with two points, and also as the slice category $\Set/(1+1)$.


id: Sh(X)
type: category
name: category of sheaves
notation: $\Sh(X)$
nlab_link: https://ncatlab.org/nlab/show/category+of+sheaves
description:
Here, we assume that the topological space $X$ is neither discrete nor indiscrete, since otherwise this category is just a product of copies of $\Set$. Another valid notation is $\Sh(X,\Set)$.


id: Sh(X,Ab)
type: category
name: category of abelian sheaves
notation: $\Sh(X,\Ab)$
nlab_link: https://ncatlab.org/nlab/show/sheaf+of+abelian+groups
description:
Here, we assume that the topological space $X$ is neither discrete nor indiscrete, since otherwise this category is just a product of copies of $\Ab$.


id: Sp
type: category
name: category of combinatorial species
notation: $\Sp$
nlab_link: https://ncatlab.org/nlab/show/species
description:
Most categorical properties are immediately inferred from <a href="/category/FinSet">$\FinSet$</a>. Notice that this category is not locally small; it is just equivalent to a locally small category.


id: Top_op
type: category
name: dual of the category of topological spaces
notation: $\Top^{\op}$
nlab_link: https://ncatlab.org/nlab/show/opposite+category
dual_structure_id: Top
description:
By definition, this category is the dual (or opposite) of the category <a href="/category/Top">$\Top$</a>.


id: Top_*
type: category
name: category of pointed topological spaces
notation: $\Top_*$
nlab_link: https://ncatlab.org/nlab/show/pointed+topological+space
description:
This category plays an important role in algebraic topology and homotopy theory. Although it may appear similar to $\Top$, adding a base point drastically changes its categorical properties. In particular, it introduces a zero object.


id: TorsFreeAb
type: category
name: category of torsion-free abelian groups
notation: $\TorsFreeAb$
nlab_link: https://ncatlab.org/nlab/show/torsion-free+module
description:
This is a typical example of a well-behaved additive category which is <i>not</i> abelian. It contains the category of free abelian groups.


id: Vect
type: category
name: category of vector spaces
notation: $\Vect_K$
nlab_link: https://ncatlab.org/nlab/show/Vect
description:
This is a special case of the category of modules over a ring, where the ring is a field. It is the prototype of a split abelian category.


id: Z
type: category
name: category of Z-functors
notation: $[\CRing, \Set]$
description:
This category is used in functorial algebraic geometry. It also provides a typical example of a functor category that is not locally small, but nevertheless relevant. Most of its properties are directly derived from the category of sets, so other functor categories $[\C, \Set]$ for large categories $\C$ will be similar.


id: Z_div
type: category
name: proset of integers w.r.t. divisibility
notation: $(\IZ,\mid)$
description:
This is a proset, not a poset, because $a$ and $-a$ divide each other, but are not equal for $a \neq 0$. Notice that this category is equivalent (but not isomorphic) to $(\IN,\mid)$.


id: real_interval
type: category
name: poset [0,1]
notation: $([0,1],\leq)$
nlab_link: https://ncatlab.org/nlab/show/interval
description:
Every poset can be regarded as a small thin category. This is a specific example. This category is locally $\aleph_1$-presentable (in fact, <i>every</i> object is $\aleph_1$-presentable), but not locally finitely presentable (in fact, only $0$ is finitely presentable).


id: walking_commutative_square
type: category
name: walking commutative square
notation: $\Square$
nlab_link: https://ncatlab.org/nlab/show/commutative+square
description:
This category consists of a commutative square: $$\begin{CD} a @>>> b \\ @VVV @VVV \\ c @>>> d \end{CD}$$ Its name comes from the fact that a functor $\Square \to \C$ is the same as a <a href="https://ncatlab.org/nlab/show/commutative+square" target="_blank">commutative square</a> in $\C$. Notice that the category is isomorphic to the product category $I \times I$ of the <a href="/category/walking_morphism">walking morphism</a> with itself. Hence, most (but not all) properties are inherited from it.


id: walking_composable_pair
type: category
name: walking composable pair
notation: $\Comp$
nlab_link: https://ncatlab.org/nlab/show/composable+pair
description:
This category can be pictured as: $$\{ 0 \to 1 \to 2 \}$$ Its name comes from the fact that a functor $\Comp \to \C$ is the same as a composable pair of morphisms in $\C$.


id: walking_coreflexive_pair
type: category
name: walking coreflexive pair
notation: $\Delta^{\leq 1}$
description:
This category is equal to the truncated simplex category $\Delta^{\leq 1}$, i.e. the full subcategory of $\Delta$ spanned by $[0] = \{0\}$ and $[1] = \{0 < 1\}$; this also explains our notation of the category and its objects. $$[0] \begin{array}{c} \xhookrightarrow{~~i~~} \\ \xtwoheadleftarrow{~~p~~} \\ \xhookrightarrow{~~j~~} \end{array} [1]$$ The morphisms $i,j$ are the two inclusions, $p$ is their unique retraction, and $ip,jp : [1] \to [1]$ are the two constant maps. The name of this category comes from the fact that a functor $\Delta^{\leq 1} \to \C$ is the same as a <a href="https://ncatlab.org/nlab/show/reflexive+pair" target="_blank">coreflexive pair</a> in $\C$. Its dual is therefore the walking reflexive pair.


id: walking_fork
type: category
name: walking fork
notation: $\Fork$
description:
This category can be pictured as: $$\{0 \to 1 \rightrightarrows 2\}$$ Its name comes from the fact that a functor $\Fork \to \C$ is the same as a <a href="https://ncatlab.org/nlab/show/fork" target="_blank">fork</a> in $\C$.


id: walking_idempotent
type: category
name: walking idempotent
notation: $\Idem$
description:
The name of this category comes from the fact that a functor $\Idem \to \C$ is the same as an <a href="https://ncatlab.org/nlab/show/idempotent" target="_blank">idempotent morphism</a> in $\C$. It can also be seen as the delooping of the monoid $\{1,e\}$ in which $e^2=e$.


id: walking_isomorphism
type: category
name: walking isomorphism
notation: $\Isom$
nlab_link: https://ncatlab.org/nlab/show/walking+isomorphism
description:
This category can be pictured as: $$\{0 \rightleftarrows 1\}$$ Its name comes from the fact that a functor $\Isom \to \C$ is the same as an isomorphism in $\C$. The walking isomorphism is actually equivalent to the trivial category.


id: walking_morphism
type: category
name: walking morphism
notation: $I$
nlab_link: https://ncatlab.org/nlab/show/interval+category
description:
This is also known as the interval category and can be pictured as: $$\{0 \to 1\}$$ It has the property that functors $I \to \C$ are the same as morphisms in $\C$, which explains its name.


id: walking_pair
type: category
name: walking parallel pair
notation: $\Pair$
nlab_link: https://ncatlab.org/nlab/show/walking+structure
description:
This category can be pictured as: $$\{0 \rightrightarrows 1 \}$$ Its name comes from the fact that a functor $\Pair \to \C$ is the same as a parallel pair of morphisms in $\C$.


id: walking_span
type: category
name: walking span
notation: $\Span$
nlab_link: https://ncatlab.org/nlab/show/span#the_walking_span
description:
This category can be pictured as: $$\{1 \leftarrow 0 \rightarrow 2\}$$ Its name comes from the fact that a functor $\Span \to \C$ is the same as a <a href="https://ncatlab.org/nlab/show/span" target="_blank">span</a> in $\C$.


id: walking_splitting
type: category
name: walking splitting
notation: $\Split$
description:
This category could also be called the "walking split idempotent" (or "walking section", "walking retraction"), but we chose a name that emphasizes that the splitting belongs to the data. Notice that the $5$ given morphisms are indeed closed under composition. For example, $p \circ ip = p$ and $ip \circ i = i$.
The walking splitting can be interpreted as a skeleton of the category of $\IF_2$-vector spaces of dimension $\leq 1$.


id: binary_coproduct_sets
type: functor
name: binary coproduct functor on sets
notation: $+$
description:
This functor maps a pair of sets $(X,Y)$ to their coproduct $X + Y$. It is an example of a right-invertible left adjoint functor which is not a reflector.


id: binary_product_sets
type: functor
name: binary product functor on sets
notation: $\times$
description:
This functor maps a pair of sets $(X,Y)$ to their product $X \times Y$. It is an example of a right-invertible right adjoint functor which is not a coreflector.


id: continuous_functions
type: functor
name: functor of continuous functions
notation: $C$
description:
This functor maps a topological space $X$ to the commutative $\IR$-algebra $C(X)$ of continuous functions $X \to \IR$. A continuous map $f : X \to Y$ is mapped to the algebra homomorphism $f^* : C(Y) \to C(X)$, $u \mapsto u \circ f$.


id: countable_copower_sets
type: functor
name: countable copower functor on sets
notation: $\IN \times (-)$
description:
This functor maps a set $X$ to the product $\IN \times X$, which can also be seen as the copower $\IN \otimes X = \coprod_{n \in \IN} X$. It is an example of a polynomial functor.


id: diagonal_sets
type: functor
name: binary diagonal functor on the category of sets
notation: $\Delta$
nlab_link: https://ncatlab.org/nlab/show/diagonal+functor
description:
Every category $\C$ has a (binary) diagonal functor $\Delta : \C \to \C^2$, $X \mapsto (X,X)$. Here, we specify that $\C$ is the category of sets.


id: discrete_topology
type: functor
name: discrete topology functor
notation: $D$
nlab_link: https://ncatlab.org/nlab/show/discrete+and+indiscrete+topology
description:
This functor maps a set $X$ to the discrete topological space $D(X) \coloneqq (X, P(X))$ in which every subset is open.


id: doubling_sets
type: functor
name: doubling functor on sets
notation: $2(-)$
description:
This functor maps a set $X$ to its double $2 X \coloneqq X + X$. It is a simple example of a polynomial functor.


id: enveloping_group
type: functor
name: enveloping group functor
notation: $F_{\Mon,\Grp}$
nlab_link: https://ncatlab.org/nlab/show/free+functor
description:
This functor maps a monoid $M$ to the group $F(M)$ that is equipped with a universal homomorphism $i_M : M \to F(M)$. It is called the (universal) enveloping group or the group completion of $M$; in the commutative case, it is known as the Grothendieck group of $M$. As a possible construction of $F(M)$, take the free group on generators $\underline{m}$ for $m \in M$ subject to the relations $\underline{1} = 1$ and $\underline{m \cdot n} = \underline{m} \cdot \underline{n}$.


id: forget_addition
type: functor
name: forgetful functor from rings to monoids
notation: $U_{\Ring,\Mon}$
nlab_link: https://ncatlab.org/nlab/show/forgetful+functor
description:
This functor maps a ring to its underlying multiplicative monoid, which as "forgotten" the addition of the ring.


id: forget_group_pointed
type: functor
name: forgetful functor from groups to pointed sets
notation: $U_{\Grp,\Set_*}$
nlab_link: https://ncatlab.org/nlab/show/forgetful+functor
description:
This functor maps a group $G$ to its underlying pointed set $U_{\Grp,\Set_*}(G)$, whose base point is the identity element of $G$. It is an example of an essentially surjective functor which is not right-invertible.


id: forget_inverses
type: functor
name: forgetful functor from groups to monoids
notation: $U_{\Grp,\Mon}$
nlab_link: https://ncatlab.org/nlab/show/forgetful+functor
description:
This functor maps a group to its underlying monoid. We view groups as structured sets $(X,m,e,i)$ (consisting of a set, a multiplication, a neutral element, and an inverse operation), and monoids as structured sets $(X,m,e)$. This forgetful functor precisely maps $(X,m,e,i)$ to $(X,m,e)$. From this point of view, it does <i>not</i> merely forget a property; it forgets an operation. This perspective is useful in contexts where the inverse operation is no longer reducible to a property, for example, the forgetful functor from topological groups to topological monoids.


id: free_group
type: functor
name: free group functor
notation: $F_{\Grp}$
nlab_link: https://ncatlab.org/nlab/show/free+functor
description:
This functor maps a set $X$ to the free group $F_{\Grp}(X)$ on that set. In the proofs, we abbreviate $F \coloneqq F_{\Grp}$.


id: group_units
type: functor
name: group of units functor
notation: $(-)^{\times}$
nlab_link: https://ncatlab.org/nlab/show/group+of+units
description:
This functor maps a monoid $M$ to its group of units $M^{\times}$, consisting of pairs $(a,b) \in M^2$ satisfying $ab=ba=1$. Equivalently, it takes the submonoid of invertible elements of $M$, equipped with the inverse operation.


id: id_Set
type: functor
name: identity functor on the category of sets
notation: $\id_{\Set}$
nlab_link: https://ncatlab.org/nlab/show/identity+functor
description:
Every category $\C$ has an identity functor $\id_{\C} : \C \to \C$. Here, we specify that $\C$ is the category of sets.


id: indiscrete_topology
type: functor
name: indiscrete topology functor
notation: $I$
nlab_link: https://ncatlab.org/nlab/show/discrete+and+indiscrete+topology
description:
This functor maps a set $X$ to the indiscrete topological space $I(X) \coloneqq (X, \{\varnothing,X\})$ in which only the empty set and $X$ are open.


id: modulo-p
type: functor
name: modulo p functor
notation: $T^p$
description:
This functor maps an abelian group $A$ to the quotient $T^p(A) \coloneqq A/pA$, where $p$ is a fixed prime number. This group can also be represented as $A \otimes \IZ/p$.


id: monoid_ring
type: functor
name: monoid ring functor
notation: $\IZ[-]$
nlab_link: https://ncatlab.org/nlab/show/group+algebra
description:
This functor maps a monoid $M$ to the monoid ring $\IZ[M]$, which consists of finite sums of elements in $M$.


id: opposite_category
type: functor
name: opposite category functor
notation: $(-)^{\op}$
nlab_link: https://ncatlab.org/nlab/show/opposite+category
description:
This functor maps a small category $\C$ to its opposite category $\C^{\op}$ and a functor $F : \C \to \D$ to the opposite functor $F^{\op} : \C^{\op} \to \D^{\op}$.


id: opposite_monoid
type: functor
name: opposite monoid functor
notation: $(-)^{\op}$
nlab_link: https://ncatlab.org/nlab/show/opposite+magma
description:
This functor maps a monoid $M$ to its opposite monoid $M^{\op}$ which has the multiplication $a *^{\op} b \coloneqq a * b$. A monoid homomorphism $f : M \to N$ is also a monoid homomorphism $f^{\op} : M^{\op} \to N^{\op}$.


id: p-torsion
type: functor
name: p-torsion functor
notation: $T_p$
description:
This functor maps an abelian group $A$ to its $p$-torsion subgroup $$T_p(A) \coloneqq \{a \in A : pa = 0\},$$ where $p$ is a fixed prime number. This group can also be represented as $\HomInternal(\IZ/p,A)$.


id: pi_0
type: functor
name: path components functor
notation: $\pi_0$
nlab_link: https://ncatlab.org/nlab/show/connected+space
description:
This functor maps a topological space $X$ to its set $\pi_0(X)$ of path components. Thus, $\pi_0(X) = U(X) / {\sim}$, where $U(X)$ is the underlying set and $x \sim y$ when there is a path from $x$ to $y$.


id: pi_1
type: functor
name: fundamental group functor
notation: $\pi_1$
nlab_link: https://ncatlab.org/nlab/show/fundamental+group
description:
The fundamental group $\pi_1(X,x_0)$ of a pointed topological space $(X,x_0)$ is the group of homotopy classes of loops at $x_0$. The group operation is concatenation of paths. For example, we have $\pi_1(S^1,1) \cong \IZ$ (see Hatcher's <a href="https://pi.math.cornell.edu/~hatcher/AT/ATpage.html" target="_blank">Algebraic Topology</a>, Theorem 1.7).


id: power_set_contravariant
type: functor
name: contravariant power set functor
notation: $P_{\forall}$
nlab_link: https://ncatlab.org/nlab/show/power+set
description:
This functor $P_{\forall}$ maps a set $X$ to its power set $P(X)$ and a map of sets $f : X \to Y$ to the induced preimage operator $f^* : P(Y) \to P(X)$.


id: power_set_covariant
type: functor
name: covariant power set functor
notation: $P_{\exists}$
nlab_link: https://ncatlab.org/nlab/show/power+set
description:
This functor $P_{\exists}$ maps a set $X$ to its power set $P(X)$ and a map of sets $f : X \to Y$ to the induced image operator $f_* : P(X) \to P(Y)$.


id: sequences_sets
type: functor
name: sequences functor on sets
notation: $(-)^{\IN}$
description:
This functor maps a set $X$ to the countable power $X^{\IN}$, i.e. the set of sequences in $X$. It is an example of a polynomial functor. It is also an example of a monadic functor for which the crude monadicity theorem does not apply.


id: torsion
type: functor
name: torsion functor
notation: $T$
nlab_link: https://ncatlab.org/nlab/show/torsion+subgroup
description:
This functor maps an abelian group $A$ to its torsion subgroup $$T(A) \coloneqq \{a \in A : \exists n \geq 1 \, (na = 0)\}.$$


id: trivial_groups
type: functor
name: trivial functor from the category of groups
notation: $!$
description:
Every category $\C$ has a unique functor $! : \C \to 1$ into the trivial category. Here, we specify that $\C$ is the category of groups. It is a basic example of a full functor which is not faithful.


id: trivial_sets
type: functor
name: trivial functor from the category of sets
notation: $!$
description:
Every category $\C$ has a unique functor $! : \C \to 1$ into the trivial category. Here, we specify that $\C$ is the category of sets.


id: walking_isomorphism_object_inclusion
type: functor
name: walking isomorphism object inclusion
notation: $\iota$
description:
This is the natural embedding of the trivial category with a single object $0$ into the walking isomorphism given by two objects $0,1$ and an isomorphism $0 \to 1$. This is the simplest example of an equivalence of categories which is not an isomorphism.


id: empty-map
type: morphism
name: map from the empty set
notation: $!$
nlab_link: https://ncatlab.org/nlab/show/empty+set
description:
For every set $X$ there is a unique map $! : \varnothing \hookrightarrow X$, the empty map. In this entry, we require that $X$ is non-empty.


id: id_X
type: morphism
name: identity map of a set
notation: $\id_X$
nlab_link: https://ncatlab.org/nlab/show/identity+morphism
description:
Every object of a category has an identity morphism. In this case, we take a set $X$ and consider its identity morphism $\id_X : X \to X$ in $\Set$.


id: integer-rational-embedding
type: morphism
name: embedding of integer into rational numbers
notation: $\iota_{\IZ,\IQ}$
nlab_link: https://ncatlab.org/nlab/show/localization+of+a+commutative+ringlization
description:
The inclusion $\iota_{\IZ,\IQ} : \IZ \hookrightarrow \IQ$ is a typical example of a localization. It is a standard example of a mono- and epimorphism which is not an isomorphism in the category of (commutative) rings.


id: terminal-map
type: morphism
name: map into the singleton set
notation: $!$
description:
For every set $X$ there is a unique map $! : X \to 1$, where $1$ is the singleton set. In this entry, we require that $X$ has at least $2$ elements.


# structure_comments

structure_id: CRing
comment:
Regular monomorphisms are discussed in <a href="https://math.stackexchange.com/questions/695685" target="_blank">MSE/695685</a>, but probably they cannot be classified.


structure_id: FiltVect
comment:
The forgetful functor $U : \FiltVect \to \Vect$ has a left adjoint that equips a vector space with the trivial filtration. It also has a right adjoint that equips a vector space with the maximal filtration. In particular, it follows that $U$ preserves limits and colimits.


structure_id: FiltVect
comment:
The filtration functor $F^n : \FiltVect \to \Vect$ has a left adjoint that equips a vector space $V$ with the filtration $F^n(V)=V$ and $F^{n+1}(V)=0$. In particular, it follows that $F^n$ preserves limits. However, it does not have a right adjoint. In fact, $F^n$ does not preserve coequalizers. On the other hand, $F^n$ does preserve filtered colimits; this can be verified using the concrete construction of filtered colimits.


structure_id: Fld
comment:
Limits and colimits are discussed in <a href="https://math.stackexchange.com/questions/359352" target="_blank">MSE/359352</a>.


structure_id: LRS_R
comment:
Monomorphisms $X \to Y$ can be formally described by the condition that the diagonal morphism $X \to X \times_Y X$ is an isomorphism. Since the fiber product of locally ringed spaces has a concrete description (<a href="https://math.stackexchange.com/questions/1033675" target="_blank">MSE/1033675</a>), we thus have a description of the monomorphisms, albeit a very complicated one. At least, we can say that every monomorphism is injective.


structure_id: Meas
comment:
The thread <a href="https://math.stackexchange.com/questions/5024471/">MSE/5024471</a> asks for the finitely presentable objects of this category.


structure_id: Sch_R
comment:
Monomorphisms are discussed at <a href="https://mathoverflow.net/questions/56591" target="_blank">MO/56591</a>. At least the case of morphisms of locally finite type is understood.


structure_id: Sch_R
comment:
Regular monomorphisms are discussed at <a href="https://mathoverflow.net/questions/66279" target="_blank">MO/66279</a>.


structure_id: Sch_R
comment:
Epimorphisms are discussed at <a href="https://mathoverflow.net/questions/56564" target="_blank">MO/56564</a>. Probably they cannot be classified.


structure_id: Sh(X)
comment:
It is likely that none of the currently remaining unknown properties (locally finitely presentable, ℵ₁-accessible, etc.) are satisfied for a <i>generic</i> space $X$, but we need to make this precise by adding additional requirements to $X$. Maybe we need to create separate entries for specific spaces $X$.


structure_id: Sh(X)
comment:
See <a href="https://math.stackexchange.com/questions/5140378" target="_blank">MSE/5140378</a> in particular for conditions when $\Sh(X)$ is locally finitely presentable.


structure_id: Sh(X,Ab)
comment:
It is likely that neither of the currently remaining unknown properties (finitary algebraic, locally finitely presentable, CSP, etc.) are satisfied for a <i>generic</i> space $X$, but we need to make this precise by adding additional requirements to $X$. Maybe we need to create separate entries for specific spaces $X$.


structure_id: forget_topology
comment:
This functor has exactly two right inverses (up to isomorphism), see <a href="https://math.stackexchange.com/questions/4368730" target="_blank">MSE/4368730</a>.


# properties

id: Barr-coexact
type: category
relation: is
dual_property_id: Barr-exact
description:
A category is <i>Barr-coexact</i> if its dual category is Barr-exact, i.e. if it is coregular and every cocongruence is effective.


id: CIP
type: category
relation: satisfies
dual_property_id: CSP
description:
A category satisfies <i>CIP</i> ("coproducts inject into products") if it has zero morphisms, products, coproducts, and for every family of objects $(X_i)_{i \in I}$ the canonical morphism $$\textstyle \alpha : \coprod_i X_i \to \prod_{i \in I} X_i$$ defined by $p_j \circ \alpha \circ \iota_i = \delta_{i,j}$ is a monomorphism. This is no standard terminology. This property has been added to clarify relationships between other properties, in particular those concerning the commutation between limits and colimits.


id: CSP
type: category
relation: satisfies
dual_property_id: CIP
description:
A category satisfies <i>CSP</i> ("coproducts surject onto products") if it has zero morphisms, products, coproducts, and for every family of objects $(X_i)_{i \in I}$ the canonical morphism $$\textstyle \alpha : \coprod_i X_i \to \prod_{i \in I} X_i$$ defined by $p_j \circ \alpha \circ \iota_i = \delta_{i,j}$ is an epimorphism. This is no standard terminology. This property has been added to clarify relationships between other properties, in particular those concerning the commutation between limits and colimits.


id: Cauchy complete
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/Cauchy+complete+category
dual_property_id: Cauchy complete
description:
A category is <i>Cauchy complete</i> if every idempotent splits. That is, every endomorphism $e : X \to X$ with $e^2 = e$ may be written as $e = i \circ p$ for some morphisms $p : X \to Y$ and $i : Y \to X$ with $p \circ i = \id_Y$. Equivalently, the pair $e,\id_X : X \rightrightarrows X$ has an equalizer (or a coequalizer).


id: Grothendieck abelian
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/Grothendieck+category
description:
A category is <i>Grothendieck abelian</i> if it is abelian, has coproducts (and is therefore cocomplete), a generator, and filtered colimits commute with finite limits. These categories play an important role in homological algebra.


id: Malcev
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/Malcev+category
dual_property_id: co-Malcev
description:
A category is <i>Malcev</i> when it has finite limits and every internal reflexive relation is an internal equivalence relation. That is, if $R \subseteq X^2$ is a subobject with $\Delta_X \subseteq R$, then $R$ is symmetric and transitive.


id: abelian
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/abelian+category
dual_property_id: abelian
description:
A category is <i>abelian</i> if it is additive, every morphism has a kernel and a cokernel, and every monomorphism and epimorphism is normal. Equivalently, it is additive, has equalizers and coequalizers, and it is mono-regular and epi-regular. As opposed to other types of categories (such as monoidal categories), being abelian turns out to be a mere property.


id: accessible
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/accessible+category
dual_property_id: coaccessible
description:
Let $\kappa$ be a regular cardinal. A category is <i>$\kappa$-accessible</i> if it has $\kappa$-filtered colimits and there is a (small) set $G$ of $\kappa$-presentable objects such that every object is a $\kappa$-filtered colimit of objects in $G$. A category is <i>accessible</i> if it is $\kappa$-accessible for some regular cardinal $\kappa$.


id: additive
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/additive+category
dual_property_id: additive
description:
A category is <i>additive</i> if it is preadditive and has finite products (equivalently, finite coproducts). Note that in the context of finite products, the preadditive structure <a href="/content/preadditive_structure_unique">is unique</a>.


id: ℵ₁-accessible
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/accessible+category
description:
This is the special case of the notion of a $\kappa$-accessible category, where $\kappa = \aleph_1$ is the first uncountable cardinal. Concretely, a category is $\aleph_1$-accessible when it has $\aleph_1$-filtered colimits and there is a small set $G$ of $\aleph_1$-presentable objects such that every object is a $\aleph_1$-filtered colimit of objects in $G$.


id: ℵ₁-cofiltered limits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/filtered+colimit
dual_property_id: ℵ₁-filtered colimits
description:
A category has <i>$\aleph_1$-cofiltered limits</i> if it has limits of diagrams indexed by small $\aleph_1$-cofiltered categories. This is the special case of having small $\kappa$-cofiltered limits for $\kappa=\aleph_1$, the first uncountable cardinal.


id: ℵ₁-cofiltered
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/cofiltered+category
dual_property_id: ℵ₁-filtered
description:
A category is <i>$\aleph_1$-cofiltered</i> if every countable diagram admits a cone. This is the special case of the notion of a $\kappa$-cofiltered category for $\kappa = \aleph_1$, the first uncountable cardinal. An equivalent characterization is given as the dual of Theorem 2.2 at the <a href="https://ncatlab.org/nlab/show/filtered+category">nLab</a>.


id: ℵ₁-filtered colimits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/filtered+colimit
dual_property_id: ℵ₁-cofiltered limits
description:
A category has <i>$\aleph_1$-filtered colimits</i> if it has colimits of diagrams indexed by small $\aleph_1$-filtered categories. This is the special case of having small $\kappa$-filtered colimits for $\kappa=\aleph_1$, the first uncountable cardinal.


id: ℵ₁-filtered
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/filtered+category
dual_property_id: ℵ₁-cofiltered
description:
A category is <i>$\aleph_1$-filtered</i> if every countable diagram admits a cocone. This is the special case of the notion of a $\kappa$-filtered category for $\kappa = \aleph_1$, the first uncountable cardinal. An equivalent characterization is given as Theorem 2.2 at the <a href="https://ncatlab.org/nlab/show/filtered+category">nLab</a>.


id: ℵ₂-small copowers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/copower
dual_property_id: ℵ₂-small powers
description:
A category has <i>$\aleph_2$-small copowers</i> if for every object $X$ and every set $I$ with cardinality $< \aleph_2$, the copower $I \otimes X$ exists. In particular, this includes copowers of the form $\aleph_1 \otimes X$. Here, $\aleph_1$ denotes the first uncountable cardinal, and $\aleph_2$ the next larger cardinal.
Notice that finite copowers are just $\aleph_0$-small copowers, and countable copowers are $\aleph_1$-small copowers.


id: ℵ₂-small coproducts
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/coproduct
dual_property_id: ℵ₂-small products
description:
A category has <i>$\aleph_2$-small coproducts</i> if it has coproducts of families of objects indexed by sets of cardinality $< \aleph_2$. In particular, this includes coproducts indexed by sets of cardinality $\aleph_1$. Here, $\aleph_1$ denotes the first uncountable cardinal, and $\aleph_2$ the next larger cardinal.
Notice that finite coproducts are just $\aleph_0$-small coproducts, and countable coproducts are $\aleph_1$-small coproducts.


id: ℵ₂-small powers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/powering
dual_property_id: ℵ₂-small copowers
description:
A category has <i>$\aleph_2$-small powers</i> if for every object $X$ and every set $I$ of cardinality $< \aleph_2$, the power $X^I$ exists. In particular, this includes powers of the form $X^{\aleph_1}$. Here, $\aleph_1$ denotes the first uncountable cardinal, and $\aleph_2$ the next larger cardinal.
Notice that finite powers are just $\aleph_0$-small powers, and countable powers are $\aleph_1$-small powers.


id: ℵ₂-small products
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/cartesian+product
dual_property_id: ℵ₂-small coproducts
description:
A category has <i>$\aleph_2$-small products</i> if it has products indexed by sets of cardinality $< \aleph_2$. In particular, this includes products indexed by sets of cardinality $\aleph_1$. Here, $\aleph_1$ denotes the first uncountable cardinal, and $\aleph_2$ the next larger cardinal.
Notice that finite products are just $\aleph_0$-small products, and countable products are $\aleph_1$-small products.


id: balanced
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/balanced+category
dual_property_id: balanced
description:
A category is <i>balanced</i> if every morphism which is a monomorphism and an epimorphism must be an isomorphism.


id: binary copowers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/copower
dual_property_id: binary powers
description:
A category has <i>binary copowers</i> when for every object $X$ and every binary set $I$ the coproduct $X \sqcup X$ exists. These objects might also be called <i>doubles</i>.


id: binary coproducts
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/coproduct
dual_property_id: binary products
description:
A category has <i>binary coproducts</i> if every pair $A,B$ of objects has a coproduct $A \sqcup B$.


id: binary powers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/powering
dual_property_id: binary copowers
description:
A category has <i>binary powers</i> when for every object $X$ the product $X \times X$  exists. These objects might also be called <i>squares</i>.


id: biproducts
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/biproduct
dual_property_id: biproducts
description:
A category has <i>biproducts</i> when it has zero morphisms, finite products (denoted $\times$), finite coproducts (denoted $\oplus$), and for every finite family of objects $A_1,\dotsc,A_n$ the canonical morphism $$\mu : A_1 \oplus \cdots \oplus A_n \to A_1 \times \cdots \times A_n$$ is an isomorphism. Such a category is also called <i>semi-additive</i>, and it is automatically enriched over commutative monoids: the sum of $f,g : A \rightrightarrows B$ is defined as: $$A \xrightarrow{(f,g)} B \times B \xrightarrow{\mu^{-1}} B \oplus B \xrightarrow{\nabla} B$$


id: cartesian closed
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/cartesian+closed+category
dual_property_id: cocartesian coclosed
description:
A category is <i>cartesian closed</i> if all finite products and exponentials $[X,Y]$ exist, defined by the adjunction $\Hom(T,[X,Y]) \cong \Hom(T \times X,Y)$.


id: cartesian filtered colimits
type: category
relation: has
dual_property_id: cocartesian cofiltered limits
description:
In a category $\C$, which we assume to have filtered colimits and finite products, we say that <i>filtered colimits are cartesian</i> if for every finite set $I$ the product functor $\prod : \C^I \to \C$ preserves filtered colimits. Equivalently, for every $X \in \C$ the functor $X \times - : \C \to \C$ preserves filtered colimits.
This is no standard terminology, it has been suggested in <a href="https://mathoverflow.net/questions/510240" target="_blank">MO/510240</a>. We have added it to the database since it clarifies the relationship between many related properties.


id: co-Malcev
type: category
relation: is
dual_property_id: Malcev
description:
A category is <i>co-Malcev</i> when its dual is Malcev, i.e., it has finite colimits and if $X \sqcup X \twoheadrightarrow R$ is a coreflexive corelation, then it is cosymmetric and cotransitive.
This terminology is not standard, but we have added it to properly formulate the interesting theorem that the dual of an elementary topos is Malcev, i.e., that every elementary topos is co-Malcev.
o settle this property, we often use that $\C$ is co-Malcev if and only if the category of representable functors $\C \to \Set^+$ is Malcev.


id: cocartesian coclosed
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/cocartesian+coclosed+category
dual_property_id: cartesian closed
description:
A category is <i>cocartesian coclosed</i> if its dual category is cartesian closed, i.e. if all finite coproducts and coexponentials $\Coexp(X,Y)$ exist, defined by the adjunction $\Hom(\Coexp[X,Y],T) \cong \Hom(Y,T \sqcup X)$.


id: cocartesian cofiltered limits
type: category
relation: has
dual_property_id: cartesian filtered colimits
description:
In a category $\C$, which we assume to have cofiltered limits and finite coproducts, we say that <i>cofiltered limits are cocartesian</i> if for every finite set $I$ the coproduct functor $\coprod : \C^I \to \C$ preserves cofiltered limits. Equivalently, for every $X \in \C$ the functor $X \sqcup - : \C \to \C$ preserves cofiltered limits.
This is no standard terminology, its dual has been suggested in <a href="https://mathoverflow.net/questions/510240" target="_blank">MO/510240</a>. We have added it to the database since it clarifies the relationship between many related properties.


id: codistributive
type: category
relation: is
dual_property_id: distributive
description:
A category is <i>codistributive</i> if it has finite coproducts, finite products, and for every object $A$ the functor $- \sqcup A$ preserves finite products. Concretely, for every finite family of objects $(B_i)$ the canonical morphism $A \sqcup \prod_i B_i \to \prod_i (A \sqcup B_i)$ must be an isomorphism.


id: coequalizers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/coequalizer
dual_property_id: equalizers
description:
A <i>coequalizer</i> of a pair of morphisms $f,g : A \rightrightarrows B$ is an object $C$ with a morphism $c : B \to C$ such that $c \circ f = c \circ g$ and which is universal with respect to this property. This property refers to the existence of coequalizers.


id: coextensive
type: category
relation: is
dual_property_id: extensive
description:
A category $\C$ is <i>coextensive</i> when it has finite products and for all objects $A,B \in \C$ the product functor $A/\C \times B/\C \to (A \times B)/\C$ is an equivalence of categories. The prototypical example is the category of commutative rings.


id: cofiltered limits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/filtered+limit
dual_property_id: filtered colimits
description:
A category has <i>cofiltered limits</i> if it has limits of diagrams indexed by small <a href="/category-property/cofiltered">cofiltered</a> categories. This is actually equivalent to having directed limits.


id: cofiltered-limit-stable epimorphisms
type: category
relation: has
dual_property_id: filtered-colimit-stable monomorphisms
description:
A category has <i>cofiltered-limit-stable epimorphisms</i> if it has cofiltered limits and for every cofiltered diagram of epimorphisms $(X_i \to Y_i)$ also their limit $\lim_i X_i \to \lim_i Y_i$ is an epimorphism.


id: cofiltered
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/cofiltered+category
dual_property_id: filtered
description:
A category is <i>cofiltered</i> if every finite diagram admits a cone. Equivalently, it is inhabited, for every two objects $x,y$ there is a span $x \leftarrow p \rightarrow y$ (not necessarily universal), and every parallel pair $x \rightrightarrows y$ is equalized by some morphism $e \to x$ (not necessarily universal).


id: cogenerating set
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/cogenerator
dual_property_id: generating set
description:
A set of objects $S$ is called a <i>cogenerating set</i> if for every pair of parallel morphisms $f,g : A \rightrightarrows B$, $f = g$ holds if and only if for every morphism $h : B \to G$ with $G \in S$ we have $h \circ f = h \circ g$. Equivalently, the functor $(\Hom(-,G))_{G \in S} : \C^{\op} \to (\Set^+)^S$ is faithful. This property refers to the existence of a cogenerating set.


id: cogenerator
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/cogenerator
dual_property_id: generator
description:
An object $Q$ of a category is called a <i>cogenerator</i> if for every pair of parallel morphisms $f,g : A \rightrightarrows B$ the equation $f = g$ holds if for every morphism $h : B \to Q$ we have $h \circ f = h \circ g$. Equivalently, the functor $\Hom(-,Q) : \C^{\op} \to \Set^+$ is faithful. This property refers to the existence of a cogenerator. By definition, $Q$ is a cogenerator if and only if $\{Q\}$ is a <a href="/category-property/cogenerating_set">cogenerating set</a>.


id: cokernels
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/cokernel
dual_property_id: kernels
description:
A category has <i>cokernels</i> if it has zero morphisms and every morphism $f : A \to B$ has a cokernel, i.e. a coequalizer of $f$ with the zero morphism $0_{A,B} : A \to B$.


id: connected colimits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/connected+colimit
dual_property_id: connected limits
description:
A category has <i>connected colimits</i> if it has colimits of diagrams indexed by connected small categories.


id: connected limits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/connected+limit
dual_property_id: connected colimits
description:
A category has <i>connected limits</i> if it has limits of diagrams indexed by connected small categories.


id: connected
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/connected+category
dual_property_id: connected
description:
A category is <i>connected</i> if it is inhabited and every two objects can be joined via a zig-zag path of morphisms. Equivalently, $\C$ is connected if $\C \simeq \coprod_{i \in I} \C_i$ implies $\C_i \simeq 0$ for some $i$.


id: conormal
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/normal+epimorphism
dual_property_id: normal
description:
A category is <i>conormal</i> if it has zero morphisms and every epimorphism is a cokernel of some morphism (in which case case it is also called a <i>normal epimorphism</i>). The assumption of having zero morphisms makes it possible to talk about cokernels.


id: copowers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/copower
dual_property_id: powers
description:
If $X$ is an object and $I$ is a set, the <i>copower</i> is defined as the coproduct $I \otimes X \coloneqq \coprod_{i \in I} X$. This property refers to the existence of copowers.


id: coproducts
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/coproduct
dual_property_id: products
description:
Given a family of objects $(A_i)_{i \in I}$, a <i>coproduct</i> $\coprod_{i \in I} A_i$ is defined as an object with morphisms $i_i : A_i \to \coprod_{i \in I} A_i$ satisfying the following universal property: For every object $T$ and every family of morphisms $(f_i : A_i \to T)_{i \in I}$ there is a unique morphism $f : \coprod_{i \in I} A_i \to T$ such that $f \circ i_i = f_i$ for all $i \in I$. This property refers to the existence of <i>small coproducts</i>, i.e., coproducts of small families of objects.


id: coquotients of cocongruences
type: category
relation: has
dual_property_id: quotients of congruences
description:
A <i>cocongruence</i> (or internal equivalence corelation) on an object $X$ of a category is a parallel pair $i_1, i_2 : X \rightrightarrows E$ which is jointly epimorphic, and such that for every object $T$, the image of $({-} \circ i_1, {-} \circ i_2) : \Hom(E, T) \to \Hom(X, T)^2$ is an equivalence relation. The category has <i>coquotients of cocongruences</i> if for each such cocongruence, there exists an equalizer of $i_1$ and $i_2$. Note that in the case of a category with binary copowers, the corresponding quotients of $X + X$ are also commonly referred to as cocongruences, or as internal equivalence corelations.


id: core-thin
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/gaunt+category
dual_property_id: core-thin
description:
A category is <i>core-thin</i> if it satisfies the following equivalent conditions: <ol> <li>Its <a href="https://ncatlab.org/nlab/show/core+groupoid" target="_blank">core</a> is thin.</li> <li>Parallel isomorphisms are equal.</li> <li>Every automorphism is the identity.</li> </ol>


id: coreflexive equalizers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/reflexive+coequalizer
dual_property_id: reflexive coequalizers
description:
A <i>coreflexive equalizer</i> is a limit of a diagram consisting of a parallel pair of morphisms with a common retraction (left inverse), which is the same concept as an equalizer of such a parallel pair. This property refers to the existence of coreflexive equalizers.


id: coregular
type: category
relation: is
dual_property_id: regular
description:
A category is <i>coregular</i> when its dual is regular, i.e. it is finitely cocomplete, for every morphism $Y \to X$ its cokernel pair $X \rightrightarrows X \sqcup_Y X$ has an equalizer, and regular monomorphisms are stable under pushouts.


id: cosifted limits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/sifted+colimit
dual_property_id: sifted colimits
description:
A category has <i>cosifted limits</i> if it has limits of diagrams indexed by small <a href="/category-property/cosifted">cosifted</a> categories.


id: cosifted
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/sifted+category
dual_property_id: sifted
description:
A category $\C$ is <i>cosifted</i> if it is inhabited and the diagonal functor $\Delta : \C \to \C \times \C$ is initial, i.e. if it is non-empty and for any two objects $X,Y \in \C$ the category of spans $$X \leftarrow Z \rightarrow Y$$ is connected. Equivalently, a small category $\C$ is cosifted if $\colim : \Set^{{\C}^\op} \to \Set$ preserves finite products. This property is a weaker notion than being cofiltered.


id: counital
type: category
relation: is
dual_property_id: unital
description:
A category is <i>counital</i> if its dual is unital, i.e., it has a zero object, finite colimits, and for all objects $X,Y$ the two morphisms $(\id_X;0) : X \sqcup Y \twoheadrightarrow X$ and $(0;\id_Y) : X \sqcup Y \twoheadrightarrow Y$ are jointly strongly monomorphic. When binary products exist, the canonical morphism $X \sqcup Y \to X \times Y$ therefore must be a <a href="https://ncatlab.org/nlab/show/strong+monomorphism" target="_blank">strong monomorphism</a>.


id: countable copowers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/copower
dual_property_id: countable powers
description:
A category has <i>countable copowers</i> when for every object $X$ and every countable set $I$ the copower $I \otimes X$ exists.


id: countable powers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/powering
dual_property_id: countable copowers
description:
A category has <i>countable powers</i> when for every object $X$ and every countable set $I$ the power $X^I$ exists.


id: countably codistributive
type: category
relation: is
dual_property_id: countably distributive
description:
A category is <i>countably codistributive</i> if it has finite coproducts, countable products, and for every object $A$ the functor $A \sqcup -$ preserves countable products. Concretely, for every countable family of objects $(B_i)$ the canonical morphism $A \sqcup \prod_i B_i \to \prod_i (A \sqcup B_i)$ must be an isomorphism.


id: countably distributive
type: category
relation: is
dual_property_id: countably codistributive
description:
A category is <i>countably distributive</i> if it has finite products, countable coproducts, and for every object $A$ the functor $A \times -$ preserves countable coproducts. Concretely, for every countable family of objects $(B_i)$ the canonical morphism $\coprod_i (A \times B_i) \to A \times \coprod_i B_i$ must be an isomorphism.


id: direct
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/direct+category
dual_property_id: inverse
description:
A category is <i>direct</i> if it contains no infinite sequence of non-identity morphisms of the form $$\cdots \to A_2 \to A_1 \to A_0.$$ For example, a poset is direct iff it is well-founded.


id: directed colimits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/directed+colimit
dual_property_id: directed limits
description:
A category has <i>directed colimits</i> if it has colimits of diagrams indexed by directed (small) posets. This is actually equivalent to having filtered colimits. Directed colimits are (somewhat confusingly) also known as <i>inverse limits</i>.


id: directed limits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/directed+limit
dual_property_id: directed colimits
description:
A category has <i>directed limits</i> if it has limits of diagrams indexed by codirected (small) posets. This is actually equivalent to having cofiltered limits.


id: discrete
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/discrete+category
dual_property_id: discrete
description:
A category is <i>discrete</i> when every morphism is an identity morphism. Thus, a discrete category is merely a collection of objects.


id: disjoint coproducts
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/disjoint+coproduct
dual_property_id: disjoint products
description:
A category has <i>disjoint coproducts</i> if it has coproducts, the coproduct inclusions $A_i \to \coprod_{i \in I} A_i$ are monomorphisms, and the pullback of the inclusions $A_i \to \coprod_{i \in I} A_i$ and $A_j \to \coprod_{i \in I} A_i$ for $i \neq j$ exists and is given by the initial object $0$.


id: disjoint finite coproducts
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/disjoint+coproduct
dual_property_id: disjoint finite products
description:
A category has <i>disjoint finite coproducts</i> if it has finite coproducts, for every pair of objects $A,B$ the coproduct inclusions $A \rightarrow A+B \leftarrow B$ are monomorphisms, and the pullback $A \times_{A + B} B$ exists and is given by the initial object $0$.


id: disjoint finite products
type: category
relation: has
dual_property_id: disjoint finite coproducts
description:
A category has <i>disjoint finite products</i> if it has finite products, for every pair of objects $A,B$ the product projections $A \leftarrow A \times B \rightarrow B$ are epimorphisms, and the pushout $A \sqcup_{A \times B} B$ exists and is given by the terminal object $1$.
This terminology does not seem to be common, but we have added it as a dual for the more commonly known property of having disjoint finite coproducts.


id: disjoint products
type: category
relation: has
dual_property_id: disjoint coproducts
description:
A category has <i>disjoint products</i> if it has products, the product projections $\prod_{i \in I} A_i \to A_i$ are epimorphisms, and the pushout of the projections $\prod_{i \in I} A_i \to A_i$ and $\prod_{i \in I} A_i \to A_j$ for $i \neq j$ exists and is given by the terminal object $1$.
This terminology does not seem to be common, but we have added it as a dual for the more commonly known property of having disjoint coproducts.


id: distributive
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/distributive+category
dual_property_id: codistributive
description:
A category is <i>distributive</i> if it has finite products, finite coproducts, and for every object $A$ the functor $A \times -$ preserves finite coproducts. Concretely, for every finite family of objects $(B_i)$ the canonical morphism $\coprod_i (A \times B_i) \to A \times \coprod_i B_i$ must be an isomorphism.


id: effective cocongruences
type: category
relation: has
dual_property_id: effective congruences
description:
A cocongruence $f, g : X \rightrightarrows E$ (see definition <a href="/category-property/coquotients_of_cocongruences">here</a>) is <i>effective</i> if it is the cokernel pair of some morphism, i.e. if there is a morphism $h : Y \to X$ such that we have a cocartesian square $$\begin{CD} Y @> h >> X \\ @V h VV @VV f V \\ X @>> g > E. \end{CD}$$ A category <i>has effective cocongruences</i> if every cocongruence in the category is effective.


id: effective congruences
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/congruence
dual_property_id: effective cocongruences
description:
A congruence $f, g : E \rightrightarrows X$ (see definition <a href="/category-property/quotients_of_congruences">here</a>) is <i>effective</i> if it is the kernel pair of some morphism, i.e. if there is a morphism $h : X \to Y$ such that we have a cartesian square $$\begin{CD} E @> f >> X \\ @V g VV @VV h V \\ X @>> h > Y. \end{CD}$$ A category <i>has effective congruences</i> if every congruence in the category is effective.


id: elementary topos
type: category
relation: is an
nlab_link: https://ncatlab.org/nlab/show/topos
description:
An <i>elementary topos</i> is a cartesian closed category with finite limits and a subobject classifier.


id: epi-regular
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/regular+epimorphism
dual_property_id: mono-regular
description:
A category is <i>epi-regular</i> when every epimorphism is regular, i.e. the coequalizer of a pair of morphisms. Notice that this is not standard terminology, <a href="https://math.stackexchange.com/questions/5031588" target="_blank">apparently</a> the literature has no name for this yet. A <i>preadditive</i> category is epi-regular iff it is conormal. The notion of a conormal category is reserved for categories with zero morphisms, while epi-regular applies to all categories.


id: equalizers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/equalizer
dual_property_id: coequalizers
description:
An <i>equalizer</i> of a pair of morphisms $f,g : A \rightrightarrows B$ is an object $E$ with a morphism $e : E \to A$ such that $f \circ e = g \circ e$ and which is universal with respect to this property. This property refers to the existence of equalizers.


id: essentially countable
type: category
relation: is
dual_property_id: essentially countable
description:
A category is <i>essentially countable</i> if it is equivalent to a countable category. This means that up to isomorphism there is only a countable collection of objects, and for every pair of objects $A,B$ the collection $\Hom(A,B)$ is countable.


id: essentially discrete
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/discrete+category
dual_property_id: essentially discrete
description:
A category is <i>essentially discrete</i> if it is equivalent to a discrete category. Equivalently, it is a thin groupoid. Notice that the nLab calls this property simply "discrete". In contrast to being discrete, this property is invariant under equivalences of categories. An essentially discrete category is the same as a <i>setoid</i> (a set equipped with an equivalence relation).


id: essentially finite
type: category
relation: is
dual_property_id: essentially finite
description:
A category is <i>essentially finite</i> if it is equivalent to a finite category. Equivalently, there are only finitely many objects up to isomorphism, and the collection of morphisms between any two objects is isomorphic to a finite set. In contrast to being finite, this property is invariant under equivalences of categories.


id: essentially small
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/small+category
dual_property_id: essentially small
description:
A category is <i>essentially small</i> when it is equivalent to a small category. In particular, there is a set of objects such that every object is isomorphic to an object in this set. In contrast to the property of being small, being essentially small is invariant under equivalences of categories.


id: exact cofiltered limits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/commutativity+of+limits+and+colimits
dual_property_id: exact filtered colimits
description:
In a category $\C$, which we assume to have cofiltered limits and finite colimits, we say that <i>cofiltered limits are exact</i> if the following equivalent conditions are satisfied: <ol> <li>For every finite category $\I$ the functor $\colim : [\I, \C] \to \C$ preserves cofiltered limits.</li> <li>For every small cofiltered category $\J$ the functor $\lim : [\J,\C] \to \C$ preserves finite colimits.</li> <li>For every diagram $X : \I \times \J \to \C$, where $\I$ is finite and $\J$ is small cofiltered, the canonical morphism $\colim_i \lim_j X(i,j) \to \lim_j \colim_i X(i,j)$ is an isomorphism.</li> </ol>


id: exact filtered colimits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/commutativity+of+limits+and+colimits
dual_property_id: exact cofiltered limits
description:
In a category $\C$, which we assume to have filtered colimits and finite limits, we say that <i>filtered colimits are exact</i> if the following equivalent conditions are satisfied: <ol> <li>For every finite category $\I$ the functor $\lim : [\I, \C] \to \C$ preserves filtered colimits.</li> <li>For every small filtered category $\J$ the functor $\colim : [\J,\C] \to \C$ preserves finite limits.</li> <li>For every diagram $X : \I \times \J \to \C$, where $\I$ is finite and $\J$ is small filtered, the canonical morphism $\colim_j \lim_i X(i,j) \to \lim_i \colim_j X(i,j)$ is an isomorphism.</li> </ol>


id: extensive
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/extensive+category
dual_property_id: coextensive
description:
A category $\C$ is <i>extensive</i> when it has finite coproducts and for all objects $A,B \in \C$ the coproduct functor $\C/A \times \C/B \to \C/(A+B)$ is an equivalence of categories. Equivalently, pullbacks of finite coproduct inclusions along arbitrary morphisms exist and finite coproducts are disjoint and stable under pullback.


id: filtered colimits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/filtered+colimit
dual_property_id: cofiltered limits
description:
A category has <i>filtered colimits</i> if it has colimits of diagrams indexed by small <a href="/category-property/filtered">filtered</a> categories. This is actually equivalent to having directed colimits.


id: filtered-colimit-stable monomorphisms
type: category
relation: has
dual_property_id: cofiltered-limit-stable epimorphisms
description:
A category has <i>filtered-colimit-stable monomorphisms</i> if it has filtered colimits and for every filtered diagram of monomorphisms $(X_i \to Y_i)$ also their colimit $\colim_i X_i \to \colim_i Y_i$ is a monomorphism.


id: filtered
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/filtered+category
dual_property_id: cofiltered
description:
A category is <i>filtered</i> if every finite diagram admits a cocone. Equivalently, it is inhabited, for every two objects $x,y$ there is a cospan $x \rightarrow s \leftarrow y$ (not necessarily universal), and every parallel pair $x \rightrightarrows y$ is coequalized by some morphism $y \to c$ (not necessarily universal). This is the special case of the notion of a $\kappa$-filtered category for $\kappa = \aleph_0$.


id: finitary algebraic
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/algebraic+category
description:
A category is <i>finitary algebraic</i> if it is equivalent to the category of models of a one-sorted finitary algebraic theory. This is also known as a variety of one-sorted finitary algebras.


id: finite copowers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/copower
dual_property_id: finite powers
description:
A category has <i>finite copowers</i> when for every object $X$ and every finite set $I$ the copower $I \otimes X$ exists. Equivalently, for every $n \in \IN$ the copower $n \otimes X$ exists.


id: finite coproducts
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/finite+coproduct
dual_property_id: finite products
description:
A category has <i>finite coproducts</i> if it has coproducts for finite families of objects. Equivalently, it has an initial object and binary coproducts.


id: finite powers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/powering
dual_property_id: finite copowers
description:
A category has <i>finite powers</i> when for every object $X$ and every finite set $I$ the power $X^I$ exists. Equivalently, for every $n \in \IN$ the power $X^n$ exists.


id: finite products
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/finite+product
dual_property_id: finite coproducts
description:
A category has <i>finite products</i> if it has products for finite families of objects. Equivalently, it has a terminal object and binary products.


id: finitely accessible
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/accessible+category
description:
A category is <i>finitely accessible</i> if it has filtered colimits and there is a (small) set $G$ of finitely presentable objects such that every object is a filtered colimit of objects in $G$.


id: gaunt
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/gaunt+category
dual_property_id: gaunt
description:
A category is <i>gaunt</i> when every isomorphism $f : A \to B$ must be the identity (in particular, $A = B$). This is the "skeletal variant" of being core-thin.


id: generalized variety
type: category
relation: is a
description:
A category is a <i>generalized variety</i> if it has sifted colimits and there is a (small) set $G$ of strongly finitely presentable objects such that every object is a sifted colimit of objects from $G$. Generalized varieties are like locally strongly finitely presentable categories but without colimits. The relation is similar as between finitely accessible and locally finitely presentable categories. This notion is defined in <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01, Def. 3.6]</a>.


id: generating set
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/separator
dual_property_id: cogenerating set
description:
A set of objects $S$ is called a <i>generating set</i> if for every pair of parallel morphisms $f,g : A \rightrightarrows B$, $f = g$ holds if and only if for every morphism $h : G \to A$ with $G \in S$ we have $f \circ h = g \circ h$. Equivalently, the functor $(\Hom(G,-))_{G \in S} : \C \to (\Set^+)^S$ is faithful. This property refers to the existence of a generating set.


id: generator
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/separator
dual_property_id: cogenerator
description:
An object $G$ of a category is called a <i>generator</i> if for every pair of parallel morphisms $f,g : A \rightrightarrows B$ the equation $f = g$ holds if for every morphism $h : G \to A$ we have $f \circ h = g \circ h$. Equivalently, the functor $\Hom(G,-) : \C \to \Set^+$ is faithful. This property refers to the existence of a generator. By definition, $G$ is a generator if and only if $\{G\}$ is a <a href="/category-property/generating_set">generating set</a>.


id: infinitary codistributive
type: category
relation: is
dual_property_id: infinitary distributive
description:
A category is <i>infinitary codistributive</i> if it has finite coproducts, all products, and for every object $A$ the functor $A \sqcup -$ preserves all products. Concretely, for every family of objects $(B_i)$ the canonical morphism $A \sqcup \prod_i B_i \to \prod_i (A \sqcup B_i)$ must be an isomorphism.


id: infinitary coextensive
type: category
relation: is
dual_property_id: infinitary extensive
description:
A category $\C$ is <i>infinitary coextensive</i> when it has products and for all families of objects $(A_i)_{i \in I}$ the product functor $\prod_{i \in I} A_i / \C/A_i \to \prod_{i \in I} A_i / \C$ is an equivalence of categories.
This terminology does not seem to be common, but we have added it as a dual for the more commonly known property of being infinitary extensive.


id: infinitary distributive
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/distributive+category
dual_property_id: infinitary codistributive
description:
A category is <i>infinitary distributive</i> if it has finite products, all coproducts, and for every object $A$ the functor $A \times -$ preserves all coproducts. Concretely, for every family of objects $(B_i)$ the canonical morphism $\coprod_i (A \times B_i) \to A \times \coprod_i B_i$ must be an isomorphism.


id: infinitary extensive
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/extensive+category
dual_property_id: infinitary coextensive
description:
A category $\C$ is <i>infinitary extensive</i> when it has coproducts and for all families of objects $(A_i)_{i \in I}$ the coproduct functor $\prod_{i \in I} \C/A_i \to \C/(\coprod_{i \in I} A_i)$ is an equivalence of categories. Equivalently, pullbacks of coproduct inclusions along arbitrary morphisms exist and coproducts are disjoint and stable under pullback.


id: inhabited
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/inhabited+set
dual_property_id: inhabited
description:
A category is <i>inhabited</i> if it has at least one object. In classical logic, this is equivalent to being <i>non-empty</i> (which is a double negation).


id: initial object
type: category
relation: has an
nlab_link: https://ncatlab.org/nlab/show/initial+object
dual_property_id: terminal object
description:
An <i>initial object</i> is an object that has a unique morphism to every object in the category. This property refers to the existence of an initial object.


id: inverse
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/inverse+category
dual_property_id: direct
description:
A category is <i>inverse</i> if its dual is direct, i.e., if it contains no infinite sequence of non-identity morphisms of the form $$A_0 \to A_1 \to A_2 \to \cdots.$$


id: kernels
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/kernel
dual_property_id: cokernels
description:
A category has <i>kernels</i> if it has zero morphisms and every morphism $f : A \to B$ has a kernel, i.e. an equalizer of $f$ with the zero morphism $0_{A,B} : A \to B$.


id: left cancellative
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/cancellative+category
dual_property_id: right cancellative
description:
A category is <i>left cancellative</i> if for every morphism $f : A \to B$ and every parallel pair of morphisms $g,h : T \rightrightarrows A$ with $f \circ g = f \circ h$ we have $g = h$. Equivalently, every morphism is a monomorphism.


id: locally ℵ₁-presentable
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+presentable+category
description:
This is the special case of the notion of locally $\kappa$-presentable categories, where $\kappa = \aleph_1$ is the first uncountable cardinal.


id: locally cocartesian coclosed
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+cocartesian+coclosed+category
dual_property_id: locally cartesian closed
description:
A category is <i>locally cocartesian coclosed</i> if its dual is locally cartesian closed, i.e. if each of its coslice categories is cocartesian coclosed.


id: locally essentially small
type: category
relation: is
dual_property_id: locally essentially small
description:
A category is <i>locally essentially small</i> when for every pair of objects $A,B$ the collection of morphisms $A \to B$ is isomorphic to a set. (Here, we work with a set-theoretic foundation in which there are sets and collections. Categories are based on collections of objects and morphisms.) Equivalently, the category is equivalent to a locally small category. In contrast to being locally small, this condition is invariant under equivalences of categories. This is why we have added it to the database. For instance, every algebraic category is locally essentially small, but not necessarily locally small. This indicates that this is the "right" notion to work with.


id: locally finite
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+finite+category
dual_property_id: locally finite
description:
A category is <i>locally finite</i> or <i>Hom-finite</i> when for all objects $A,B$ the collection $\Hom(A,B)$ is finite. (We do not assume that this collection is an actual set. Therefore, this property is invariant under equivalences of categories.)


id: locally finitely multi-presentable
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+multipresentable+category
description:
A category is <i>locally finitely multi-presentable</i> if it satisfies one of the following equivalent conditions: <ol> <li>It is finitely accessible and multi-cocomplete.</li> <li>It is finitely accessible and has connected limits.</li> <li>It is equivalent to the category of models of a small (finite limit, coproduct)-sketch.</li> </ol> For equivalence of conditions above, see Thm. 4.30, Thm. 4.32, and the remark below in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>.


id: locally finitely presentable
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+finitely+presentable+category
description:
A category is <i>locally finitely presentable</i> if it satisfies one of the following equivalent conditions: <ol> <li>It is finitely accessible and cocomplete.</li> <li>It is finitely accessible and complete.</li> <li>It is equivalent to the category of finite-limit-preserving functors to $\Set$ from a small category with finite limits.</li> <li>It is equivalent to the category of models of a small finite-limit sketch.</li> </ol> For equivalence of conditions above, see Cor. 2.47, Thm. 1.46, and Cor. 1.52 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. This is the same as being locally $\aleph_0$-presentable.


id: locally multi-presentable
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+multipresentable+category
description:
Let $\kappa$ be a regular cardinal. A category is <i>locally $\kappa$-multi-presentable</i> if it satisfies one of the following equivalent conditions: <ol> <li>It is $\kappa$-accessible and multi-cocomplete.</li> <li>It is $\kappa$-accessible and has connected limits.</li> <li>It is equivalent to the category of models of a small ($\kappa$-limit, coproduct)-sketch.</li> </ol> For equivalence of conditions above, see Thm. 4.30, Thm. 4.32, and the remark below in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. A category is <i>locally multi-presentable</i> if it is locally $\kappa$-multi-presentable for some $\kappa$.


id: locally presentable
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+presentable+category
dual_property_id: locally copresentable
description:
Let $\kappa$ be a regular cardinal. A category is <i>locally $\kappa$-presentable</i> if it satisfies one of the following equivalent conditions: <ol> <li>It is $\kappa$-accessible and cocomplete.</li> <li>It is $\kappa$-accessible and complete.</li> <li>It is equivalent to the category of $\kappa$-limit-preserving functors to $\Set$ from a small category with $\kappa$-limits.</li> <li>It is equivalent to the category of models of a small $\kappa$-limit sketch.</li> </ol> For equivalence of conditions above, see Cor. 2.47, Thm. 1.46, and Cor. 1.52 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. A category is <i>locally presentable</i> if it is locally $\kappa$-presentable for some regular cardinal $\kappa$.


id: locally small
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+small+category
dual_property_id: locally small
description:
A category is <i>locally small</i> when for every pair of objects $A,B$ the collection of morphisms $A \to B$ is a set. Here, we work with a set-theoretic foundation in which there are sets and collections. Categories are based on collections of objects and morphisms.


id: locally strongly finitely presentable
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/locally+strongly+finitely+presentable+category
description:
A category is <i>locally strongly finitely presentable</i> if it is cocomplete and there is a set $G$ of strongly finitely presentable objects such that every object is a sifted colimit of objects from $G$. There are several equivalent conditions: <ol> <li>It is equivalent to the category of models of a many-sorted finitary algebraic theory.</li> <li>It is equivalent to the category of finite-product-preserving functors to $\Set$ from a small category with finite products (=Lawvere theory).</li> <li>It is equivalent to the category of models of a small finite-product sketch.</li> <li>It is equivalent to the Eilenberg&ndash;Moore category of a finitary (=filtered-colimit-preserving) monad on $\Set^S$ for some set $S$.</li> <li>It is equivalent to the Eilenberg&ndash;Moore category of a sifted-colimit-preserving monad on $\Set^S$ for some set $S$. (cf. [<a href="https://doi.org/10.2168/LMCS-8(3:14)2012" target="_blank">KR12</a>, Proposition 3.3])</li> </ol> A category satisfying this property is simply called a <i>variety</i> (of algebras) by some authors, although one should be aware that this term is sometimes used only for the one-sorted case.


id: mono-regular
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/regular+monomorphism
dual_property_id: epi-regular
description:
A category is <i>mono-regular</i> when every monomorphism is regular, i.e. the equalizer of a pair of morphisms. Notice that this is not standard terminology, <a href="https://math.stackexchange.com/questions/5031588" target="_blank">apparently</a> the literature has no name for this yet. A <i>preadditive</i> category is mono-regular iff it is normal. The notion of a normal category is reserved for categories with zero morphisms, while mono-regular applies to all categories.


id: multi-algebraic
type: category
relation: is
description:
A category is <i>multi-algebraic</i> if it satisfies one of the following equivalent conditions: <ol> <li>It is a multi-cocomplete generalized variety, that is, it has multi-colimits and sifted colimits of all small diagrams, and there is a (small) set $G$ of strongly finitely presentable objects such that every object is a sifted colimit of objects from $G$.</li> <li>It is equivalent to the category of models of a small (finite product, coproduct)-sketch, shortly small <i>FPC-sketch</i>.</li> <li>It is equivalent to the category of multi-finite-product-preserving functors to $\Set$ from a small category with multi-finite-products (<i>multi-algebraic theory</i>). Here, multi-finite-products means multi-limits of finite discrete diagrams.</li> <li>It is equivalent to the category of models of a small multi-finite-product sketch.</li> </ol> Multi-algebraic categories are like locally strongly finitely presentable categories but only with multi-colimits. The relation is similar as between locally finitely multi-presentable and locally finitely presentable categories. For equivalence of conditions above, see [<a href="https://doi.org/10.1016/S0022-4049(01)00015-9" target="_blank">AR01a</a>, Lem. 1] and [<a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">AR01b</a>, Thm. 4.4]. This notion was originally introduced by <a href="https://doi.org/10.1007/BF01224953" target="_blank">Diers</a>.


id: multi-cocomplete
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/multilimit
dual_property_id: multi-complete
description:
A <i>multi-colimit</i> of a diagram $D : \S \to \C$ is a set $I$ of cocones under $D$ such that every cocone under $D$ uniquely factors through a unique cocone belonging to $I$. This property refers to the existence of multi-colimits of small diagrams. Note that any diagram with no cocone admits a multi-colimit, which is the empty set of cocones.


id: multi-complete
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/multilimit
dual_property_id: multi-cocomplete
description:
A <i>multi-limit</i> of a diagram $D : \S \to \C$ is a set $I$ of cones over $D$ such that every cone over $D$ uniquely factors through a unique cone belonging to $I$. This property refers to the existence of multi-limits of small diagrams. Note that any diagram with no cone admits a multi-limit, which is the empty set of cones.


id: multi-initial object
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/multilimit
dual_property_id: multi-terminal object
description:
This property refers to the existence of a multi-colimit of the empty diagram. A category has a multi-initial object if and only if the collection of all connected components is isomorphic to a set, and each connected component has a initial object.


id: multi-terminal object
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/multilimit
dual_property_id: multi-initial object
description:
This property refers to the existence of a multi-limit of the empty diagram. A category has a multi-terminal object if and only if the collection of all connected components is isomorphic to a set, and each connected component has a terminal object.


id: natural numbers object
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/natural+numbers+object
description:
A <i>natural numbers object</i> (NNO) in a category with finite products is a triple $$(N,\, z : 1 \to N,\, s : N \to N)$$ satisfying the following universal property: for all $f : A \to X$, $g : X \to X$ there is a unique $\Phi : A \times N \to X$ such that $\Phi(a,z)=f(a)$ and $\Phi(a,s(n)) = g(\Phi(a,n))$ in element notation.
This concept is an abstraction of the set of natural numbers, which indeed provide a NNO for the category of sets. We have used the parametrized definition here which is more natural (sic!) for categories that are not cartesian closed (cf. <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Remark 2.5.3).


id: normal
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/normal+monomorphism
dual_property_id: conormal
description:
A category is <i>normal</i> if it has zero morphisms and every monomorphism is a kernel of some morphism (in which case case it is also called a <i>normal monomorphism</i>). The assumption of having zero morphisms makes it possible to talk about kernels.


id: pointed
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/pointed+category
dual_property_id: pointed
description:
A category is <i>pointed</i> when it has a zero object, i.e. an object which is both initial and terminal.


id: powers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/powering
dual_property_id: copowers
description:
If $X$ is an object and $I$ is a set, the <i>power</i> is defined as the product $X^I \coloneqq \prod_{i \in I} X$. This property refers to the existence of powers.


id: preadditive
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/Ab-enriched+category
dual_property_id: preadditive
description:
A category is <i>preadditive</i> when it is locally essentially small* and each hom-set carries the structure of an abelian group such that the composition is bilinear. Notice that "preadditive" is an extra structure. The property here just says that some preadditive structure exists.
*We demand this instead of the more common "locally small" to ensure that preadditive categories are invariant under equivalences of categories.


id: products
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/cartesian+product
dual_property_id: coproducts
description:
Given a family of objects $(A_i)_{i \in I}$, a <i>product</i> $\prod_{i \in I} A_i$ is defined as an object with morphisms $p_i : \prod_{i \in I} A_i \to A_i$ satisfying the following universal property: For every object $T$ and every family of morphisms $(f_i : T \to A_i)_{i \in I}$ there is a unique morphism $f : T \to \prod_{i \in I} A_i$ such that $p_i \circ f = f_i$ for all $i \in I$. This property refers to the existence of <i>small products</i>, i.e., products of small families of objects.


id: pullbacks
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/pullback
dual_property_id: pushouts
description:
A category $\C$ has <i>pullbacks</i> if every cospan of morphisms $X \rightarrow S \leftarrow Y$ has a pullback $X \times_S Y$. This is also known as a fiber product. Equivalently, the slice category $\C/S$ has binary products.


id: pushouts
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/pushout
dual_property_id: pullbacks
description:
A category $\C$ has <i>pushouts</i> if every span of morphisms $X \leftarrow S \rightarrow Y$ has a pushout $X \sqcup_S Y$. This is also known as a fiber coproduct. Equivalently, the coslice category $S/\C$ has binary coproducts.


id: quasitopos
type: category
relation: is a
nlab_link: https://ncatlab.org/nlab/show/quasitopos
description:
A <i>quasitopos</i> is a category which is finitely complete, finitely cocomplete, locally cartesian closed, and has a regular subobject classifier. This gives the category properties similar to those of an elementary topos; a major difference is that a quasitopos need not be balanced.
Note that some authors use a strong subobject classifier in place of a regular subobject classifier in the definition, i.e. a morphism $\top : 1 \to \Omega$ which classifies strong monomorphisms. For the equivalence of these definitions, see <a href="https://math.stackexchange.com/q/4335533" target="_blank">MSE/4335533</a>.


id: quotient object classifier
type: category
relation: has a
dual_property_id: subobject classifier
description:
A category $\C$ has a <i>quotient object classifier</i> if its dual has a subobject classifier. This means that it has finite colimits and an epimorphism* $\top : \Psi \twoheadrightarrow 0$ such that for every epimorphism $e : A \twoheadrightarrow B$ there is a unique morphism $\psi_e : \Psi \to A$ such that $$\begin{CD} \Psi @>{\top}>> 0 \\ @V{\psi_e}VV @VV{!}V \\ A @>>{e}> B \end{CD}$$ is a pushout diagram. Equivalently, the functor $\Quot : \C \to \Set^+$ is representable. *Every morphism $\Psi \to 0$ is a split epimorphism anyway.


id: quotient-trivial
type: category
relation: is
dual_property_id: subobject-trivial
description:
A category is <i>quotient-trivial</i> if every epimorphism is an isomorphism. Equivalently, the poset of quotients of any object is trivial. This is no standard terminology. We have added it to the database since it clarifies the relationship between several related properties.


id: quotients of congruences
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/congruence
dual_property_id: coquotients of cocongruences
description:
A <i>congruence</i> (or internal equivalence relation) on an object $X$ of a category is a parallel pair $p_1, p_2 : E \rightrightarrows X$ which is jointly monomorphic, and such that for every object $T$, the image of $(p_1 \circ {-}, p_2 \circ {-}) : \Hom(T, E) \to \Hom(T, X)^2$ is an equivalence relation. The category has <i>quotients of congruences</i> if for each such congruence, there exists a coequalizer of $p_1$ and $p_2$. Note that in the case of a category with binary powers, the corresponding subobjects of $X \times X$ are also commonly referred to as congruences, or as internal equivalence relations.


id: reflexive coequalizers
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/reflexive+coequalizer
dual_property_id: coreflexive equalizers
description:
A <i>reflexive coequalizer</i> is a colimit of a diagram consisting of a parallel pair of morphisms with a common section (right inverse), which is the same concept as a coequalizer of such a parallel pair. This property refers to the existence of reflexive coequalizers.


id: regular quotient object classifier
type: category
relation: has a
dual_property_id: regular subobject classifier
description:
A category $\C$ has a <i>regular quotient object classifier</i> if its dual has a regular subobject classifier. This means that it has finite colimits and a regular epimorphism* $\top : \Psi \twoheadrightarrow 0$ such that for every regular epimorphism $e : A \twoheadrightarrow B$ there is a unique morphism $\psi_e : \Psi \to A$ such that $$\begin{CD} \Psi @>{\top}>> 0 \\ @V{\psi_e}VV @VV{!}V \\ A @>>{e}> B \end{CD}$$ is a pushout diagram. Equivalently, the functor $\Quot_{\reg} : \C \to \Set^+$ is representable. *Every morphism $\Psi \to 0$ is a split epimorphism anyway.


id: regular subobject classifier
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/subobject+classifier
dual_property_id: regular quotient object classifier
description:
A category $\C$ has a <i>regular subobject classifier</i> if it has finite limits and a regular monomorphism* $\top : 1 \hookrightarrow \Omega$ such that for every regular monomorphism $m : A \hookrightarrow B$ there is a unique morphism $\chi_m : B \to \Omega$ such that $$\begin{CD} A @>{m}>> B \\ @V{!}VV @VV{\chi_m}V \\ 1 @>>{\top}> \Omega \end{CD}$$ is a pullback diagram. Equivalently, the functor $\Sub_{\reg} : \C^{\op} \to \Set^+$ is representable. *Every morphism $1 \to \Omega$ is a split monomorphism and hence regular anyway.


id: regular-quotient-trivial
type: category
relation: is
dual_property_id: regular-subobject-trivial
description:
A category is <i>regular-quotient-trivial</i> if every regular epimorphism is an isomorphism. Equivalently, the poset of regular quotients of any object is trivial. Or as another equivalent condition, if a parallel pair $f, g : X \rightrightarrows Y$ has a coequalizer, then $f = g$. This is not standard terminology. We have added it to the database since it clarifies the relationship between several related properties.


id: regular-subobject-trivial
type: category
relation: is
dual_property_id: regular-quotient-trivial
description:
A category is <i>regular-subobject-trivial</i> if every regular monomorphism is an isomorphism. Equivalently, the poset of regular subobjects of any object is trivial. Or as another equivalent condition, whenever a parallel pair $f, g : X \rightrightarrows Y$ has an equalizer, then $f = g$. This is not standard terminology. We have added it to the database since it clarifies the relationship between several related properties.


id: regular
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/regular+category
dual_property_id: coregular
description:
A category is <i>regular</i> when it is finitely complete, for every morphism $X \to Y$ its kernel pair $X \times_Y X \rightrightarrows X$ has a coequalizer, and regular epimorphisms are stable under pullbacks.


id: right cancellative
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/cancellative+category
dual_property_id: left cancellative
description:
A category is <i>right cancellative</i> if for every morphism $f : A \to B$ and every parallel pair of morphisms $g,h : B \rightrightarrows T$ with $g \circ f = h \circ f$ we have $g = h$. Equivalently, every morphism is an epimorphism.


id: semi-strongly connected
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/strongly+connected+category
dual_property_id: semi-strongly connected
description:
A category is <i>semi-strongly connected</i> if it is inhabited and for every two objects $A,B$ there is a morphism $A \to B$ or there is a morphism $B \to A$. Notice that this is stronger than being connected, and that posets with this property are precisely the inhabited totally ordered sets.


id: sequential colimits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/sequential+colimit
dual_property_id: sequential limits
description:
A category has <i>sequential colimits</i> if it has colimits of diagrams of the form: $\bullet \to \bullet \to \bullet \to \cdots$.


id: sequential limits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/sequential+limit
dual_property_id: sequential colimits
description:
A category has <i>sequential limits</i> if it has limits of diagrams of the form $\cdots \bullet \to \bullet \to \bullet$.


id: sifted colimits
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/sifted+colimit
dual_property_id: cosifted limits
description:
A category has <i>sifted colimits</i> if it has colimits of diagrams indexed by small <a href="/category-property/sifted">sifted</a> categories.


id: sifted
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/sifted+category
dual_property_id: cosifted
description:
A category $\C$ is <i>sifted</i> if it is inhabited and the diagonal functor $\Delta : \C \to \C \times \C$ is final, i.e. if it is non-empty and for any two objects $X,Y \in \C$ the category of cospans $$X \rightarrow Z \leftarrow Y$$ is connected. Equivalently, a small category $\C$ is sifted if $\colim : \Set^{\C} \to \Set$ preserves finite products. This property is a weaker notion than being filtered.


id: skeletal
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/skeleton
dual_property_id: skeletal
description:
A category is <i>skeletal</i> when isomorphic objects are already equal. Every category is equivalent to a skeletal category (using the axiom of choice).


id: small
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/small+category
dual_property_id: small
description:
A category is <i>small</i> when the collection of objects and the collection of morphisms are sets, i.e. small.


id: split abelian
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/split+exact+sequence
dual_property_id: split abelian
description:
A category is <i>split abelian</i> if it is abelian and every short exact sequence splits. Equivalently, every object is projective. Equivalently, every object is injective.


id: strict initial object
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/strict+initial+object
dual_property_id: strict terminal object
description:
A <i>strict initial object</i> is an initial object $0$ such that every morphism $A \to 0$ is an isomorphism. This property refers to the existence of a strict initial object.


id: strict terminal object
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/strict+terminal+object
dual_property_id: strict initial object
description:
A <i>strict terminal object</i> is a terminal object $1$ such that every morphism $1 \to A$ is an isomorphism. This property refers to the existence of a strict terminal object.


id: strongly connected
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/strongly+connected+category
dual_property_id: strongly connected
description:
A category is <i>strongly connected</i> if it is inhabited and for every two objects $A,B$ there is a morphism $A \to B$. In other words, each hom-set is inhabited. Notice that when a terminal object $1$ exists, this property means that every object $A$ admits a morphism $1 \to A$.


id: subobject classifier
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/subobject+classifier
dual_property_id: quotient object classifier
description:
A category $\C$ has a <i>subobject classifier</i> if it has finite limits and a monomorphism* $\top : 1 \hookrightarrow \Omega$ such that for every monomorphism $m : A \hookrightarrow B$ there is a unique morphism $\chi_m : B \to \Omega$ such that $$\begin{CD} A @>{m}>> B \\ @V{!}VV @VV{\chi_m}V \\ 1 @>>{\top}> \Omega \end{CD}$$ is a pullback diagram. Equivalently, the functor $\Sub : \C^{\op} \to \Set^+$ is representable. *Every morphism $1 \to \Omega$ is a split monomorphism anyway.


id: subobject-trivial
type: category
relation: is
dual_property_id: quotient-trivial
description:
A category is <i>subobject-trivial</i> if every monomorphism is an isomorphism. Equivalently, the poset of subobjects of any object is trivial. This is no standard terminology. We have added it to the database since it clarifies the relationship between several related properties.


id: terminal object
type: category
relation: has a
nlab_link: https://ncatlab.org/nlab/show/terminal+object
dual_property_id: initial object
description:
A <i>terminal object</i> (or <i>final object</i>) is an object that has a unique morphism from every object in the category. This property refers to the existence of a terminal object.


id: thin
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/thin+category
dual_property_id: thin
description:
A category is <i>thin</i> when between any pair of objects there is at most one morphism. Such categories correspond to preordered collections.


id: trivial
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/terminal+category
dual_property_id: trivial
description:
A category is <i>trivial</i> if it is equivalent to the <a href="/category/1">trivial category</a>. Equivalently, there is an initial object $0$ such that for every object $A$ the unique morphism $0 \to A$ is an isomorphism. Notice that we do not demand that the category is <i>isomorphic</i> to the trivial category. As a consequence, every inhabited indiscrete category is trivial in our sense.


id: unital
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/unital+category
dual_property_id: counital
description:
A category is <i>unital</i> if it has a zero object, finite limits, and for all objects $X,Y$ the two morphisms $(\id_X,0) : X \hookrightarrow X \times Y$ and $(0,\id_Y) : Y \hookrightarrow X \times Y$ are jointly strongly epimorphic. This means: there is no proper subobject of $X \times Y$ that contains $X$ and $Y$. When binary coproducts exist, the canonical morphism $X \sqcup Y \to X \times Y$ therefore must be a <a href="https://ncatlab.org/nlab/show/strong+epimorphism" target="_blank">strong epimorphism</a>.


id: well-copowered
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/well-powered+category
dual_property_id: well-powered
description:
A category is <i>well-copowered</i> if the collection of quotients of any object is isomorphic to a set.


id: well-powered
type: category
relation: is
nlab_link: https://ncatlab.org/nlab/show/well-powered+category
dual_property_id: well-copowered
description:
A category is <i>well-powered</i> if the collection of subobjects of any object is isomorphic to a set.


id: wide pullbacks
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/wide+pullback
dual_property_id: wide pushouts
description:
A category $\C$ has <i>wide pullbacks</i> if for every object $S$ the slice category $\C/S$ has arbitrary small products.


id: wide pushouts
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/wide+pushout
dual_property_id: wide pullbacks
description:
A category $\C$ has <i>wide pushouts</i> if for every object $S$ the coslice category $S/\C$ has arbitrary small coproducts.


id: zero morphisms
type: category
relation: has
nlab_link: https://ncatlab.org/nlab/show/zero+morphism
dual_property_id: zero morphisms
description:
A category has <i>zero morphisms</i> if for every pair of objects $A,B$ there is a distinguished morphism $0_{A,B} : A \to B$, called the zero morphism, such that we have $f \circ 0_{A,B} = 0_{A,C}$ and $0_{B,C} \circ g = 0_{A,C}$ for all morphisms $f : B \to C$ and $g : A \to B$. The zero morphisms are unique if they exist, hence this is actually a property of the category.


id: comonadic
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/comonadic+functor
dual_property_id: monadic
description:
A functor $F : \C \to \D$ is <i>comonadic</i> when there is a comonad $T$ on $\D$ such that $F$ is equivalent to the forgetful functor $U^T : \CoAlg(T) \to \D$.


id: conservative
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/conservative+functor
dual_property_id: conservative
description:
A functor $F : \C \to \D$ is <i>conservative</i> when it is isomorphism-reflecting: If $f$ is a morphism in $\C$ such that $F(f)$ is an isomorphism, then $f$ is an isomorphism.


id: coreflector
type: functor
relation: is a
nlab_link: https://ncatlab.org/nlab/show/coreflective+subcategory
dual_property_id: reflector
description:
A functor $F : \C \to \D$ is a <i>coreflector</i> if it is right adjoint to a functor $G : \D \to \C$ which is fully faithful. Hence, $G$ is equivalent to the inclusion of a full coreflective subcategory. The condition that $G$ is fully faithful can also be expressed by the condition that the unit $\eta : \id_{\D} \to F \circ G$ is an isomorphism (Prop. 3.4 at the <a href="https://ncatlab.org/nlab/show/adjoint+functor#FullyFaithfulAndInvertibleAdjoints" target="_blank">nLab</a>).


id: coregular
type: functor
relation: is
dual_property_id: regular
description:
A functor $F : \C \to \D$ is <i>coregular</i> when it preserves finite colimits and regular monomorphisms. This notion is used in particular when $\C$ and $\D$ are coregular categories.


id: dominant
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/dominant+functor
dual_property_id: dominant
description:
A functor $F : \C \to \D$ is <i>dominant</i> when every object $Y \in \D$ there is an object $X \in \C$ such that $Y$ is a retract of $F(X)$, i.e. there is a split monomorphism $Y \hookrightarrow F(X)$.


id: equivalence
type: functor
relation: is an
nlab_link: https://ncatlab.org/nlab/show/equivalence+of+categories
dual_property_id: equivalence
description:
A functor $F : \C \to \D$ is an <i>equivalence</i> when there is a functor $G : \D \to \C$ with $F \circ G \cong \id_{\D}$ and $G \circ F \cong \id_{\C}$. In contrast to an isomorphism of categories, these compositions are only required to be <i>isomorphic</i> to the identities.


id: essentially injective
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/essentially+injective+functor
dual_property_id: essentially injective
description:
A functor $F : \C \to \D$ is <i>essentially injective</i> if the implication $$F(A) \cong F(B) \implies A \cong B$$ holds for all objects $A,B \in \C$. This is a condition solely on the objects themselves, i.e. it is <i>not</i> required that every isomorphism between $F(A)$ and $F(B)$ is induced by an isomorphism between $A$ and $B$ (cf. <a href="/functor-property/full_on_isomorphisms">full on isomorphisms</a>). An equivalent condition is that $F$ induces an injective map on isomorphism classes.


id: essentially surjective
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/essentially+surjective+functor
dual_property_id: essentially surjective
description:
A functor $F : \C \to \D$ is <i>essentially surjective</i> (on objects) when every object $Y \in \D$ is isomorphic to $F(X)$ for some object $X \in \C$.


id: faithful
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/faithful+functor
dual_property_id: faithful
description:
A functor $F$ is <i>faithful</i> when it is injective on Hom-sets: If $F(f)=F(g)$ for two morphisms $f,g : A \rightrightarrows B$, then $f=g$.


id: full on isomorphisms
type: functor
relation: is
dual_property_id: full on isomorphisms
description:
A functor $F : \C \to \D$ is <i>full on isomorphisms</i> when, for every pair of objects $A,B \in \C$, every isomorphism $F(A) \to F(B)$ is induced by an isomorphism $A \to B$.


id: full
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/full+functor
dual_property_id: full
description:
A functor $F$ is <i>full</i> when it is surjective on Hom-sets: Every morphism $F(A) \to F(B)$ is induced by a morphism $A \to B$.


id: fully faithful
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/full+and+faithful+functor
dual_property_id: fully faithful
description:
A functor $F : \C \to \D$ is <i>fully faithful</i> when it is full and faithful. This means that for all objects $A,B \in \C$ the map $\Hom(A,B) \to \Hom(F(A),F(B))$, $f \mapsto F(f)$ is a bijection.


id: isomorphism
type: functor
relation: is an
nlab_link: https://ncatlab.org/nlab/show/equivalence+of+categories#Isomorphism
dual_property_id: isomorphism
description:
A functor $F : \C \to \D$ is an <i>isomorphism</i> when there is a functor $G : \D \to \C$ with $F \circ G = \id_{\D}$ and $G \circ F = \id_{\C}$. In contrast to an equivalence of categories, these compositions are required to be <i>equal</i> to the identities.


id: left adjoint
type: functor
relation: is a
nlab_link: https://ncatlab.org/nlab/show/left+adjoint
dual_property_id: right adjoint
description:
A functor $F : \C \to \D$ is a <i>left adjoint</i> when there is a functor $G : \D \to \C$ such that there are natural bijections $\Hom(F(A),B) \cong \Hom(A,G(B))$.


id: left-invertible
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/inverse+functor
dual_property_id: left-invertible
description:
A <i>left inverse</i> of a functor $F : \C \to \D$ is a functor $G : \D \to \C$ satisfying $G \circ F \cong \id_{\C}$. We do not require $G \circ F = \id_{\C}$ here, which is often too strict. A functor is called <i>left-invertible</i> when it has a left inverse.


id: monadic
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/monadic+functor
dual_property_id: comonadic
description:
A functor $F : \C \to \D$ is <i>monadic</i> when there is a monad $T$ on $\D$ such that $F$ is equivalent to the forgetful functor $U^T : \Alg(T) \to \D$.


id: preserves binary coproducts
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+colimit
dual_property_id: preserves binary products
description:
A functor $F : \C \to \D$ <i>preserves binary coproducts</i> when for every pair of objects $A,B \in \C$ whose coproduct $A \sqcup B$ exists, also the coproduct $F(A) \sqcup F(B)$ exists and such that the canonical morphism $F(A) \sqcup F(B) \to F(A \sqcup B)$ is an isomorphism.


id: preserves binary products
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+limit
dual_property_id: preserves binary coproducts
description:
A functor $F : \C \to \D$ <i>preserves binary products</i> when for every pair of objects $A,B \in \C$ whose product $A \times B$ exists, also the product $F(A) \times F(B)$ exists and such that the canonical morphism $F(A \times B) \to F(A) \times F(B)$ is an isomorphism.


id: preserves coequalizers
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+colimit
dual_property_id: preserves equalizers
description:
A functor $F$ <i>preserves coequalizers</i> when for every parallel pair of morphisms $f,g : A \rightrightarrows B$ whose coequalizer $p : B \to Q$ exists, also $F(p) : F(B) \to F(Q)$ is a coequalizer of $F(f),F(g) : F(A) \rightrightarrows F(B)$.


id: preserves coproducts
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+colimit
dual_property_id: preserves products
description:
A functor $F$ <i>preserves coproducts</i> when for every family of objects $(A_i)$ in the source whose coproduct $\prod_i A_i$ exists, also the coproduct $\coprod_i F(A_i)$ exists in the target and such that the canonical morphism $\coprod_i F(A_i) \to F(\coprod_i A_i)$ is an isomorphism.


id: preserves coreflexive equalizers
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/reflexive+coequalizer
dual_property_id: preserves reflexive coequalizers
description:
A functor $F$ <i>preserves coreflexive equalizers</i> when for every coreflexive parallel pair of morphisms $f,g : A \rightrightarrows B$ whose equalizer $i : E \to A$ exists, also $F(i) : F(E) \to F(A)$ is an equalizer of $F(f),F(g) : F(A) \rightrightarrows F(B)$.


id: preserves epimorphisms
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/epimorphism
dual_property_id: preserves monomorphisms
description:
A functor $F : \C \to \D$ <i>preserves epimorphisms</i> if every epimorphism in $\C$ is mapped to an epimorphism in $\D$.
This property is useful to rule out some adjunctions.


id: preserves equalizers
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+limit
dual_property_id: preserves coequalizers
description:
A functor $F$ <i>preserves equalizers</i> when for every parallel pair of morphisms $f,g : A \rightrightarrows B$ whose equalizer $i : E \to A$ exists, also $F(i) : F(E) \to F(A)$ is an equalizer of $F(f),F(g) : F(A) \rightrightarrows F(B)$.


id: preserves finite coproducts
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+colimit
dual_property_id: preserves finite products
description:
A functor $F$ <i>preserves finite coproducts</i> when for every family of objects $(A_i)$ in the source whose coproduct $\prod_i A_i$ exists, also the coproduct $\coprod_i F(A_i)$ exists in the target and such that the canonical morphism $\coprod_i F(A_i) \to F(\coprod_i A_i)$ is an isomorphism.


id: preserves finite products
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+limit
dual_property_id: preserves finite coproducts
description:
A functor $F$ <i>preserves finite products</i> when for every finite family of objects $(A_i)$ in the source whose product $\prod_i A_i$ exists, also the product $\prod_i F(A_i)$ exists in the target and such that the canonical morphism $F(\prod_i A_i) \to \prod_i F(A_i)$ is an isomorphism.


id: preserves initial objects
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+colimit
dual_property_id: preserves terminal objects
description:
A functor $F$ <i>preserves initial objects</i> when it maps every initial object to an initial object. It is not assumed that the source category has a initial object.


id: preserves monomorphisms
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/monomorphism
dual_property_id: preserves epimorphisms
description:
A functor $F : \C \to \D$ <i>preserves monomorphisms</i> if every monomorphism in $\C$ is mapped to a monomorphism in $\D$.
This property is useful to rule out some adjunctions.


id: preserves products
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+limit
dual_property_id: preserves coproducts
description:
A functor $F$ <i>preserves products</i> when for every family of objects $(A_i)$ in the source whose product $\prod_i A_i$ exists, also the product $\prod_i F(A_i)$ exists in the target and such that the canonical morphism $F(\prod_i A_i) \to \prod_i F(A_i)$ is an isomorphism.


id: preserves reflexive coequalizers
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/reflexive+coequalizer
dual_property_id: preserves coreflexive equalizers
description:
A functor $F$ <i>preserves reflexive coequalizers</i> when for every reflexive parallel pair of morphisms $f,g : A \rightrightarrows B$ whose coequalizer $p : B \to Q$ exists, also $F(p) : F(B) \to F(Q)$ is a coequalizer of $F(f),F(g) : F(A) \rightrightarrows F(B)$.


id: preserves regular epimorphisms
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/regular+epimorphism
dual_property_id: preserves regular monomorphisms
description:
A functor $F : \C \to \D$ <i>preserves regular epimorphisms</i> if every regular epimorphism in $\C$ is mapped to a regular epimorphism in $\D$.


id: preserves regular monomorphisms
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/regular+monomorphism
dual_property_id: preserves regular epimorphisms
description:
A functor $F : \C \to \D$ <i>preserves regular monomorphisms</i> if every regular monomorphism in $\C$ is mapped to a regular monomorphism in $\D$.


id: preserves terminal objects
type: functor
relation: preserves
nlab_link: https://ncatlab.org/nlab/show/preserved+limit
dual_property_id: preserves initial objects
description:
A functor $F$ <i>preserves terminal objects</i> when it maps every terminal object to a terminal object. It is not assumed that the source category has a terminal object.


id: reflector
type: functor
relation: is a
nlab_link: https://ncatlab.org/nlab/show/reflective+subcategory
dual_property_id: coreflector
description:
A functor $F : \C \to \D$ is a <i>reflector</i> if it is left adjoint to a functor $G : \D \to \C$ which is fully faithful. Hence, $G$ is equivalent to the inclusion of a full reflective subcategory. The condition that $G$ is fully faithful can also be expressed by the condition that the counit $\varepsilon : F \circ G \to \id_{\D}$ is an isomorphism (Prop. 3.4 at the <a href="https://ncatlab.org/nlab/show/adjoint+functor#FullyFaithfulAndInvertibleAdjoints" target="_blank">nLab</a>).


id: regular
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/regular+functor
dual_property_id: coregular
description:
A functor $F : \C \to \D$ is <i>regular</i> when it preserves finite limits and regular epimorphisms. This notion is used in particular when $\C$ and $\D$ are regular categories.


id: representable
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/representable+functor
description:
A functor $F : \C \to \D$ is <i>representable</i> if $\C$ is locally small, $\D = \Set$, and there is an object $A \in \C$ with $F \cong \Hom(A,-)$.


id: right adjoint
type: functor
relation: is a
nlab_link: https://ncatlab.org/nlab/show/right+adjoint
dual_property_id: left adjoint
description:
A functor $F : \C \to \D$ is a <i>right adjoint</i> when there is a functor $G : \D \to \C$ such that there are natural bijections $\Hom(G(A),B) \cong \Hom(A,F(B))$.


id: right-invertible
type: functor
relation: is
nlab_link: https://ncatlab.org/nlab/show/inverse+functor
dual_property_id: right-invertible
description:
A <i>right inverse</i> of a functor $F : \C \to \D$ is a functor $G : \D \to \C$ satisfying $F \circ G \cong \id_{\D}$. We do not require $F \circ G = \id_{\D}$ here, which is often too strict. A functor is called <i>right-invertible</i> when it has a right inverse.


id: epimorphism
type: morphism
relation: is an
nlab_link: https://ncatlab.org/nlab/show/epimorphism
dual_property_id: monomorphism
description:
A morphism $f : A \to B$ is an <i>epimorphism</i> if it is right-cancellative, i.e. if $g \circ f = h \circ f$ for two morphisms $g,h : B \rightrightarrows T$, then $g = h$. In many concrete categories appearing in practice, these are are a bit harder to understand than monomorphisms; in particular, epimorphisms usually are not to be confused with surjective structure-preserving maps.


id: isomorphism
type: morphism
relation: is an
nlab_link: https://ncatlab.org/nlab/show/isomorphism
dual_property_id: isomorphism
description:
A morphism $f : A \to B$ is an <i>isomorphism</i> if it has an inverse, i.e. a morphism $f^{-1} : B \to A$ satisfying $f \circ f^{-1} = \id_B$ and $f^{-1} \circ f = \id_A$.


id: monomorphism
type: morphism
relation: is a
nlab_link: https://ncatlab.org/nlab/show/monomorphism
dual_property_id: epimorphism
description:
A morphism $f : A \to B$ is a <i>monomorphism</i> if it is left-cancellative, i.e. if $f \circ g = f \circ h$ for two morphisms $g,h : T \rightrightarrows A$, then $g = h$. In many concrete categories appearing in practice, these are injective structure-preserving maps.


id: regular epimorphism
type: morphism
relation: is a
nlab_link: https://ncatlab.org/nlab/show/regular+epimorphism
dual_property_id: regular monomorphism
description:
A morphism $f : A \to B$ is a <i>regular epimorphism</i> if it is the coequalizer of a pair of morphisms $g,h : C \rightrightarrows A$. In many categories appearing in practice, this is the same as a quotient.


id: regular monomorphism
type: morphism
relation: is a
nlab_link: https://ncatlab.org/nlab/show/regular+monomorphism
dual_property_id: regular epimorphism
description:
A morphism $f : A \to B$ is a <i>regular monomorphism</i> if it is the equalizer of a pair of morphisms $g,h : B \rightrightarrows C$. In many categories appearing in practice, this is the same as an embedding.


id: split epimorphism
type: morphism
relation: is a
nlab_link: https://ncatlab.org/nlab/show/split+epimorphism
dual_property_id: split monomorphism
description:
A morphism $f : A \to B$ is a <i>split epimorphism</i> if there is a morphism $g : B \to A$ with $f \circ g = \id_B$.


id: split monomorphism
type: morphism
relation: is a
nlab_link: https://ncatlab.org/nlab/show/split+monomorphism
dual_property_id: split epimorphism
description:
A morphism $f : A \to B$ is a <i>split monomorphism</i> if there is a morphism $g : B \to A$ with $g \circ f = \id_A$.


# property_assignments

structure_id: Ab
property_id: abelian
type: category
proof:
This is standard, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. VIII.


structure_id: Ab
property_id: split abelian
type: category
proof:
The short exact sequence $0 \xrightarrow{} \IZ \xrightarrow{p} \IZ \xrightarrow{} \IZ/p \xrightarrow{} 0$ does not split.


structure_id: Ab
property_id: CSP
type: category
proof:
The canonical homomorphism $\bigoplus_{n \geq 0} \IZ \to \prod_{n \geq 0} \IZ$ is not surjective, hence no epimorphism.


structure_id: Ab_fg
property_id: abelian
type: category
proof:
This follows from the fact for abelian groups and the fact that subgroups of finitely generated abelian groups are also finitely generated.


structure_id: Ab_fg
property_id: essentially countable
type: category
proof:
Every finitely generated abelian group is isomorphic to a group of the form $\IZ^n / U$, where $n \in \IN$ and $U$ is a subgroup of $\IZ^n$. Since $\IZ^n$ is Noetherian as a $\IZ$-module, $U$ is finitely generated, hence the category $\Ab_\fg$ has only countably many objects up to isomorphism. Furthermore, for any objects $A \cong \IZ^n / U$ and $B \cong \IZ^m / T$, the hom-set $\Hom(A,B)$ is countable. Indeed, precomposition with the quotient map yields an injection $\Hom(A,B) \hookrightarrow \Hom(\IZ^n, B) \cong B^n$, and $B^n$ is countable.


structure_id: Ab_fg
property_id: ℵ₁-accessible
type: category
proof:
The inclusion $\Ab_{\fg} \hookrightarrow \Ab$ is closed under $\aleph_1$-filtered colimits by <a href="https://mathoverflow.net/questions/400763/" target="_blank">MO/400763</a>. In particular, $\Ab_{\fg}$ has $\aleph_1$-filtered colimits. Since $\Ab_{\fg}$ is essentially small, there is a set $G$ such that every f.g. abelian group is isomorphic to one in $G$. So trivially it is also a $\aleph_1$-filtered colimit of such objects (take the constant diagram). Finally, every object is $\Ab_{\fg} = \Ab_{\fp}$ is finitely presentable in $\Ab$ and hence also in $\Ab_{\fg}$, a fortiori $\aleph_1$-presentable.


structure_id: Ab_fg
property_id: cogenerator
type: category
proof:
Let $Q$ be a finitely generated abelian group. By their well-known classification, we have $Q = F \oplus T$ for a free abelian group $F$ and a finite abelian group $T$. Let $p$ be a prime number which does not divide the order of $T$. Then $\Hom(\IZ/p, Q) = 0$, but $\IZ/p \neq 0$. Therefore, $Q$ is no cogenerator.


structure_id: Alg(R)
property_id: Malcev
type: category
proof:
This follows in the same way as for <a href="/category/Grp">$\Grp$</a>, see also Example 2.2.5 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>.


structure_id: Alg(R)
property_id: balanced
type: category
proof:
Take a prime ideal $P \subseteq R$ and consider the $R$-algebra $A \coloneqq R/P$ (which is an integral domain). Then the inclusion $A \hookrightarrow Q(A)$ is a counterexample.


structure_id: Alg(R)
property_id: semi-strongly connected
type: category
proof:
This is because already the full subcategory <a href="/category/CAlg(R)">$\CAlg(R)$</a> of commutative algebras is not semi-strongly connected.


structure_id: Alg(R)
property_id: cogenerating set
type: category
proof:
We apply <a href="/content/missing_cogenerating_sets">this lemma</a> to the collection of $R$-algebras which are fields: If $F$ is an $R$-algebra that is also a field and $A$ is a non-trivial $R$-algebra, any algebra homomorphism $F \to A$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables over some residue field of $R$ has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


structure_id: Alg(R)
property_id: codistributive
type: category
proof:
If $\sqcup$ denotes the coproduct of $R$-algebras (see <a href="https://math.stackexchange.com/questions/625874" target="_blank">MSE/625874</a> for their description) and $A$ is an $R$-algebra, the canonical morphism $A \sqcup R^2 \to (A \sqcup R)^2 = A^2$ is usually no isomorphism. For example, for $A = R[X]$ the coproduct on the LHS is not commutative, it has the algebra presentation $\langle X,E : E^2=E \rangle$.


structure_id: Alg(R)
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \Alg(R) \to \Set$ and the relation $S \subseteq U^2$ defined by $S(A) \coloneqq \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $R[X]$ and $S$ by $R \langle X,Y \rangle / \langle XY-X^2 \rangle$. It is clear that $S$ is reflexive, but not symmetric.


structure_id: Alg(R)
property_id: coregular
type: category
proof:
We just need to tweak the proof for <a href="/category/Ring">$\Ring$</a>. Since $R \neq 0$, there is an infinite field $K$ with a homomorphism $R \to K$. Since $K$ is infinite, we may choose some $\lambda \in K \setminus \{0,1\}$. Let $B \coloneqq M_2(K)$ and $A \coloneqq K \times K$. Then $A \to B$, $(x,y) \mapsto \diag(x,y)$ is a regular monomorphism: A direct calculation shows that a matrix is diagonal iff it commutes with $M \coloneqq \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & \lambda \end{smallmatrix}\bigr)$, so that $A \to B$ is the equalizer of the identity $B \to B$ and the conjugation $B \to B$, $X \mapsto M X M^{-1}$. Consider the homomorphism $A \to K$, $(a,b) \mapsto a$. We claim that $K \to K \sqcup_A B$ is not a monomorphism, because in fact, the pushout $K \sqcup_A B$ is zero: Since $A \to K$ is surjective with kernel $0 \times K$, the pushout is $B/\langle 0 \times K \rangle$, which is $0$ because $B$ is simple (<a href="https://math.stackexchange.com/questions/22629" target="_blank">proof</a>) or via a direct calculation with elementary matrices.


structure_id: Alg(R)
property_id: regular quotient object classifier
type: category
proof:
We may copy the proof for <a href="/category/CAlg(R)">$\CRing$</a> (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\Alg(R)$ would produce one in the reflective subcategory $\CAlg(R)$ by Lemma 1 <a href="/content/subcategories">here</a> (dualized).


structure_id: Alg(R)
property_id: cocartesian cofiltered limits
type: category
proof:
Consider the ring $A = R[X]$ and the sequence of rings $B_n = R[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $R[[Y]]$. Every element in the coproduct of rings $R[X] \sqcup R[[Y]]$ has a finite "free product" length. Now consider the elements  $$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n.$$ Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.


structure_id: Alg(R)
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/CRing">$\CAlg(R)$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the forgetful functor $\CAlg(R) \to \Alg(R)$. It preserves epimorphisms by <a href="https://math.stackexchange.com/questions/5133488" target="_blank">MSE/5133488</a>.


structure_id: Alg(R)
property_id: effective cocongruences
type: category
proof:
The counterexample is similar to the one for <a href="/category/Ring">$\Ring$</a>: Let $X \coloneqq R[p] / (p^2-p)$ with cocongruence $E \coloneqq R \langle p, q \rangle / (p^2-p, q^2-q, pq-q, qp-p)$.


structure_id: B
property_id: locally finite
type: category
proof:
There is a faithful functor $\IB \to \FinSet$ and <a href="/category/FinSet">$\FinSet$</a> is locally finite.


structure_id: BN
property_id: locally cartesian closed
type: category
proof:
The slice category $B\IN / *$ is isomorphic to the poset $(\IN,\geq)$ (not to $(\IN,\leq)$). This category is thin and and semi-strongly connected, <a href="/category-implication/sequential_implies_lcc">hence</a> cartesian closed.


structure_id: BN
property_id: ℵ₁-accessible
type: category
proof:
A proof can be found <a href="/content/aleph1-filtered-colimits-in-deloopings">here</a> as Proposition 2.


structure_id: BN
property_id: sequential limits
type: category
proof:
Assume that the sequence $\cdots \xrightarrow{1}  \bullet \xrightarrow{1} \bullet \xrightarrow{1} \bullet$ has a limit. This is a (universal) sequence of natural numbers $n_0,n_1,\dotsc$ satisfying $n_i = n_{i+1} + 1$. But then $n_i = n_0 - i$, and in particular $n_{n_0 + 1} = - 1$, a contradiction.


structure_id: BOn
property_id: left cancellative
type: category
proof:
It is well-known that ordinal addition satisfies $\alpha + \beta = \alpha + \gamma \implies \beta = \gamma$.


structure_id: BOn
property_id: well-copowered
type: category
proof:
This follows from the description of epimorphisms as finite ordinals, see <a href="https://math.stackexchange.com/questions/5029605" target="_blank">MO/5029605</a>.


structure_id: BOn
property_id: locally cartesian closed
type: category
proof:
The slice category $B\On / *$ is isomorphic to the poset $(\On,\geq)$ (not to $(\On,\leq)$). This category is thin and and semi-strongly connected, <a href="/category-implication/sequential_implies_lcc">hence</a> cartesian closed.


structure_id: BOn
property_id: ℵ₁-cofiltered
type: category
proof:
In fact, it is $\kappa$-cofiltered for every cardinal $\kappa$. By the dual of Theorem 2.2 at the <a href="https://ncatlab.org/nlab/show/filtered+category">nLab</a> it suffices to prove any set of objects has a cone (which is trivial in a one-object category) and that any set of parallel morphisms is equalized by some morphism. Here, this means that for every set of ordinals $A$ there is some ordinal $\beta$ such that $\alpha + \beta$ for $\alpha \in A$ does not depend on $\alpha$. Take $\beta$ to be any ordinal larger than $\sup(A)$ of the form $\omega^\gamma$. It is well-known that $\omega^\gamma$ has the property that $\alpha + \omega^\gamma = \omega^\gamma$ for all $\alpha < \omega^\gamma$ (Kunen's <i>Set Theory</i>, Exercise I.9.53), from which the claim follows.


structure_id: BOn
property_id: ℵ₁-filtered colimits
type: category
proof:
A proof can be found <a href="/content/aleph1-filtered-colimits-in-deloopings">here</a> as Proposition 3.


structure_id: BOn
property_id: balanced
type: category
proof:
Every finite ordinal is both a mono- and an epimorphism (see below), but only $0$ is an isomorphism.


structure_id: BOn
property_id: sequential colimits
type: category
proof:
Assume that the sequence $\bullet \xrightarrow{1} \bullet \xrightarrow{1} \cdots$ has a colimit. This mounts to a (universal) sequence of ordinals $\alpha_n$ with $\alpha_n = \alpha_{n+1} + 1$. But then $\alpha_{n+1} < \alpha_n$, contradicting the fact that $\alpha_0$ is well-ordered.


structure_id: BOn
property_id: pushouts
type: category
proof:
Assume that $1,\omega$ have a pushout. This is a (universal) pair of ordinals $\alpha,\beta$ with $\alpha + 1 = \beta + \omega$. But $\beta + \omega$ is a limit ordinal, while $\alpha + 1$ is not.


structure_id: BOn
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
The epimorphisms are the finite ordinals (see below), but the limit of the sequential diagram $\cdots \xrightarrow{1} * \xrightarrow{1} *$ is the ordinal $\omega$ by <a href="https://math.stackexchange.com/questions/5129138">MSE/5129138</a>.


structure_id: Ban
property_id: locally ℵ₁-presentable
type: category
proof:
Example 1.48 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>.


structure_id: Ban
property_id: cartesian filtered colimits
type: category
proof:
If $X$ is a Banach space and $(Y_i)$ is a filtered diagram of Banach spaces, the canonical map $\colim_i (X \times Y_i) \to X \times \colim_i Y_i$ is the completion of the canonical map in the category of normed vector spaces with non-expansive linear maps. Now the claim follows directly from <a href="/category/Met">$\Met$</a>.


structure_id: Ban
property_id: cocartesian cofiltered limits
type: category
proof:
If $X$ is a Banach space and $(Y_i)$ is a cofiltered diagram of Banach spaces, the canonical map $X \oplus \lim_i Y_i \to \lim_i (X \oplus Y_i)$ is an isomorphism: Since the forgetful functor $\Ban \to \Vect$ preserves finite coproducts and all limits, and $\Vect$ has the claimed property (see <a href="/category-implication/biproducts_cartesian_filtered_colimits">here</a>), the canonical map is bijective. It remains to show that it is isometric. For $(x,y) \in X \oplus \lim_i Y_i$ the norm in the domain is $|x| + \sup_i |y_i|$, and the norm in the codomain is $\sup_i (|x| + |y_i|)$, and these clearly agree.


structure_id: Ban
property_id: regular
type: category
proof:
It suffices to prove that regular epimorphisms are stable under pullbacks. We will use their classification via open unit balls below. So let $f : X \to Y$ be a regular epimorphism and let $g : T \to Y$ be any morphism. We need to show that the projection $X \times_Y T \to X$ is a regular epimorphism. Let $t \in T$ be an element of norm $<1$. Since $g$ is a linear contraction, $g(t)$ has norm $<1$. Since $f$ is a regular epimorphism, there is some $x \in X$ with norm $<1$ and $f(x) = g(t)$. Then $(x,t) \in X \times_Y T$ is a preimage of $t$ with norm $\max(|x|,|t|) < 1$.


structure_id: Ban
property_id: coregular
type: category
proof:
It suffices to prove that regular monomorphisms are stable under pushouts. We will use their classification as isometric linear maps below. So let $i : X \to Y$ be an isometric linear map and let $f : X \to T$ be any morphism. We need to show that the linear contraction $\iota : T \to T \sqcup_X Y$ is isometric as well. The pushout can be constructed as the quotient of the direct sum $T \oplus Y$, equipped with the $1$-norm, modulo the closure of the subspace containing all $(-f(x),i(x))$ for $x \in X$. Using that $i$ is an isometry, it is easily checked that this subspace is already closed. For $t \in T$ the norm of $\iota(t) = [(t,0)]$ is the infimum of the norms of $(t,0) + (-f(x),i(x)) = (t - f(x), i(x))$ for $x \in X$. By taking $x=0$ we see that the infimum is $\leq |t|$. On the other hand, for any $x \in X$ we compute $$\begin{align*} |(t - f(x), i(x))| & = |t - f(x)| + |i(x)| \\ & = |t - f(x)| + |x| \\ & \geq |t - f(x)| + |f(x)| \geq |t|. \end{align*}$$ This shows that the infimum is precisely $|t|$.


structure_id: Ban
property_id: effective congruences
type: category
proof:
Let $R \hookrightarrow X \times X$ be a congruence in $\Ban$. We may assume that it is an inclusion of vector spaces. Since the forgetful functor $\Ban \to \Vect$ preserves finite limits, the congruence yields a congruence in $\Vect$, which we understand well: It is fully classified by the subspace $U \coloneqq \{x \in X : (x,0) \in R\}$ via $R = \{(x,y) \in X \times X : x-y \in U\}$.

Since $R \hookrightarrow X \times X$ is a linear contraction, the norm $|{-}|_R$ on $R$ satisfies $$\max(|x|,|y|) \leq |(x,y)|_R.$$ We claim that this is an equality, i.e. that $R \hookrightarrow X \times X$ is an isometry.

Since the reflexivity morphism $X \to R$, $x \mapsto (x,x)$ is a split monomorphism, it is isometric, i.e. $$|(x,x)|_R = |x|.$$ Since the transitivity morphism $R \times_X R \to R$ is a linear contraction and the fiber product carries the $\max$-norm, we have $$|(x,z)|_R \leq \max(|(x,y)|_R, |(y,z)|_R)$$ whenever $(x,y) \in R$ and $(y,z) \in R$.

Now let $(x,z) \in R$ and define $y \coloneqq \frac{1}{2} (x + z)$. Then $$(x,y) = \tfrac{1}{2} (x,x) + \tfrac{1}{2} (x,z)$$ and likewise $$(y,z) = \tfrac{1}{2} (z,z) + \tfrac{1}{2} (x,z)$$ are contained in $R$. We have $$|(x,y)_R| \leq \tfrac{1}{2} |x| + \tfrac{1}{2} |(x,z)|_R$$ and likewise $$|(y,z)|_R \leq \tfrac{1}{2} |z| + \tfrac{1}{2} |(x,z)|_R$$ Together with the transitivity bound we obtain $$\begin{align*} |(x,z)|_R & \leq \max(|(x,y)|_R, |(y,z)|_R) \\ & \leq \tfrac{1}{2} \max(|x|,|z|) + \tfrac{1}{2} |(x,z)|_R, \end{align*}$$ which proves the claim $|(x,z)|_R \leq \max(|x|,|z|)$.

Since $R \hookrightarrow X \times X$ is an isometry, $R$ is a closed subspace of $X \times X$. It follows that also $U$ is a closed subspace of $X$. Finally, $R$ is the kernel pair of the projection $X \to X/U$ since this is true by construction on the level of vector spaces and the kernel pair carries the $\max$-norm, just like $R$.


structure_id: Ban
property_id: balanced
type: category
proof:
The linear map $\IC \to \IC$, $x \mapsto x/2$ with the usual norm on $\IC$ is a counterexample. It is bijective, hence a mono- and epimorphism, but not isometric and therefore no isomorphism.


structure_id: Ban
property_id: CSP
type: category
proof:
By using the concrete description of products and coproducts, for the constant family $X_n = \IC$ equipped with the usual norm, the canonical morphism $\coprod_n X_n \to \prod_n X_n$ becomes the canonical inclusion map $\ell^1 \hookrightarrow \ell^\infty$. This is not an epimorphism (i.e., has no dense image) since the closure of the image is precisely $c_0$. So for example, $(1,1,\dotsc)$ is not contained in the closure of the image.


structure_id: Ban
property_id: regular subobject classifier
type: category
proof:
If $\Omega$ is a regular subobject classifier, then by the classification of regular monomorphisms, $\Hom(X,\Omega)$ is isomorphic to the set of closed subspaces of $X$ for any Banach space $X$. For $X = \IC$ this implies that there are exactly two vectors in $\Omega$ with norm $\leq 1$, which is absurd. (For $\Omega = 0$ there is just one, and for $\Omega \neq 0$ there are infinitely many.)


structure_id: Ban
property_id: regular quotient object classifier
type: category
proof:
Assume that $\Psi$ is a regular quotient object classifier in $\Ban$. For a Banach space $X$ let $S(X)$ be the set of closed subspaces of $X$. Then for any Banach space $X$ the map $$\Hom(\Psi,X) \to S(X), \quad f \mapsto \overline{\im(f)}$$ is bijective. In particular, there is a morphism $f : \Psi \to X$ with dense image, i.e. an epimorphism. But this contradicts the already established fact that $\Ban$ is well-copowered (since it is locally presentable). Alternatively, we may use that for every cardinal $\kappa$ the Banach space $\ell^2(\kappa^+)$ has no dense subset of size $\kappa$.


structure_id: Ban
property_id: unital
type: category
proof:
The canonical morphism $X \sqcup Y \to X \times Y$ is injective, and therefore a monomorphism. Hence, if it were also a strong epimorphism, it would be an isomorphism. However, for example, when $X = Y = \IC$, the norms do not agree.


structure_id: Ban
property_id: counital
type: category
proof:
The canonical morphism $X \sqcup Y \to X \times Y$ is surjective, and therefore an epimorphism. Hence, if it were also a strong monomorphism, it would be an isomorphism. However, for example, when $X = Y = \IC$, the norms do not agree.


structure_id: Ban
property_id: filtered-colimit-stable monomorphisms
type: category
proof:
The proof is similar to <a href="/category/Met">$\Met$</a>. For $n \geq 1$ let $X_n$ be the Banach space with underlying vector space $\IC$ and the norm $|x|_n \coloneqq \frac{1}{n} |x|$. For $n \leq m$ the identity map provides a morphism $X_n \to X_m$, which is clearly a monomorphism (also an epimorphism by the way, but an isomorphism iff $n=m$). Let $X$ be the colimit of all $X_n$ in the category of semi-normed vector spaces. It is constructed as the colimit in the category of vector spaces with the semi-norm $|x| \coloneqq \inf \{|x|_m : n \leq m \}$ for $x \in X_n$. So clearly, the semi-norm is zero. Hence, the colimit in the category of normed vector spaces is $0$. The colimit in the category of Banach spaces is its completion, which is also $0$. Thus, the monomorphisms $X_1 \hookrightarrow X_n$ become the zero map $X_1 \to 0$ in the colimit, which is not a monomorphism.


structure_id: Ban
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We show that epimorphisms are not stable under sequential limits. Let $X_n = Y_n = \IC$ for all $n \geq 0$, equipped with the usual norm. The transition morphism $Y_{n+1} \to Y_n$ is the identity, and the transition morphism $X_{n+1} \to X_n$ is $x \mapsto x/2$. The morphisms $X_n \to Y_n$, $x \mapsto x/2^n$ are compatible with the transitions, and they are surjective, hence epimorphisms. Now we check $\lim_n X_n = 0$: An element $(x_n) \in \lim_n X_n$ is a family of complex numbers satisfying $x_n = x_{n+1}/2$ <i>and</i> $\sup_n |x_n| < \infty$. But then $x_n = 2^n x_0$ and this can only be bounded when $x_0=0$. Hence, $0 = \lim_n X_n \to \lim_n Y_n = \IC$ is no epimorphism.


structure_id: CAlg(R)
property_id: balanced
type: category
proof:
Take a prime ideal $P \subseteq R$ and consider the commutative $R$-algebra $A \coloneqq R/P$ (which is an integral domain). Then the inclusion $A \hookrightarrow Q(A)$ is a counterexample.


structure_id: CAlg(R)
property_id: cogenerating set
type: category
proof:
We apply <a href="/content/missing_cogenerating_sets">this lemma</a> to the collection of commutative $R$-algebras which are fields: If $F$ is a commutative $R$-algebra that is also a field and $A$ is a non-trivial commutative $R$-algebra, any algebra homomorphism $F \to A$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables over some residue field of $R$ has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


structure_id: CAlg(R)
property_id: countably codistributive
type: category
proof:
The canonical homomorphism $A \otimes_R R^{\IN} \to A^{\IN}$ is given by $a \otimes (r_n)_n \mapsto (r_n a)_n$ and does not have to be surjective: Since $R \neq 0$, there is a commutative $R$-algebra $K$ which is a field. Now take $A \coloneqq K[X]$ and consider the sequence $(X^n)_{n} \in A^{\IN}$.


structure_id: CAlg(R)
property_id: semi-strongly connected
type: category
proof:
Choose a maximal ideal $\m$ of $R$, so $K \coloneqq R/\m$ is a field. If $\CAlg(R)$ is semi-strongly connected, then also $\CAlg(K)$ is semi-strongly connected. This has been disproven in <a href="https://math.stackexchange.com/questions/5129689" target="_blank">MSE/5129689</a>.


structure_id: CAlg(R)
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \CAlg(R) \to \Set$ and the relation $S \subseteq U^2$ defined by $S(A) \coloneqq \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $R[X]$ and $S$ by $R[X,Y] / \langle XY-X^2 \rangle$. It is clear that $S$ is reflexive, but not symmetric.


structure_id: CAlg(R)
property_id: regular quotient object classifier
type: category
proof:
The strategy is similar to the one for <a href="/category/CRing">$\CRing$</a>: Assume that $P \to R$ is a regular quotient object classifier. If $J$ denotes the kernel of $P \to R$, every ideal $I \subseteq A$ of any commutative $R$-algebra has the form $I = \langle \varphi(J) \rangle$ for a unique homomorphism $\varphi : P \to A$. If $\sigma : A \to A$ is an automorphism with $\sigma(I)=I$, then uniqueness gives us $\sigma \circ \varphi = \varphi$, which means that $\varphi(J)$ lies in $A^{\sigma}$, the fixed algebra of $\sigma$. But then $I$ is generated by elements in $A^{\sigma} \cap I$. If $K$ is a residue field of $R$, this fails for $A = K[X,Y]$, $I = \langle X,Y \rangle$, $\sigma(X)=Y$, $\sigma(Y)=X$. The fixed algebra is the subalgebra of symmetric polynomials, which is $K[X+Y,XY]$. So $\langle X,Y \rangle$ is generated by symmetric polynomials without constant term, which implies $\langle X,Y \rangle \subseteq \langle X+Y,XY \rangle$ in $K[X,Y]$. But reducing an equation like $X = a(X,Y) \cdot (X+Y) + b(X,Y) \cdot (XY)$ modulo $\langle X^2,Y^2,XY \rangle$ yields a contradiction.


structure_id: CAlg(R)
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
Let $K$ be a field over $R$. Consider the sequence of projections $\cdots \to K[X]/\langle X^2 \rangle \to K[X]/\langle X \rangle$ and the constant sequence $\cdots \to K[X] \to K[X]$. The surjective homomorphisms $K[X] \to K[X]/\langle X^n \rangle$ induce the inclusion $K[X] \hookrightarrow K[[X]]$ in the limit, where $K[[X]]$ is the algebra of formal power series. It is clearly not surjective, but this is not sufficient, we need to argue that it is not an epimorphism in $\CAlg(R)$, or equivalently, in $\CRing$. For a proof, see <a href="https://math.stackexchange.com/questions/2391187" target="_blank">MSE/2391187</a>.


structure_id: CMon
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \CMon \to \Set$ and the relation $R \subseteq U^2$ defined by $R(A) \coloneqq \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by the free monoid on a single generator and $R$ by the free commutative monoid on two generators $x,y$ subject to the relation $xy=x^2$. It is clear that $R$ is reflexive, but not symmetric.


structure_id: CMon
property_id: coregular
type: category
proof:
We can show this analogously to the case of commutative rings <a href="https://math.stackexchange.com/a/3746890" target="_blank">MSE/3746890</a>. Consider the commutative monoid $\IN^2$ and its submonoid $U \coloneqq \{(m,n)\mid m\ge n\}$ with the inclusion $i : U\hookrightarrow\IN^2$. Then, the pushout of $i$ along itself is $\langle x,y,z : x+y=x+z \rangle$, and the equalizer of the cokernel pair of $i$ is $D \coloneqq \{(m,n)\mid m=0 \implies n=0 \}$. If the category $\CMon$ were coregular, the canonical inclusion $j : U \hookrightarrow D$ would have to be an epimorphism. However, it is not: let $I \coloneqq \{0,1\}$ be the two-element commutative monoid with $1+1=1$, and let $u,v : D \rightrightarrows I$ be the morphisms defined by $u^{-1}(0)=\{(0,0)\}$ and $v^{-1}(0)=\{(0,0),(1,2)\}$; then we have $u\circ j = v\circ j$.


structure_id: CMon
property_id: regular quotient object classifier
type: category
proof:
If $P \in \CMon$ is a regular quotient object classifier, this means that every surjective homomorphism of commutative monoids $A \to B$ is the cokernel of a unique homomorphism $P \to A$. But there are many surjective homomorphisms which are no cokernels at all: Consider the Boolean monoid $(\{0,1\},\vee)$ with $1 \vee 1 = 1$ and the surjective homomorphism $f : (\IN,+) \to (\{0,1\},\vee)$ defined by $f(0)=0$ and $f(n)=1$ for $n \geq 1$. It has trivial kernel, but is no isomorphism, so it cannot be a cokernel.


structure_id: CMon
property_id: CSP
type: category
proof:
First of all, epimorphisms in $\CMon$ are preserved and reflected by the forgetful functor to $\Mon$ (see below). Furthermore, if $M \to N$ is an epimorphism in $\Mon$ and $M$ is infinite, then $\card(N) \leq \card(M)$ (see <a href="https://mathoverflow.net/questions/510431/" target="_blank">MO/510431</a>). This implies that in $\CMon$ the canonical homomorphism $\bigoplus_{n \geq 0} \IN \to \prod_{n \geq 0} \IN$ is not an epimorphism because its domain is countable and its codomain is uncountable.


structure_id: CRing
property_id: coextensive
type: category
proof:
[Sketch] A ring homomorphism $f : A \times B \to R$ yields the idempotent element $e \coloneqq f(1,0) \in R$, so that $R \cong eR \times (1-e)R$. Then $f$ decomposes into the ring homomorphisms $f_A : A \to eR$, $f_A(a) \coloneqq f(a,0)$ and $f_B : B \to (1-e)R$, $f_B(b) \coloneqq f(0,b)$.


structure_id: CRing
property_id: cogenerating set
type: category
proof:
We apply <a href="/content/missing_cogenerating_sets">this lemma</a> to the collection of fields: If $F$ is a field and $R$ is a non-trivial commutative ring, any ring homomorphism $F \to R$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


structure_id: CRing
property_id: countably codistributive
type: category
proof:
The canonical homomorphism $\IQ \otimes \IZ^{\IN} \to (\IQ \otimes \IZ)^{\IN} = \IQ^{\IN}$ is not an isomorphism: its image consists of those sequences of rational numbers whose denominators can be bounded.


structure_id: CRing
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \CRing \to \Set$ and the relation $R \subseteq U^2$ defined by $R(A) \coloneqq \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $\IZ[X]$ and $R$ by $\IZ[X,Y] / \langle XY-X^2 \rangle$. It is clear that $R$ is reflexive, but not symmetric.


structure_id: CRing
property_id: regular quotient object classifier
type: category
proof:
Assume that $P \to \IZ$ is a regular quotient object classifier. If $J$ denotes its kernel, this means that every ideal $I \subseteq A$ of any commutative ring has the form $I = \langle \varphi(J) \rangle$ for a unique homomorphism $\varphi : P \to A$. If $\sigma : A \to A$ is an automorphism with $\sigma(I)=I$, then uniqueness gives us $\sigma \circ \varphi = \varphi$, which means that $\varphi(J)$ lies in $A^{\sigma}$, the fixed ring of $\sigma$. But then $I$ is generated by elements in the fixed ring. This fails for $A = \IZ[X]$, $I = \langle X \rangle$, $\sigma(X)=-X$. The fixed ring is $\IZ[X^2]$, and if $I$ was generated by elements $f \in \IZ[X^2] \cap I$, they would be multiples of $X^2$, but $X$ is not a multiple of $X^2$.


structure_id: CRing
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
For a prime $p$ consider the sequence of projections $\cdots \to \IZ/p^2 \to \IZ/p$ and the constant sequence $\cdots \to \IZ \to \IZ$. The surjective homomorphisms $\IZ \to \IZ/p^n$ induce the homomorphism $\IZ \to \IZ_p$ in the limit, where $\IZ_p$ is the ring of $p$-adic integers. It is not surjective since $\IZ_p$ is uncountable, but this is not sufficient (at least, for this category): We need to use <a href="https://stacks.math.columbia.edu/tag/04W0" target="_blank">SP/04W0</a> to conclude that it is no epimorphism in $\CRing$.


structure_id: Cat
property_id: locally small
type: category
proof:
There is a forgetful functor $\Cat \to \Set \times \Set$, $\C \mapsto (\Ob(\C),\Mor(\C))$, and $\Set \times \Set$ is locally small.


structure_id: Cat
property_id: cartesian closed
type: category
proof:
See p. 98 in <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>.


structure_id: Cat
property_id: generator
type: category
proof:
The <a href="/category/walking_morphism">walking morphism</a> $I$ is a generator: Assume that $F,G : \C \rightrightarrows \D$ are functors that agree when being precomposed with any functor from $I$. This means that $F(f) = G(f)$ for all morphisms $f : X \to Y$ in $\C$. By comparing the domains and applying this to $f = \id_X$, we see that $F(X) = G(X)$ for all objects $X$. And we just saw that $F,G$ also agree on morphisms.


structure_id: Cat
property_id: infinitary extensive
type: category
proof:
[Sketch] This is straight forward from the fact that <a href="/category/Set">$\Set$</a> is infinitary extensive: A functor $\C \to \coprod_i \D_i$ yields full subcategories $\C_i \subseteq \C$ (the preimages of $\D_i)$ with $\C = \coprod_i \C_i$.


structure_id: Cat
property_id: balanced
type: category
proof:
Since we know that <a href="/category/Mon">$\Mon$</a> is not balanced, there is a monoid map $M \to N$ which is a monomorphism and an epimorphism which is not an isomorphism. Then $B(M) \to B(N)$ has the corresponding properties.


structure_id: Cat
property_id: cogenerating set
type: category
proof:
Assume that $S$ is a cogenerating set in $\Cat$. Then one checks that the set of monoids $\{\End(X) : X \in \C \in S\}$ is a cogenerating set in <a href="/category/Mon">$\Mon$</a>, which we know does not exist.


structure_id: Cat
property_id: regular
type: category
proof:
See Example 3.14 at the <a href="https://ncatlab.org/nlab/show/regular+category" target="_blank">nLab</a>.


structure_id: Cat
property_id: coregular
type: category
proof:
We already know that <a href="/category/Mon">$\Mon$</a> is not coregular, in fact there is a regular monomorphism $M \to N$ of monoids and a morphism $M \to K$ such that $K \to K \sqcup_M N$ is not a monomorphism. The delooping functor $B : \Mon \to \Cat$ has a left adjoint (<a href="https://math.stackexchange.com/questions/574745" target="_blank">MSE/574745</a>), hence it preserves regular monomorphisms. It also preserves pushouts (<a href="https://math.stackexchange.com/questions/5130854" target="_blank">MSE/5130854</a>), and it reflects monomorphisms since it is faithful. Therefore, $B(M) \to B(N)$ provides the desired counterexample of a non-stable regular monomorphism of categories.


structure_id: Cat
property_id: co-Malcev
type: category
proof:
We can adapt the proof from <a href="/category/Mon">$\Mon$</a> as follows: Consider the functor $U : \Cat \to \Set^+$ sending a category $\C$ to the (large) set $\{(x,u) : x \in \Ob(\C) ,\, u \in \End(x) \}$. It is represented by $B \IN$, the one-object category associated to the free monoid in one generator. Consider the relation $R \subseteq U^2$ consisting of those pairs $((x,u),(y,v))$ where $x = y$ and $uv = u^2$. This also representable, namely be the one-object category associated to the monoid with the presentation $\langle u,v : uv = u^2 \rangle$. Clearly, $R$ is reflexive, but not symmetric.


structure_id: Cat
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Cat$ that maps a set to its discrete category.


structure_id: Cat
property_id: effective cocongruences
type: category
proof:
The counterexample is similar to the one for <a href="/category/Mon">$\Mon$</a>: Let $X$ be the <a href="/category/walking_idempotent">walking idempotent</a>, and let $E$ be the delooping of the monoid with presentation $$\langle p, q \mid p^2=p,\, q^2=q,\, pq=q,\, qp=p \rangle.$$ The induced relation on functors in $[X, \C]$ is that $F \sim G$ if and only if $F$ and $G$ send the object of $X$ to the same object of $\C$, and they send the idempotent of $X$ to idempotent morphisms $a, b$ in $\C$ satisfying $ab=b$, $ba=a$. From here, the proof that this gives a cocongruence on $\Cat$ which is not effective is similar to the one in $\Mon$.


structure_id: Cat
property_id: regular subobject classifier
type: category
proof:
Consider the pushout square $$\begin{CD} \{1\} @> \inc >> \{ 0 \to 1 \} \\ @V \inc VV @VV \inc V \\ \{ 1 \to 2 \} @>> \inc > \{ 0 \to 1 \to 2 \}. \end{CD}$$ Here the upper right corner is the walking morphism, the lower left corner is an isomorphic copy of the walking morphism, and the lower right corner is the walking composable pair.
We will now consider two equalizers on $\{ 0 \to 1 \to 2 \}$. The first will be the equalizer of the two paths in the walking commutative square $\{ 0 \to 1 \}^2$, i.e. the functors will be $$\begin{align*} 0 \mapsto (0,0),~ 1 \mapsto (0,1),~ 2 \mapsto (1,1); \\ 0 \mapsto (0,0),~ 1 \mapsto (1,0),~ 2 \mapsto (1,1). \end{align*}$$ This equalizer contains the unique morphism $0 \to 2$. On the other hand, if for the target we instead use the free category on the square diagram, in this case the equalizer does not contain the unique morphism $0 \to 2$. However, in both cases, the pullbacks of these equalizers to $\{ 0 \to 1 \}$ and to $\{ 1 \to 2 \}$ will be the same. Namely, the pullback of each to $\{ 0 \to 1 \}$ will be the subcategory with just the object $0$ and its identity morphism; and the pullback of each to $\{ 1 \to 2 \}$ will be the subcategory with just the object $2$ and its identity morphism.
This shows that the comparison function $$\Sub_{\reg}(\{ 0 \to 1 \to 2 \}) \to \Sub_{\reg}(\{ 0 \to 1 \}) \times_{\Sub_{\reg}(\{1\})} \Sub_{\reg}(\{ 1 \to 2 \})$$ is not injective. Therefore, $\Sub_{\reg} : \Cat^{\op} \to \Set^+$ does not preserve pullbacks, so it cannot be representable.


structure_id: CompHaus
property_id: products
type: category
proof:
By the Tychonoff product theorem, a product in <a href="/category/Top">$\Top$</a> of compact Hausdorff spaces is compact; it is also clearly Hausdorff. Since the forgetful functor from $\CompHaus$ to $\Top$ is fully faithful, this limit is reflected in $\CompHaus$ as well.


structure_id: CompHaus
property_id: equalizers
type: category
proof:
The equalizer in <a href="/category/Top">$\Top$</a> of two continuous functions $f, g : X \rightrightarrows Y$ between compact Hausdorff spaces is a closed subspace of $X$, and therefore it is also compact Hausdorff. Since the forgetful functor from $\CompHaus$ to $\Top$ is fully faithful, this limit is reflected in $\CompHaus$ as well.


structure_id: CompHaus
property_id: cocomplete
type: category
proof:
$\CompHaus$ is a reflective subcategory of $\Top$, with the reflector being the Stone-Čech compactification functor. See <a href="https://ncatlab.org/nlab/show/compact+Hausdorff+space#StoneCechCompactification" target="_blank">nLab</a> for example. Therefore, as usual, we can form colimits in $\CompHaus$ by forming colimits in <a href="/category/Top">$\Top$</a> and then applying Stone-Čech compactification.


structure_id: CompHaus
property_id: generator
type: category
proof:
The one-point space is a generator because it represents the forgetful functor to $\Set$, which is faithful.


structure_id: CompHaus
property_id: cogenerator
type: category
proof:
The unit interval $[0, 1]$ is a cogenerator: Suppose we have $f, g : X \rightrightarrows Y$ with $f \ne g$. Choose $x\in X$ such that $f(x) \ne g(x)$. Then by Urysohn's lemma, there is a continuous function $h : Y \to [0, 1]$ such that $h(f(x)) = 0$ and $h(g(x)) = 1$. Therefore, $h\circ f \ne h\circ g$.


structure_id: CompHaus
property_id: Barr-exact
type: category
proof:
The forgetful functor from $\CompHaus$ to $\Set$ is monadic; see for example <a href="https://ncatlab.org/nlab/show/compact+Hausdorff+space#compact_hausdorff_spaces_are_monadic_over_sets">nLab</a>. Therefore, by <a href="https://ncatlab.org/nlab/show/colimits+in+categories+of+algebras#exact">this result</a>, $\CompHaus$ is Barr-exact.


structure_id: CompHaus
property_id: coregular
type: category
proof:
It suffices to show that pushouts preserve (regular) monomorphisms in $\CompHaus$. Thus, suppose we have a pushout square $$\begin{CD} A @> i >> B \\ @V f VV @VV g V \\ C @>> j > D, \end{CD}$$ with $i : A \hookrightarrow B$ a monomorphism. Then for any pair of distinct elements $c, c' \in C$, by Urysohn's lemma there exists $\gamma : C \to [0, 1]$ with $\gamma(c) = 0$ and $\gamma(c') = 1$. Also, by Tietze's extension theorem, there exists $\beta : B \to [0, 1]$ such that $\beta \circ i = \gamma \circ f$. By the pushout property, there is a unique $\delta : D \to [0, 1]$ such that $\delta \circ g = \beta$ and $\delta \circ j = \gamma$. Since $\delta(j(c)) \ne \delta(j(c'))$, we conclude that $j(c) \ne j(c')$. This shows that $j$ is injective, so it is a regular monomorphism.


structure_id: CompHaus
property_id: extensive
type: category
proof:
This follows as for <a href="/category/Top">$\Top$</a> or <a href="/category/Haus">$\Haus$</a> since finite coproducts in $\CompHaus$ are formed as disjoint union spaces with the disjoint union topology.


structure_id: CompHaus
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
Suppose we have a cofiltered diagram of epimorphisms $(f_i : X_i \to Y_i)$, and $y = (y_i) \in \lim_i Y_i$. Then by lemma 1 <a href="/content/comphaus_copresentable">here</a>, the limit of $f_i^{-1}(\{ y_i \})$ is non-empty. If $x$ is in this limit, that implies that $(\lim_i f_i)(x) = y$.


structure_id: CompHaus
property_id: Malcev
type: category
proof:
This is clear since <a href="/category/FinSet">$\FinSet$</a> is not Malcev and can be interpreted as the subcategory of finite discrete spaces.


structure_id: CompHaus
property_id: natural numbers object
type: category
proof:
Let $I \coloneqq [0, 1]$. If a natural numbers object $(N, z : 1 \to N, s : N \to N)$ existed, then we could iterate the initial conditions $I\to I\times I$, $x \mapsto (x, x)$ and the recursive step function $I\times I \to I \times I$, $(x, y) \mapsto (x, xy)$ to get a continuous function $N \times I \to I \times I$ such that $(s^n(z), x) \mapsto (x, x^n)$ for $x\in I$, $n \in \IN$. The sequence $(s^n(z)) \in N$ has a convergent subnet $(s^{n_\lambda}(z))_{\lambda \in \Lambda}$, say with limit $y$. Thus, for any $x\in I$ and $\lambda \in \Lambda$, we have $(s^{n_\lambda}(z), x) \mapsto (x, x^{n_\lambda})$. Taking limits, we see $(y, x) \mapsto (x, 0)$ if $x \ne 1$ or $(y, x) \mapsto (x, 1)$ if $x = 1$. In other words, $(y, x) \mapsto (x, \delta_{x, 1})$ for all $x\in I$. However, that contradicts the fact that the composition $$\begin{align*} I & \overset{y \times \id}\longrightarrow N\times I \to I\times I \overset{p_2}\longrightarrow I, \\ x & \mapsto (y, x) \mapsto (x, \delta_{x,1}) \mapsto \delta_{x,1}, \end{align*}$$ would have to be continuous.


structure_id: CompHaus
property_id: filtered-colimit-stable monomorphisms
type: category
proof:
The proof is similar to <a href="/category/Haus">$\Haus$</a>. For $n \geq 1$ let $X_n$ be the pushout of $[1/n, 1] \hookrightarrow [0, 1]$ with itself. That is, $X_n$ is the union of two unit intervals $[0, 1] \times \{ 1 \}$ and $[0, 1] \times \{ 2 \}$ where we identify $(x,1) \equiv (x,2)$ when $x \geq 1/n$. As in the construction for $\Haus$, we see that the colimit in $\Haus$ is $[0, 1]$ where all corresponding points of both unit intervals are identified. Since this is compact Hausdorff, it also provides the colimit in $\CompHaus$. Again, the injective continuous maps $\{1,2\} \to X_n$, $i \mapsto (0,i)$ (where $\{1,2\}$ is discrete) become the constant map $0 : \{1,2\} \to [0,1]$ in the colimit, which is not a monomorphism.


structure_id: CompHaus
property_id: exact cofiltered limits
type: category
proof:
Consider the $\IN$-codirected systems $X_n \coloneqq [0, 1] \times [0, 1/n]$ with the maps $X_{n+1} \to X_n$ being inclusion maps, and $Y_n \coloneqq [0, 1+1/n]$ with the maps $Y_{n+1} \to Y_n$ also being inclusion maps. We define $f_n : X_n \to Y_n$, $(x, y) \mapsto x$ and $g_n : X_n \to Y_n$, $(x, y) \mapsto x+y$. It is straightforward to check these give morphisms of $\IN$-codirected systems in $\CompHaus$.
Now for each $n$, we claim the coequalizer of $f_n$ and $g_n$ is a singleton space. To see this, we prove the more general result that for $r, s > 0$ the coequalizer of $f, g : [0, r] \times [0, s] \rightrightarrows [0, r+s]$, $f(x,y) = x$, $g(x,y) = x+y$ is a singleton. We must show that for any $h : [0, r+s] \to T$ with $h\circ f = h\circ g$, then $h$ is constant. To this end, we show by induction on $n$ that whenever $x \in [0, r+s]$ and $x \le ns$, we have $h(x) = h(0)$. The base case $n=0$ is trivial. For the inductive step, if $x \le s$, then $f(0,x) = 0$ and $g(0,x) = x$, so $h(0) = h(x)$. Otherwise, we have $x-s \in [0,r]$ and $x-s \le (n-1)s$, so by inductive hypothesis $h(x-s) = h(0)$. Also, $f(x-s, s) = x-s$ and $g(x-s, s) = x$, so $h(x-s) = h(x)$, completing the induction. With this established, the desired result follows from the case $n \coloneqq \lceil r/s \rceil + 1$.
On the other hand, $\lim X_n \cong [0, 1] \times \{ 0 \}$; $\lim Y_n \cong [0, 1]$; and $\lim f_n = \lim g_n$, $(x, 0) \mapsto x$. Thus, the coequalizer of $\lim f_n$ and $\lim g_n$ is $[0, 1]$, showing that the limit does not preserve this coequalizer.


structure_id: Delta
property_id: locally finite
type: category
proof:
There is a faithful functor $\Delta \to \FinSet$ and <a href="/category/FinSet">$\FinSet$</a> is locally finite.


structure_id: Delta
property_id: skeletal
type: category
proof:
If $f : [n] \to [m]$ is an isomorphism, then $n + 1 = m + 1$ by comparing the cardinalities, hence $n = m$.


structure_id: Delta
property_id: coequalizers
type: category
proof:
Assume that $X \rightrightarrows Y$ are morphisms in $\FinOrd \setminus \{\varnothing\}$. Since <a href="/category/FinOrd">$\FinOrd$</a> has coequalizers, we have a coequalizer $Y \to Q$. Since $Y$ is non-empty, $Q$ is non-empty as well, and clearly $Y \to Q$ is then also the coequalizer in $\FinOrd \setminus \{\varnothing\}$.


structure_id: Delta
property_id: core-thin
type: category
proof:
The category $\FinOrd \setminus \{\varnothing\}$ is core-thin because already <a href="/category/FinOrd">$\FinOrd$</a> is core-thin.


structure_id: Delta
property_id: mono-regular
type: category
proof:
The proof for <a href="/category/FinOrd">$\FinOrd$</a> also works for $\FinSet \setminus \{\varnothing\}$.


structure_id: Delta
property_id: cosifted
type: category
proof:
Let $X,Y \in \Delta$. We may pick $x \in X$, $y \in Y$. Then there is a "point span" $X \xleftarrow{x} [0] \xrightarrow{y} Y$. Every span $X \xleftarrow{f} Z \xrightarrow{g} Y$ is connected to such a point span: Pick $z \in Z$. This defines a morphism of spans:  $$\begin{CD} X @<{f(z)}<< [0] @>{g(z)}>> Y \\ @| @VV{z}V @| \\ X @<<{f}< Z @>>{g}> Y \end{CD}$$ It remains to show that all point spans are connected to each other. Assume $x_0,x_1 \in X$ and $y \in Y$, w.l.o.g. $x_0 \leq x_1$. Define the map $f : [1] \to X$ by $f(0) = x_0$, $f(1) = x_1$, and the map $g : [1] \to Y$ by $g(0)=g(1)=y$. They are order-preserving and fit into a zig-zag of spans: $$\begin{CD} X @<{x_0}<< [0] @>{y}>> Y \\ @| @V{0}VV @| \\ X @<{f}<< [1] @>{g}>> Y \\ @| @A{1}AA @| \\ X @<{x_1}<< [0] @>{y}>> Y \end{CD}$$ This shows that the choice of $x \in X$ does not matter, and for $y \in Y$ the proof is the same.


structure_id: Delta
property_id: ℵ₁-cofiltered limits
type: category
proof:
We already know that <a href="/category/FinOrd">$\FinOrd$</a> has $\aleph_1$-cofiltered limits and that the forgetful functors to <a href="/category/FinSet">$\FinSet$</a> and <a href="/category/Set">$\Set$</a> preserve them. Therefore, it suffices to prove that a $\aleph_1$-cofiltered limit of non-empty finite sets is also non-empty. While a direct proof is possible, we can conveniently derive this from Lemma 1 <a href="/content/comphaus_copresentable">here</a> by regarding finite sets as discrete compact Hausdorff spaces.


structure_id: Delta
property_id: coreflexive equalizers
type: category
proof:
The two maps $d^0,d^1 : [0] \rightrightarrows [1]$ have a common left inverse, the unique map $s^0 : [1] \to [0]$, but are not equalized by any morphism.


structure_id: Delta
property_id: sequential colimits
type: category
proof:
We can just copy the proof for <a href="/category/FinOrd">$\FinOrd$</a> to show that the sequence of inclusions $[0] \hookrightarrow [1] \hookrightarrow [2] \hookrightarrow \cdots$ has no colimit.


structure_id: Delta
property_id: sequential limits
type: category
proof:
We can just copy the proof for <a href="/category/FinOrd">$\FinOrd$</a> to show that the sequence of truncations $\cdots \twoheadrightarrow [2] \twoheadrightarrow [1] \twoheadrightarrow [0]$ has no limit.


structure_id: Delta
property_id: pushouts
type: category
proof:
Assume that the two inclusions $\{0 < 1\} \leftarrow \{0\} \rightarrow \{0 < 2\}$ have a pushout in $\FinOrd \setminus \{\varnothing\}$. This would be a universal non-empty finite ordered set $X$ with three elements $0,1,2$ satisfying $0 \leq 1$ and $0 \leq 2$. Assume w.l.o.g. $1 \leq 2$ (the case $2 \leq 1$ is similar). The universal property yields an order-preserving map $X \to \{a < b < c\}$ with $0 \mapsto a$, $1 \mapsto c$, $2 \mapsto b$. But then $c \leq b$, which is a contradiction.


structure_id: Delta
property_id: multi-complete
type: category
proof:
We will prove that the multi-product of $I \coloneqq \{0 < 1\}$ with itself does not exist in $\FinOrd \setminus \{\varnothing\}$. Assume that it does, i.e. there is a family of spans $$X_s = (I \xleftarrow{a_s} P_s \xrightarrow{b_s} I)$$ such that every span admits a unique morphism into exactly one of these. Now consider the following three spans: $$\begin{align*} A & \coloneqq (I \xleftarrow{0} \{0\} \xrightarrow{0} I) \\ B & \coloneqq (I \xleftarrow{0} I \xrightarrow{\id} I) \\ C & \coloneqq (I \xleftarrow{\id} I \xrightarrow{0} I) \end{align*}$$ There are morphisms of spans $0 : A \to B$ and $0 : A \to C$. This shows that  $B,C$ are in the same connected component. Thus, there is some index $s$ such that $B$ and $C$ admit a unique morphism to $X_s$. The morphism $B \to X_s$ maps $1 \in I$ to an element $u \in P_s$ with $a_s(u)=0$, $b_s(u)=1$. Likewise, the morphism $C \to X_s$ maps $1 \in I$ to an element $v \in P_s$ with $a_s(v)=1$, $b_s(v)=0$. But now neither $u \leq v$ nor $v \leq u$ can be true, contradicting that $P_s$ is linearly ordered.


structure_id: FI
property_id: locally finite
type: category
proof:
There is a faithful functor $\FI \to \FinSet$ and <a href="/category/FinSet">$\FinSet$</a> is locally finite.


structure_id: FI
property_id: equalizers
type: category
proof:
We construct equalizers just like in <a href="/category/FinSet">$\FinSet$</a> and observe that the universal property still holds.


structure_id: FI
property_id: wide pullbacks
type: category
proof:
We construct wide pullbacks just like in <a href="/category/Set">$\Set$</a>, i.e., for a w.l.o.g. non-empty family of injective maps $f_i : X_i \to S$ we consider the subset $P \subseteq \prod_{i \in I} X_i$ of those tuples $x$ where $f_i(x_i) = f_j(x_j)$. Each projection $P \to X_i$ is injective, so in particular $P$ is finite, and $P \to X_i$ becomes a morphism in $\FI$. It is easy to check that the universal property still holds in $\FI$.


structure_id: FI
property_id: mono-regular
type: category
proof:
If $f : X \to Y$ is an injective map of finite sets, it is the equalizer of the two injective maps $i_1,i_2 : Y \rightrightarrows Y \sqcup_X Y$, and $Y \sqcup_X Y$ is finite.


structure_id: FI
property_id: semi-strongly connected
type: category
proof:
If $X,Y$ are two finite sets, we have $\card(X) \leq \card(Y)$ or $\card(Y) \leq \card(X)$. In the first case there will be an injection $X \to Y$, in the second case there will be an injection $Y \to X$.


structure_id: FI
property_id: locally cartesian closed
type: category
proof:
IF $X$ is a finite set, the slice category $\FI / X$ is equivalent to the poset of subsets of $X$. This is cartesian closed because $A \cap B \subseteq C$ holds if and only if $B \subseteq (X \setminus A) \cup C$, where $A,B,C \subseteq X$.


structure_id: FI
property_id: cogenerator
type: category
proof:
Let $Q$ be finite set. When $Q$ is empty, it is clearly no cogenerator. Otherwise, $Q + 1$ has at least two elements, so that there are two different morphisms $1 \rightrightarrows Q + 1$. But there is no morphism $Q + 1 \to Q$ at all. Hence, $Q$ is no cogenerator.


structure_id: FI
property_id: binary powers
type: category
proof:
Assume that two finite sets $X,Y$ have a product $P$ in this category. Elements of $P$ are the same as maps $1 \to P$, and they are automatically injective. Therefore, $P \cong \Hom(1,P) \times \Hom(1,X) \times \Hom(1,Y) \cong X \times Y$, and the projections must agree as well. But they are usually not injective. In particular, the product $X \times X$ never exists when $X$ has $>1$ elements.


structure_id: FI
property_id: sequential colimits
type: category
proof:
Let $X_n \coloneqq \{1,\dotsc,n\}$. Assume the sequence of inclusion maps $X_n \hookrightarrow X_{n+1}$ has a colimit $(f_n : X_n \to X)$ in this category. But $f_n$ must be an injective map, so that $\card(X) \geq n$ for all $n$. Since $X$ is finite, this is a contradiction.


structure_id: FS
property_id: locally finite
type: category
proof:
There is a faithful functor $\FS \to \FinSet$ and <a href="/category/FinSet">$\FinSet$</a> is locally finite.


structure_id: FS
property_id: cogenerator
type: category
proof:
We prove that $\{0,1\}$ is a cogenerator: The surjective maps $X \to \{0,1\}$ correspond to the non-empty proper subsets of $X$. If $a,b \in X$ are elements that have the same image under each surjective map $X \to \{0,1\}$, it therefore means that they lie in the same non-empty proper subsets of $X$. This implies $a=b$: If $X = \{a\}$, this is trivial. Otherwise, use the subset $\{a\}$.


structure_id: FS
property_id: coequalizers
type: category
proof:
We construct coequalizers as in <a href="/category/FinSet">$\FinSet$</a> (or $\Set$) and observe that the universal property still holds when we restrict to surjective maps.


structure_id: FS
property_id: wide pushouts
type: category
proof:
We construct wide pushouts as in <a href="/category/Set">$\Set$</a> and observe that the universal property still holds when we restrict to surjective maps. If $f_i : S \to X_i$ are surjective maps and $P$ is their wide pushout, then each $X_i \to P$ is surjective, so that in particular $P$ is finite.


structure_id: FS
property_id: epi-regular
type: category
proof:
If $f : X \to Y$ is a surjective map of finite sets, it is the coequalizer of the two projections $p_1, p_2 : X \times_Y X \rightrightarrows X$ in <a href="/category/FinSet">$\FinSet$</a>, but also in $\FS$. Notice that $p_1,p_2$ are surjective. Even though $X \times_Y X$ is not a pullback in $\FS$, we can use this finite set here.


structure_id: FS
property_id: multi-complete
type: category
proof:
Let $D : \I \to \FS$ be a small diagram. Let $L \subseteq \prod_{i \in \I} D(i)$ be the limit object of the corresponding diagram in $\Set$. Consider the set of all finite subsets $R \subseteq L$ with the property that $p_i(R) = D(i)$ for every $i \in \I$. For each of these, $(R \xrightarrow{p_i|_R} D(i))_{i \in \I}$ is a cone in $\FS$. We claim that the set of these cones is universal: Let $(f_i : T \to D(i))_{i \in \I}$ be any cone in $\FS$. Let $R \subseteq L$ be the image of the induced map of sets $f : T \to \prod_{i \in \I} D(i)$ defined by $p_i f = f_i$. Then $R$ is finite, since $T$ is finite, and we have $p_i(R) = p_i(f(T)) = f_i(T) = D(i)$. Moreover, the map $f$ corestricts to a map $f' : T \to R$ with $p_i|_R  \, f' = f_i$. Therefore, $f'$ is a morphism of cones. Conversely, let $h : T \to R$ be a morphism of cones, i.e. $p_i|_R \, h = f_i$, where $R$ is not known yet. Then $h$ equals the corestriction of the map into the product induced by the $f_i$, and $h$ is surjective as a morphism in $\FS$. This shows that both $R$ and the morphism of cones are unique.


structure_id: FS
property_id: generalized variety
type: category
proof:
Let $D : \I \to \FS$ be a sifted diagram. Consider the set $$\{\card(D(i)) : i \in \I\} \subseteq \IN.$$ Since $\IN$ is well-ordered and $\I$ is non-empty, it has a minimal element, say $\card(D(i_0))$. For every morphism $i_0 \to i$ then the map of finite sets $D(i_0) \to D(i)$ is an isomorphism since it is surjective and $\card(D(i_0)) \leq \card(D(i))$. Therefore, every sifted diagram is eventually constant. From Corollary 3 <a href="/content/sifted-colimits-in-groupoids">here</a> we deduce that $\FS$ is a generalized variety.


structure_id: FS
property_id: finitely accessible
type: category
proof:
In the previous proof we have seen that every sifted diagram is eventually constant. From Corollary 4 <a href="/content/sifted-colimits-in-groupoids">here</a> we deduce $\FS$ is finitely accessible.


structure_id: FS
property_id: ℵ₁-cofiltered limits
type: category
proof:
Let $D : \I \to \FS$ be an $\aleph_1$-cofiltered diagram. We claim that the set $$\bigl\{\card(D(i)) : i \in \I\bigr\} \subseteq \IN$$ is bounded. If not, for every $n \in \IN$ there is some $i_n \in \I$ with $\card(D(i_n)) \geq n$. Choose a cone $(j \to i_n)_{j \in \I}$. For every $n \in \IN$ the set $F(j)$ surjects onto $F(i_n)$, so that $\card(F(j)) \geq n$. This contradicts the fact that $F(j)$ is finite. Hence, the maximum $$M \coloneqq \max \bigl\{\card(D(i)) : i \in \I \bigr\} \in \IN$$ is well-defined. Let $\J \subseteq \I$ be the full subcategory of those $i \in \I$ with $\card(D(i)) = M$. Then $\J$ is non-empty, and when there is a morphism $i \to j$ with $j \in \J$ we also have $i \in \J$. Since $\I$ is cofiltered, it follows that $\J \subseteq \I$ is initial. Therefore, the limit of $D$ coincides with the limit of $D|_{\J}$. But this diagram is constant: for every $j \to j'$ in $\J$ the map $D(j) \to D(j')$ is a surjective map between finite sets of the same cardinality $M$, hence bijective. This implies that the limit of $D|_{\J}$ exists and is given by any $D(j)$ with $j \in \J$.


structure_id: FS
property_id: generator
type: category
proof:
Let $G$ be a finite set. There are at least two morphisms $G + 2 \rightrightarrows 2$, but there is no morphism $G \to G + 2$ at all. Hence, $G$ is not a generator.


structure_id: FS
property_id: connected
type: category
proof:
If $f : \varnothing \to X$ is surjective, then $X = \varnothing$, and if $f : X \to \varnothing$ is any map, then also $X = \varnothing$. This shows that $\{ \varnothing \}$ is a connected component in this category. (The other connected component consists of all non-empty finite sets.)


structure_id: FS
property_id: sequential limits
type: category
proof:
Let $X_n \coloneqq \{1,\dotsc,n\}$. We define the truncation $p_n : X_{n+1} \to X_n$ by extending the identity of $X_n$ with $p_n(n+1) \coloneqq n$. Assume the sequence of truncations $\cdots \to X_2 \to X_1$ has a limit $(f_n : X \to X_n)$ in this category. But $f_n$ is surjective, so that $\card(X) \geq n$ for all $n$. Since $X$ is finite, this is a contradiction.


structure_id: FS
property_id: pullbacks
type: category
proof:
The connected component of non-empty sets has a terminal object, $1$, and it suffices to prove that it has no products. Let $X$ be a finite set with more than $1$ element. Assume that the product $P$ of $X$ with itself exists. The diagonal $X \to P$ is a split monomorphism, hence injective, but also surjective, i.e. an isomorphism. In other words, the two projections $P \rightrightarrows X$ are equal. The universal property of $P$ now implies that every two morphisms $Y \rightrightarrows X$ are equal, which is absurd.


structure_id: FS
property_id: binary copowers
type: category
proof:
Assume that the copower $X \coloneqq 2+2$ exists. Since we have a surjective map $2 \to X$, the set $X$ has at most $2$ elements. The codiagonal $X \to 2$ shows that $X$ has at least $2$ elements. Thus, $X \cong 2$. For all finite sets $Y$ we get a bijection $\Hom(2,Y) \cong \Hom(2,Y)^2$, in particular the cardinalities are the same. For $Y=2$ this gives the contradiction $2 = 4$.


structure_id: FS
property_id: locally cocartesian coclosed
type: category
proof:
If $X$ is a finite set, the coslice category $X / \FS$ is thin and in fact equivalent to the lattice of equivalence relations on $X$. If $X$ has $\geq 3$ elements, it is not codistributive* and <a href="/category-implication/dual_distributive_criterion">hence</a> not cocartesian coclosed: For simplicity assume $X = \{a,b,c\}$. The bottom element $\bot$ corresponds to the partition $\{\{a\},\{b\},\{c\}\}$, the top element $\top$ to the partition $\{\{a,b,c\}\}$. Now consider the three equivalence relations $E_1,E_2,E_3$ corresponding to the three partitions $$\{\{a,b\},\{c\}\}, \, \{\{a,c\},\{b\}\}, \, \{\{b,c\},\{a\}\}.$$ Then $$E_1 \vee (E_2 \wedge E_3) = E_1 \vee \bot = E_1,$$ but $$(E_1 \vee E_2) \wedge (E_1 \vee E_3) = \top \wedge \top = \top.$$ *For thin categories, the properties codistributive and distributive <a href="/category-implication/distributive_duality">are equivalent</a>.


structure_id: FS
property_id: multi-initial object
type: category
proof:
If a multi-initial object exists, then the connected component consisting of non-empty finite sets has an initial object $X$. Then, any non-empty finite set cannot have a cardinality strictly greater than $X$,  which is a contradiction.


structure_id: FiltVect
property_id: preadditive
type: category
proof:
We know that $\Vect$ is preadditive with pointwise operations. It is easy to check that the sum of two filtered linear maps is again a filtered linear map. Moreover, the additive inverse of a filtered linear map is again a filtered linear map.


structure_id: FiltVect
property_id: kernels
type: category
proof:
Let $f : (V,F) \to (W,F)$ be a filtered linear map. Equip the kernel $U \subseteq V$ of the underlying linear map $f : V \to W$ with the induced filtration $F^n(U) \coloneqq U \cap F^n(V)$. Then $(U,F)$ is a filtered vector space together with a filtered linear map $\iota : (U,F) \to (V,F)$, which is clearly the kernel of $f$ in $\FiltVect$.


structure_id: FiltVect
property_id: cokernels
type: category
proof:
Let $f : (V,F) \to (W,F)$ be a filtered linear map. Let $\pi : W \to C$ be the cokernel of the underlying linear map $f : V \to W$. Equip $C$ with the induced filtration $F^n(C) \coloneqq \pi_*(F^n(W))$. Then $(C,F)$ is a filtered vector space together with a filtered linear map $\pi : (W,F) \to (C,F)$, which is clearly the cokernel of $f$ in $\FiltVect$.


structure_id: FiltVect
property_id: products
type: category
proof:
Let $(V_i,F)_{i \in I}$ be a family of filtered vector spaces. Equip the product $\prod_{i \in I} V_i$ with the filtration $F^n(\prod_{i \in I} V_i) \coloneqq \prod_{i \in I} F^n(V_i)$. It is straightforward to check that this satisfies the universal property of products in $\FiltVect$.


structure_id: FiltVect
property_id: coproducts
type: category
proof:
Let $(V_i,F)_{i \in I}$ be a family of filtered vector spaces. Equip the direct sum $\bigoplus_{i \in I} V_i$ with the filtration $F^n(\bigoplus_{i \in I} V_i) \coloneqq \bigoplus_{i \in I} F^n(V_i)$. It is straightforward to check that this satisfies the universal property of coproducts in $\FiltVect$.


structure_id: FiltVect
property_id: generator
type: category
proof:
The vector space $K$ equipped with the trivial filtration $F^n(K) \coloneqq 0$ is a generator, since it represents the forgetful functor $\FiltVect \to \Set$.


structure_id: FiltVect
property_id: cogenerator
type: category
proof:
It is straightforward to check that the vector space $K$ equipped with the maximal filtration $F^n(K) \coloneqq K$ is a cogenerator.


structure_id: FiltVect
property_id: finitely accessible
type: category
proof:
Consider a filtered vector space $(V,F)$ such that $V$ is finite-dimensional and there exists some $N \in \IN$ with $F^N(V)=0$ (and hence $F^n(V)=0$ for all $n \geq N$). We claim that such an object is finitely presentable in $\FiltVect$:

Let $(W_i,F)$ be a filtered diagram of filtered vector spaces. Its colimit is the vector space $\colim_i W_i$ equipped with the filtration $F^n(\colim_i W_i) = \colim_i F^n(W_i)$. Let $f : (V,F) \to (\colim_i W_i,F)$ be a filtered linear map. The underlying linear map $f : V \to \colim_i W_i$ factors through some morphism $W_i \to \colim_i W_i$. Moreover, $f$ maps $F^n(V)$ into $\colim_i F^n(W_i)$ for every $n \in \IZ$. This condition is automatic when $n \geq N$, so assume that $n < N$. Since $V$ is finite-dimensional, the filtration stabilizes; that is, there exists some $M \in \IN$ such that $F^M(V) = F^n(V)$ for all $n \leq M$. Thus the filtered condition only needs to be checked in the finitely many degrees $$n \in \{M,M+1,\cdots,N-1\}.$$ Since the index category is filtered, there exists some index $j \geq i$ such that, for every relevant degree $n$, the map $f|_{F^n(V)}$ factors through $F^n(W_j)$. It follows that $f$ factors as a filtered linear map $(V,F) \to (W_j,F)$ followed by the colimit inclusion. This factorization is essentially unique because the corresponding statement holds in $\Vect$.

It remains to prove that every filtered vector space $(V,F)$ is a filtered colimit of such objects. First, it is clearly the filtered colimit of the subspaces $(V',F)$, where $V' \subseteq V$ is a finite-dimensional subspace equipped with the induced filtration. Second, every filtered vector space $(V,F)$ is the filtered colimit of the spaces $(V,F_{< N})$ for $N \in \IN$, where $$F_{< N}^n(V) \coloneqq \begin{cases} F^n(V) & n < N \\ 0 & n \geq N. \end{cases}$$ Indeed, we have $F_{< N} \subseteq F_{< N+1}$, so that $\id_V : (V,F_{<N}) \to (V,F_{<N+1})$ is a filtered linear map. If $(W,F)$ is a filtered vector space together with a linear map $f : V \to W$ such that each map $f : (V,F_{<N}) \to (W,F)$ is filtered, then $f : (V,F) \to (W,F)$ is filtered as well. Indeed, for every $n \in \IZ$ we can choose some $N \in \IN$ with $n < N$, so that $F_{<N}^n(V) = F^n(V)$ is mapped into $F^n(W)$.


structure_id: FiltVect
property_id: regular
type: category
proof:
It remains to prove that regular epimorphisms are stable under pullbacks. This follows immediately from their classification below, from the fact that $F^n$ preserves limits, and from the regularity of $\Vect$.

In more detail, if $(V,F) \to (W,F)$ is a regular epimorphism and $(U,F) \to (W,F)$ is any morphism, then $(V,F) \times_{(W,F)} (U,F) \to (U,F)$ is a regular epimorphism, since $V \times_W U \to U$ is surjective and, for every $n \in \IZ$, the restricted map $$F^n(V \times_W U) = F^n(V) \times_{F^n(W)} F^n(U) \to F^n(U)$$ is surjective.


structure_id: FiltVect
property_id: coregular
type: category
proof:
It remains to prove that regular monomorphisms (as classified below) are stable under pushouts. Let $i : (U,F) \to (V,F)$ be a regular monomorphism, i.e. $i$ is injective and $F^n(U) = i^*(F^n(V))$. Let $f : (U,F) \to (W,F)$ be any morphism. We must prove that the canonical morphism $(W,F) \to (V,F) \oplus_{(U,F)} (W,F)$ is a regular monomorphism. It is certainly injective, since the forgetful functor to $\Vect$ preserves colimits and $\Vect$ is abelian, and <a href="/category-implication/dual_abelian_implies_regular">hence coregular</a>. Now suppose that $w \in W$ is an element whose image $[0,w] \in V \oplus_U W$ lies in $F^n(V \oplus_U W)$; we must show that $w \in F^n(W)$. Since, by the construction of colimits in $\FiltVect$, the subspace $F^n(V \oplus_U W)$ is the sum of the images of $F^n(V)$ and $F^n(W)$, there exist $v \in F^n(V)$ and $w' \in F^n(W)$ such that $[0,w] = [v,w']$. This means that there exists some $u \in U$ with $v = i(u)$ and $w = f(u) + w'$. Then $u \in F^n(U)$ because $i(u) \in F^n(V)$. Hence $f(u) \in F^n(W)$, and therefore $w = f(u) + w' \in F^n(W)$.


structure_id: FiltVect
property_id: balanced
type: category
proof:
Take any non-trivial vector space $V$ and consider the two filtrations defined by $F^n(V) \coloneqq 0$ and $G^n(V) \coloneqq V$ for all $n \in \IZ$. Then $\id_V : (V,F) \to (V,G)$ is both a monomorphism and an epimorphism of filtered vector spaces, but it is not an isomorphism.


structure_id: FiltVect
property_id: CSP
type: category
proof:
The canonical filtered linear map $\bigoplus_{n \geq 0} (K,0) \to \prod_{n \geq 0} (K,0)$ is not an epimorphism, since its underlying linear map is the canonical map $\bigoplus_{n \geq 0} K \to \prod_{n \geq 0} K$, which is not surjective.


structure_id: FinAb
property_id: locally finite
type: category
proof:
There is a faithful functor $\FinAb \to \FinSet$ and <a href="/category/FinSet">$\FinSet$</a> is locally finite.


structure_id: FinAb
property_id: self-dual
type: category
proof:
This is a simple special case of Pontryagin duality: The functor $\Hom(-,\IQ/\IZ)$ provides the equivalence.


structure_id: FinAb
property_id: generator
type: category
proof:
If $A,B$ are finite abelian groups whose orders are coprime, then we know that $\Hom(A,B)$ is trivial. But a generator would admit a non-trivial homomorphism to any other non-trivial finite abelian group.


structure_id: FinGrp
property_id: locally finite
type: category
proof:
There is a faithful functor $\FinGrp \to \FinSet$ and <a href="/category/FinSet">$\FinSet$</a> is locally finite.


structure_id: FinGrp
property_id: mono-regular
type: category
proof:
See Prop. 4.2 at the <a href="https://ncatlab.org/nlab/show/regular+monomorphism#Examples" target="_blank">nLab</a>. The proof also works for finite groups.


structure_id: FinGrp
property_id: conormal
type: category
proof:
Since epimorphisms are surjective (see below), this is the first isomorphism theorem for finite groups.


structure_id: FinGrp
property_id: Malcev
type: category
proof:
A direct argument is possible, but this can also be derived from the observation that $\FinGrp$ is the category of group objects in $(\FinSet,\times)$ and Example 2.2.16 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>.


structure_id: FinGrp
property_id: effective congruences
type: category
proof:
Suppose we have a congruence $f, g : E \rightrightarrows X$ in $\FinGrp$. Since the embedding $\FinGrp \hookrightarrow \Grp$ preserves finite limits, it is also a congruence in $\Grp$. We already know that <a href="/category/Grp">$\Grp$</a> has effective congruences since it is algebraic. Using <a href="/content/effective-congruence-quotients">this result</a>, we see that $E$ is the kernel pair of $X \to (X/E)_{\Grp}$ in $\Grp$. Also, the quotient $(X/E)_{\Grp}$ is finite; and the forgetful functor $\FinGrp \to \Grp$ is fully faithful <a href="https://ncatlab.org/nlab/show/reflected+limit#FullSubcategoryInclusionReflectCoLimits" target="_blank">and therefore reflects limits</a>. Thus, we conclude that $E$ is the kernel pair of $X \to (X/E)_{\Grp}$ in $\FinGrp$ as well.


structure_id: FinGrp
property_id: effective cocongruences
type: category
proof:
A proof can be found in <a href="https://mathoverflow.net/questions/511516" target="_blank">MO/511516</a>. It even shows that every cocongruence in $\FinGrp$ is trivial. In short, the proof goes like this: We know that <a href="/category/Grp">$\Grp$</a> has effective cocongruences. Using the fact that amalgamated sums of finite groups are residually finite, one can can show that every cocongruence in $\FinGrp$ is also a cocongruence in $\Grp$.


structure_id: FinGrp
property_id: regular
type: category
proof:
The category is Malcev and hence finitely complete, and it has all coequalizers. The regular epimorphisms coincide with the surjective group homomorphisms (see below), hence are clearly stable under pullbacks.


structure_id: FinGrp
property_id: ℵ₁-cofiltered limits
type: category
proof:
For <a href="/category/FinSet">$\FinSet$</a> know that the embedding $\FinSet \hookrightarrow \Set$ is closed under $\aleph_1$-cofiltered limits. From this and the fact that the forgetful functor $\Grp \to \Set$ preserves limits it follows that $\FinGrp \hookrightarrow \Grp$ is closed under $\aleph_1$-cofiltered limits.


structure_id: FinGrp
property_id: normal
type: category
proof:
Every non-normal subgroup of a finite group (such as $C_2 \hookrightarrow S_3$) provides a counterexample.


structure_id: FinGrp
property_id: binary copowers
type: category
proof:
Assume that $C_2 \sqcup C_2$ exists. This is a finite group, say of order $N$, with two involutions $u,v$ such that for every finite group $G$ with two involutions $a,b$ there is a unique homomorphism $\varphi : C_2 \sqcup C_2 \to G$ with $\varphi(u)=a$ and $\varphi(v)=2$. In particular, when $G$ is generated by $a,b$, then $\ord(G) \leq N$. But then the dihedral group $G \coloneqq D_N$ of order $2N$ yields a contradiction.


structure_id: FinGrp
property_id: generator
type: category
proof:
If $A,B$ are finite groups whose orders are coprime, then we know that $\Hom(A,B)$ is trivial. But a generator would admit a non-trivial homomorphism to any other non-trivial finite group.


structure_id: FinGrp
property_id: cogenerator
type: category
proof:
We apply <a href="/content/missing_cogenerator">this lemma</a> to the collection of finite simple groups: Any non-trivial homomorphism from a finite simple group to a finite group must be injective, and for every $n \in \IN$ there is a finite simple group of size $\geq n$ (for example, the alternating group on $n+5$ elements).


structure_id: FinOrd
property_id: locally finite
type: category
proof:
There is a faithful functor $\FinOrd \to \FinSet$ and <a href="/category/FinSet">$\FinSet$</a> is locally finite.


structure_id: FinOrd
property_id: generator
type: category
proof:
The one-point finite ordered set is a generator since it represents the forgetful functor $\FinOrd \to \Set$.


structure_id: FinOrd
property_id: coequalizers
type: category
proof:
It suffices to construct quotients by equivalence relations. Let $\sim$ be an equivalence relation on $X$, where $(X,\leq)$ is a finite ordered set. Since $X$ is finite, by induction we may assume that $\sim$ is generated by a single relation $(a,b)$. If $a=b$, there is nothing to prove. If $a < b$ and $X = \{0,1,\dotsc,n-1\}$ with the usual order, the quotient is $\{0,1,\dotsc,a,b+1,\dotsc,n-1\}$ with the usual order.


structure_id: FinOrd
property_id: mono-regular
type: category
proof:
Let $i : A \to B$ be a monomorphism of finite ordered sets. If $A$ is empty, then $i$ is clearly regular, so assume it is not. The map $i$ is injective (see below), hence order-reflecting. Define maps $u,v  : B \to A$ by $u(b) \coloneqq \max \{a \in A : i(a) \leq b \}$ and $v(b) \coloneqq \min \{a \in A : b \leq i(a) \}$. These are order-preserving and satisfy $u \circ i = v \circ i$, both sides are $\id_A$. Conversely, if $b \in B$ satisfies $u(b) = v(b) \eqqcolon a$, then $i(a) \leq b$ and $b \leq i(a)$, hence $b = i(a)$. This shows that $i$ is the equalizer of $u,v$.


structure_id: FinOrd
property_id: epi-regular
type: category
proof:
Let $f : A \to B$ be an epimorphism of finite ordered sets. It is surjective (see below). Define $u,v : B \to A$ by $u(b) \coloneqq \min(f^{-1}(b))$ and $v(b) \coloneqq \max(f^{-1}(b))$. One can easily check that $u,v$ are order-preserving maps with $f \circ u = f \circ v$ (both sides are $\id_B$). Let $h : A \to T$ be an order-preserving map with $h \circ u = h \circ v$. Then $h(a)$ only depends on $b \coloneqq f(a)$: We have $u(b) \leq a \leq v(b)$, hence $h(u(b)) \leq h(a) \leq h(v(b)) = h(u(b))$. Therefore, there is a unique map $\tilde{h} : B \to T$ with $\tilde{h}(f(a)) = h(a)$, and one easily checks that it is order-preserving. This shows that $f$ is the coequalizer of $u,v$.


structure_id: FinOrd
property_id: core-thin
type: category
proof:
Let $f : \{1 < \cdots < n \} \to \{1 < \cdots < n \}$ be an automorphism. Then $f(i)$ is the smallest element not contained in $\{f(j) : j < i\}$. From this one can deduce $f(i)=i$ by induction.


structure_id: FinOrd
property_id: ℵ₁-cofiltered limits
type: category
proof:
Let $D : \I \to \FinOrd$ be an $\aleph_1$-cofiltered diagram. Since <a href="/category/FinSet">$\FinSet$</a> is closed under $\aleph_1$-cofiltered limits in $\Set$, the limit of $D$ taken in $\Set$ is a finite set $L$. We define a partial order on $L$ in the obvious way: $x \leq y$ iff $p_i(x) \leq p_i(y)$ for all $i \in \I$. It remains to prove that this is indeed a total order. So assume that $x,y \in L$ satisfy neither $x \leq y$ nor $y \leq x$. Then there exist $i,j \in \I$ such that $p_i(x) \not\leq p_i(y)$ and $p_j(y) \not\leq p_j(x)$. Choose a span $i \leftarrow k \rightarrow j$. Then $p_k(x) \not\leq p_k(y)$ and $p_k(y) \not\leq p_k(x)$ in $D(k)$, which is impossible.


structure_id: FinOrd
property_id: sequential limits
type: category
proof:
Consider the (non-empty) ordered set $[n] \coloneqq \{0 < \cdots < n\}$ for $n \in \IN$. The forgetful functor to $\Set$ is representable, hence preserves all limits. Thus, if the diagram of truncation maps $\cdots \twoheadrightarrow [2] \twoheadrightarrow [1] \twoheadrightarrow [0]$ has a limit in $\FinOrd$, its underlying set is isomorphic to the limit taken in $\Set$, which is $\IN \cup \{\infty\}$. But this is not a finite set.


structure_id: FinOrd
property_id: sequential colimits
type: category
proof:
Consider the (non-empty) ordered set $[n] \coloneqq \{0 < \cdots < n\}$ for $n \in \IN$. Assume the sequence of inclusion maps $[0] \hookrightarrow [1] \hookrightarrow [2] \hookrightarrow \cdots$ has a colimit $(f_n : [n] \to X)$ in $\FinOrd$. Let $n_0 \geq 0$ be fixed. I claim that $f_{n_0}$ is injective, which will then yield a contradiction by taking $n_0 \geq \card(X)$. For $n \geq 0$ define $g_n : [n] \to [n_0]$ as follows. For $n \leq n_0$ it is the inclusion, and for $n \geq n_0$ it is the surjection which keeps all elements of $[n_0]$ and maps all other elements to $n_0$. Observe that $g_n$ preserves the order and $g_{n+1} |_{[n]} = g_n$. Hence, there is a unique order-preserving map $g : X \to [n_0]$ with $g \circ f_n = g_n$ for all $n$. For $n = n_0$ this shows $g \circ f_{n_0} = \id_{[n_0]}$, and $f_{n_0}$ is injective.


structure_id: FinSet
property_id: ℵ₁-cofiltered limits
type: category
proof:
Let $D : \I \to \FinSet$ be a $\aleph_1$-cofiltered diagram. Let $(p_i : P \to D(i))_{i \in \I}$ be a limit cone of the corresponding diagram in $\Set$. We only need to show that $P$ is finite. If not, then it contains a sequence of pairwise distinct elements $x_1,x_2,\dotsc$. For all $n \neq m$ there is some $i_{n,m} \in I$ with $p_{i_{n,m}}(x_n) \neq p_{i_{n,m}}(x_m)$. Since $\I$ is $\aleph_1$-cofiltered, it contains a cone $(j \to i_{m,n})$. Then $p_j(x_n) \neq p_j(x_m)$ for all $n \neq m$. This is not possible since $D(j)$ is a finite set.


structure_id: FinSet
property_id: natural numbers object
type: category
proof:
If $(N,z,s)$ is a natural numbers object, then $$1 \xrightarrow{z} N \xleftarrow{s} N$$ is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. But there is no finite set $N$ with $N \cong 1 + N$.


structure_id: FinVect_c
property_id: essentially countable
type: category
proof:
Every object is isomorphic to $K^n$ for some $n \in \IN$, and $\Hom(K^n,K^m) \cong M_{m \times n}(K)$ is a countable set.


structure_id: FinVect_c
property_id: generator
type: category
proof:
The forgetful functor $\FinVect_K \to \Set$ is faithful and represented by $K$. Hence, $K$ is a generator.


structure_id: FinVect_c
property_id: self-dual
type: category
proof:
The functor $V \mapsto V^*$ defines an equivalence of categories $\FinVect_K^{\op} \simeq \FinVect_K$. In fact, the natural map $V \to V^{**}$, $v \mapsto (\omega \mapsto \omega(v))$ is an isomorphism by standard linear algebra.


structure_id: FinVect_c
property_id: ℵ₁-accessible
type: category
proof:
The inclusion $\FinVect_K \hookrightarrow \Vect_K$ is closed under $\aleph_1$-filtered colimits by <a href="https://mathoverflow.net/a/511519/2841" target="_blank">MO/400763</a>. In particular, $\FinVect_K$ has $\aleph_1$-filtered colimits. Moreover, every object in $\FinVect_K$ is $\aleph_1$-presentable: it is isomorphic to some $K^n$, which is finitely presentable in $\Vect_K$, hence $\aleph_1$-presentable in $\Vect_K$, and therefore also $\aleph_1$-presentable in $\FinVect_K$.


structure_id: FinVect_f
property_id: essentially countable
type: category
proof:
Every object is isomorphic to $K^n$ for some $n \in \IN$, and $\Hom(K^n,K^m) \cong M_{m \times n}(K)$ is a finite, hence countable set.


structure_id: FinVect_u
property_id: essentially small
type: category
proof:
Every object is isomorphic to $K^n$ for some $n \in \IN$, and $\Hom(K^n,K^m) \cong M_{m \times n}(K)$ is a set.


structure_id: Fld
property_id: well-copowered
type: category
proof:
Epimorphisms are the purely inseparable field extensions. If $K \to L$ is purely inseparable, then for all $x \in L$ there is some $n \in \IN$ with $x^n \in L$. An element of $K$ has at most $n$ $n$th-roots. So we can bound the size of $L$.


structure_id: Fld
property_id: multi-algebraic
type: category
proof:
See Eg. 4.3(1) in <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01]</a>.


structure_id: Fld
property_id: connected
type: category
proof:
A field of characteristic $0$ cannot be connected with a field of characteristic $p > 0$. in fact, the connected components of $\Fld$ are the subcategories $\Fld_p$ of fields of characteristic $p$, where $p$ is a prime or $0$.


structure_id: Fld
property_id: balanced
type: category
proof:
Every non-trivial purely inseparable field extension, such as $\IF_p(X^p) \to \IF_p(X)$, provides a counterexample by the descriptions of special morphisms below.


structure_id: Fld
property_id: core-thin
type: category
proof:
If this category was core-thin, Galois theory would not exist. Specifically, the conjugation $\IC \to \IC$, $z \mapsto \overline{z}$ is a non-trivial automorphism.


structure_id: Fld
property_id: multi-terminal object
type: category
proof:
Every field has a non-trivial extension, for instance, the rational function field over itself in one variable. Hence, a multi-terminal object never exists.


structure_id: Fld
property_id: pushouts
type: category
proof:
By <a href="https://math.stackexchange.com/questions/359352/" target="_blank">MSE/359352</a>, the pushout $E \sqcup_K F$ of two field homomorphisms $E \leftarrow K \rightarrow F$ exists if and only if the tensor product $E \otimes_K F$ has a "fieldification": this means that the nilradical is a prime ideal whose quotient ring is a field. This is quite rare: Consider $E = K(X)$, $F = K(Y)$. Then $E \otimes_K F$ is isomorphic to $K[X,Y] (K[X]-\{0\})^{-1} (K[Y]-\{0\})^{-1}$, which is an integral domain but not a field: for example, $X-Y$ has no inverse.


structure_id: Fld
property_id: generator
type: category
proof:
Assume that $G$ is a generator, say of characteristic $p$. Then for all $q \neq p$ all homomorphisms between two fields of characteristic $q$ would be equal, which is absurd.


structure_id: Fld
property_id: cogenerating set
type: category
proof:
We apply <a href="/content/missing_cogenerating_sets">this lemma</a> to the collection of fields: Any homomorphism of fields is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


structure_id: Fld
property_id: binary powers
type: category
proof:
Assume that the product $P \coloneqq \IQ(\sqrt{2}) \times \IQ(\sqrt{2})$ exists. This field is isomorphic to a subfield of $\IQ(\sqrt{2})$, hence $P \cong \IQ$ or $P \cong \IQ(\sqrt{2})$. In the first case, the two projections $P \rightrightarrows \IQ(\sqrt{2})$ must be equal, which means that every two homomorphisms $K \rightrightarrows \IQ(\sqrt{2})$ are equal, which is absurd (take $K = \IQ(\sqrt{2})$ and its two automorphisms). In the second case, the projections induce for every field $K$ a bijection $\Hom(K,\IQ(\sqrt{2})) \cong \Hom(K,\IQ(\sqrt{2}))^2$, which however fails for $K = \IQ(\sqrt{2})$: the left hand side has $2$ elements, the right hand side has $4$ elements. A more general result about products in $\Fld$ can be found at <a href="https://math.stackexchange.com/questions/359352" target="_blank">MSE/359352</a>.


structure_id: Fld
property_id: locally cartesian closed
type: category
proof:
Assume that $K$ is a field such that $\Fld / K$ is cartesian closed. This slice category is equivalent to the poset of subfields of $K$. This poset is a lattice, and our assumption implies that it is distributive (see <a href="/category-implication/distributive_criterion">here</a>). But this is quite rare: Consider $K = \IQ(\sqrt{2}, \sqrt{3})$. By Galois theory, the lattice of subfields is isomorphic to the diamond lattice $M_3$ which is not distributive. Specifically, $(\IQ(\sqrt{2}) \wedge \IQ(\sqrt{6})) \vee (\IQ(\sqrt{3}) \wedge \IQ(\sqrt{6})) = \IQ \vee \IQ = \IQ$, while $(\IQ(\sqrt{2}) \vee \IQ(\sqrt{3})) \wedge \IQ(\sqrt{6}) = \IQ(\sqrt{2},\sqrt{3}) \wedge \IQ(\sqrt{6}) = \IQ(\sqrt{6})$.


structure_id: Fld
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
Inside of $\IF_p(X)$ consider the descending sequence of subfields $$\IF_p(X) \supseteq \IF_p(X^p) \supseteq \IF_p(X^{p^2}) \supseteq \cdots,$$ whose intersection is $\IF_p$. Each $\IF_p(X^{p^n}) \hookrightarrow \IF_p(X)$ is purely inseparable, hence an epimorphism, but in the limit we get $\IF_p \hookrightarrow \IF_p(X)$, which is not even algebraic.


structure_id: FreeAb
property_id: additive
type: category
proof:
The embedding $\FreeAb \hookrightarrow \Ab$ is closed under (finite) direct sums, and <a href="/category/Ab">$\Ab$</a> is additive.


structure_id: FreeAb
property_id: regular
type: category
proof:
This follows formally from the fact that <a href="/category/Ab">$\Ab$</a> is regular and $\FreeAb$ is closed under subobjects and finite products: By Prop. 2.5 in the <a href="https://ncatlab.org/nlab/show/regular+category">nLab</a> it suffices to prove that there are pullback-stable (reg epi, mono)-factorizations. Every homomorphism $f : A \to B$ in $\FreeAb$ factors as $f = i \circ p : A \twoheadrightarrow C \hookrightarrow B$, where $C$ is a subgroup, hence free, and $A \to C$ is surjective. Clearly, surjective homomorphisms are pullback-stable. It remains to show that they coincide with the regular epimorphisms.
(1) If $f : A \to B$ is surjective, it is the coequalizer of $A \times_B A \rightrightarrows A$ in $\Ab$. Since $A \times_B A$ is free abelian, $f$ is also a coequalizer in $\FreeAb$.
(2) If $f : A \to B$ is a regular epimorphism in $\FreeAb$, consider the factorization $f = i \circ p$ as above. Since $f$ is an extremal epimorphism, $i$ must be an isomorphism, so that $f$ is surjective.


structure_id: FreeAb
property_id: countable powers
type: category
proof:
Assume that the power $P = \prod_{n \geq 0} \IZ$ exists in this category. The projections $P \to \IZ$ induce an isomorphism of abelian groups $\Hom(\IZ, P) \to \prod_n \Hom(\IZ, \IZ)$, hence $P \cong \IZ^{\IN}$. But it is <a href="https://groupprops.subwiki.org/wiki/Baer-Specker_group_is_not_free_abelian" target="_blank">well-known</a> that $\IZ^{\IN}$ is not free abelian.


structure_id: FreeAb
property_id: ℵ₁-filtered colimits
type: category
proof:
It is shown in <a href="https://mathoverflow.net/questions/511426" target="_blank">MO/511426</a> that the $\aleph_1$-filtered diagram of countable subgroups of $\IZ^{\IN}$ does not have a colimit in $\FreeAb$.


structure_id: FreeAb
property_id: accessible
type: category
proof:
The question if this category is accessible is undecidable in ZFC. See <a href="https://math.stackexchange.com/questions/720885" target="_blank">MSE/720885</a>.


structure_id: Grp
property_id: mono-regular
type: category
proof:
See Prop. 4.2 at the <a href="https://ncatlab.org/nlab/show/regular+monomorphism#Examples" target="_blank">nLab</a>.


structure_id: Grp
property_id: Malcev
type: category
proof:
See Example 2.2.4 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>.


structure_id: Grp
property_id: cogenerator
type: category
proof:
We apply <a href="/content/missing_cogenerator">this lemma</a> to the collection of simple groups: Any non-trivial homomorphism from a simple group to a group must be injective, and for every infinite cardinal $\kappa$ there is a simple group of size $\geq \kappa$ (for example, the alternating group on $\kappa$ elements).


structure_id: Grp
property_id: coregular
type: category
proof:
This is because injective group homomorphisms are not stable under pushouts, see e.g. <a href="https://math.stackexchange.com/questions/601463/" target="_blank">MSE/601463</a> or <a href="https://math.stackexchange.com/questions/5088032" target="_blank">MSE/5088032</a>.


structure_id: Grp
property_id: counital
type: category
proof:
The canonical morphism $F_2 = \IZ \sqcup \IZ \to \IZ \times \IZ$ is not a monomorphism since $F_2$ is not abelian.


structure_id: Grp
property_id: CSP
type: category
proof:
The canonical homomorphism $\coprod_{n \geq 0} \IZ \to \prod_{n \geq 0} \IZ$ is not surjective because its domain is countable and its codomain is uncountable. Hence it is no epimorphism.


structure_id: Grp
property_id: regular quotient object classifier
type: category
proof:
Assume that $\Grp$ has a (regular) quotient object classifier, i.e. a group $P$ such that every surjective homomorphism $G \to H$ is the cokernel of a unique homomorphism $\varphi : P \to G$. Equivalently, every normal subgroup $N \subseteq G$ is $\langle \langle \varphi(P) \rangle \rangle$ for a unique homomorphism $\varphi : P \to G$, where $\langle \langle - \rangle \rangle$ denotes the normal closure. If $c_g : G \to G$ denotes the conjugation with $g \in G$, then the images of $\varphi$ and $c_g \circ \varphi$ have the same normal closures, so the homomorphisms must be equal. In other words, $\varphi$ factors through the center $Z(G)$. But then every normal subgroup of $G$, in particular $G$ itself, would be contained in $Z(G)$, which is wrong for every non-abelian group $G$.


structure_id: Grp
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Ab">$\Ab$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the forgetful functor $\Ab \to \Grp$ which indeed preserves epimorphisms.


structure_id: Grp
property_id: cocartesian cofiltered limits
type: category
proof:
For cofiltered diagrams of groups $(H_i)$ and a group $G$ the canonical homomorphism $$\textstyle \alpha : G \sqcup \lim_i H_i \to \lim_i (G \sqcup H_i)$$ is injective, but often fails to be surjective because the components of an element in the image have bounded <i>free product length</i> (the number of factors appearing in the reduced form). Specifically, consider the free groups $G = \langle y \rangle$ and $H_n = \langle x_1,\dotsc,x_n \rangle$ for $n \in \IN$ with the truncation maps $H_{n+1} \to H_n$, $x_{n+1} \mapsto 1$. Define $$p_n \coloneqq x_1 \, y \, x_2 \, y \, \cdots \, x_{n-1} \, y \, x_n \, y^{-(n-1)} \in G \sqcup H_n.$$ If we substitute $x_{n+1}=1$ in $p_{n+1}$, we get $p_n$. Thus, we have $p = (p_n) \in \lim_n (G \sqcup H_n)$. This element does not lie in the image of $\alpha$ since the free product length of $p_n$ (which is well-defined) is $2n$, which is unbounded.


structure_id: Grp_c
property_id: generator
type: category
proof:
The countable group $\IZ$ is a generator because it represents the forgetful functor $\Grp_\c \to \Set$.


structure_id: Grp_c
property_id: finite products
type: category
proof:
This is because <a href="/category/Grp">$\Grp$</a> has finite (in fact, all) products, and $\Grp_\c \hookrightarrow \Grp$ is closed under finite products. This is because a finite product of countable sets is again countable.


structure_id: Grp_c
property_id: equalizers
type: category
proof:
One can use the same construction as in <a href="/category/Grp">$\Grp$</a> since a subgroup of a countable group is again countable.


structure_id: Grp_c
property_id: coequalizers
type: category
proof:
One can use the same construction as in <a href="/category/Grp">$\Grp$</a> since a quotient of a countable group is again countable.


structure_id: Grp_c
property_id: countable coproducts
type: category
proof:
This is because <a href="/category/Grp">$\Grp$</a> has countable (in fact, all) coproducts, and $\Grp_\c \hookrightarrow \Grp$ is closed under countable coproducts. This is because a countable union of countable sets is again countable.


structure_id: Grp_c
property_id: mono-regular
type: category
proof:
This can be deduced from the corresponding property of <a href="/category/Grp">$\Grp$</a> as follows: Let $i : K \hookrightarrow G$ be a monomorphism in $\Grp_\c$, i.e. an injective homomorphism of countable groups. Since $\Grp$ is mono-regular, there is a group $H$ and two homomorphisms $f,g : G \rightrightarrows H$ with $i = \eq(f,g)$. Let $H' \subseteq H$ be the subgroup generated by $\im(f) \cup \im(g)$. Since $G$ is countable, $H'$ is countable as well, and $f,g$ corestrict to homomorphisms $f', g' : G \rightrightarrows H'$. Hence, $i = \eq(f',g')$.


structure_id: Grp_c
property_id: conormal
type: category
proof:
If $f : G \to H$ is an epimorphism in $\Grp_\c$, i.e. a surjective homomorphism of countable groups, then $f$ is the cokernel of $K \hookrightarrow G$ in $\Grp$, where $K$ is the kernel of $f$. Since $K$ is countable, it is also the cokernel in $\Grp_\c$.


structure_id: Grp_c
property_id: regular
type: category
proof:
We already know that the category is finitely complete, and that it has all coequalizers. The regular epimorphisms coincide with the surjective group homomorphisms (see below), hence are clearly stable under pullbacks.


structure_id: Grp_c
property_id: effective congruences
type: category
proof:
A congruence on a countable group $G$ has the form $\{(g,h) \in G^2 : g^{-1} h \in N \}$ for some normal subgroup $N \subseteq G$. It is the kernel pair of the projection $p : G \twoheadrightarrow G/N$ in $\Grp$, but also in $\Grp_\c$ since $G/N$ is countable.


structure_id: Grp_c
property_id: effective cocongruences
type: category
proof:
Let $G + G \twoheadrightarrow H$ be a cocongruence in $\Grp_\c$. Since $\Grp_\c \hookrightarrow \Grp$ is closed under finite colimits, this is the same as a cocongruence in $\Grp$ where $G,H \in \Grp$ happen to be countable groups. Since we already know that <a href="/category/Grp">$\Grp$</a> has effective cocongruences, the cocongruence is the cokernel pair of some homomorphism of groups $K \to H$. If $K' \subseteq H$ denotes the image of $K$, it is then also the cokernel pair of the inclusion $K' \hookrightarrow H$, and $K'$ is countable.


structure_id: Grp_c
property_id: normal
type: category
proof:
Every non-normal subgroup of a countable group (such as $C_2 \hookrightarrow S_3$) provides a counterexample.


structure_id: Grp_c
property_id: countable powers
type: category
proof:
Since the forgetful functor $\Grp_\c \to \Set$ is representable, it preserves products. Therefore, if the power $\IZ^{\IN}$ exists in $\Grp_\c$, its underlying set must be the ordinary cartesian product, which however is uncountable.


structure_id: Grp_c
property_id: ℵ₂-small copowers
type: category
proof:
Assume that the copower $G \coloneqq I \otimes \IZ$ exists in $\Grp_\c$. We will prove that $I$ is countable; in particular, $\aleph_1 \otimes \IZ$ does not exist. The copower is a countable group $G$ with a universal family of elements $(x_i)_{i \in I}$. For every $i \in I$ there is a homomorphism $G \to \IZ$ defined by $x_i \mapsto 1$ and $x_j \mapsto 0$ for $j \neq i$. In particular, $x_i \neq x_j$ whenever $i \neq j$. Thus, $x : I \to G$ is an injective map, which proves that $I$ is countable.


structure_id: Grp_c
property_id: coregular
type: category
proof:
Pushouts of injective homomorphisms between countable groups do not need to be injective, see <a href="https://math.stackexchange.com/questions/5088032" target="_blank">MSE/5088032</a>.


structure_id: Grp_c
property_id: cogenerator
type: category
proof:
Assume that a cogenerator $Q$ exists in $\Grp_\c$. There are only countably many finitely generated subgroups of $Q$. But there are continuum many finitely generated simple groups; this follows from Corollary 1.5 in <a href="https://arxiv.org/abs/1807.06478" target="_blank">Finitely generated infinite simple groups of homeomorphisms of the real line</a> by J. Hyde and Y. Lodha. Hence, there is a finitely generated (and hence countable) simple group $H$ which does not embed into $Q$. Since $H$ is simple, any homomorphism $H \to Q$ must be trivial then. But then $\id_H, 1 : H \rightrightarrows H$ are not separated by a homomorphism $H \to Q$.


structure_id: Grp_c
property_id: ℵ₁-cofiltered limits
type: category
proof:
The proof is similar to <a href="/category/Set_c">$\Set_\c$</a>. Since the forgetful functor $\Grp_\c \to \Set$ is representable, it preserves limits. Moreover, $\Grp \to \Set$ preserves limits. Therefore, it suffices to construct an $\aleph_1$-cofiltered diagram of countable groups whose $\Grp$-limit is not countable. Let $R$ be an uncountable set (such as $\IR$) and consider the poset $P_{<\aleph_1}(R)$ of countable subsets of $R$. It is $\aleph_1$-filtered, so its opposite is $\aleph_1$-cofiltered. Define a diagram $X : P_{<\aleph_1}(R)^{\op} \to \Grp_\c$ by mapping a countable subset $C \subseteq R$ to the free abelian group $F(C)$ on $C$ and an inclusion $C \subseteq D$ to the canonical epimorphism $F(D) \to F(C)$ which sends each element of $C$ to itself and sends all elements of $D \setminus C$ to $0$. For $r \in R$ we define $\varepsilon(r) \in \lim(X)$ by $$\varepsilon(r)_C = \begin{cases} r & r \in C \\ 0 & r \notin C \end{cases}$$ It is easily checked that this indeed lies in the limit. Moreover, $\varepsilon(r) = \varepsilon(r')$ implies $r=r'$, since evaluating at $C \coloneqq \{r\}$ yields $r = r'$. Hence, $\lim(X)$ is uncountable.


structure_id: Haus
property_id: equalizers
type: category
proof:
This follows from the corresponding fact for <a href="/category/Top">$\Top$</a> since subspaces of Hausdorff spaces are again Hausdorff.


structure_id: Haus
property_id: products
type: category
proof:
This follows from the corresponding fact for <a href="/category/Top">$\Top$</a> since products of Hausdorff spaces are again Hausdorff.


structure_id: Haus
property_id: cocomplete
type: category
proof:
This follows since $\Haus$ is a reflective subcategory of <a href="/category/Top">$\Top$</a>, which is cocomplete. For the reflector, see e.g. the <a href="https://ncatlab.org/nlab/show/Hausdorff+space#HausdorffReflection" target="_blank">nLab</a>. Explicitly, we construct the colimit of Hausdorff spaces by applying the reflector to the colimit of the underlying topological spaces.


structure_id: Haus
property_id: infinitary extensive
type: category
proof:
This follows exactly as for <a href="/category/Top">$\Top$</a> since Hausdorff spaces are closed under taking subspaces and coproducts in $\Top$.


structure_id: Haus
property_id: well-copowered
type: category
proof:
Every epimorphism has dense image, so it suffices to prove that if a Hausdorff space $X$ has a dense subset $D \subseteq X$, we can bound the cardinality of $X$ in terms of the cardinality of $D$. For $x \in X$ let $\alpha(x)$ be the set of all $U \cap D$, where $U$ is an open neighborhood of $x$. This defines a map $\alpha : X \to P(P(D))$, and we claim that it is injective: this is because $\{x\}$ is the intersection of all $\overline{U} = \overline{U \cap D}$, where $U$ runs through the open neighborhoods of $x$.


structure_id: Haus
property_id: effective cocongruences
type: category
proof:
As the proof at <a href="https://mathoverflow.net/a/509582/2841" target="_blank">MO/509548</a> shows, in fact any coreflexive corelation on $X$ in $\Haus$ is of the form $X +_S X$ for a closed subset $S$ of $X$. Such a cocongruence is clearly effective.


structure_id: Haus
property_id: balanced
type: category
proof:
The inclusion $\IQ \hookrightarrow \IR$ is a counterexample; it is an epimorphism since $\IQ$ is dense in $\IR$.


structure_id: Haus
property_id: Malcev
type: category
proof:
This is clear since <a href="/category/Set">$\Set$</a> is not Malcev and can be interpreted as the subcategory of discrete spaces (which are Hausdorff).


structure_id: Haus
property_id: regular subobject classifier
type: category
proof:
Assume that there is a regular subobject classifier $\Omega$. By the classification of regular monomorphisms, we would have an isomorphism between $\Hom(X,\Omega)$ and the set of closed subsets of $X$ for any Hausdorff space $X$. If we take $X = 1$ we see that $\Omega$ has two points. Since $\Omega$ is Hausdorff, $\Omega \cong 1 + 1$ must be discrete. But then $\Hom(X,\Omega)$ is isomorphic to the set of all clopen subsets of $X$, of which there are usually far fewer than closed subsets (consider $X = [0,1]$).


structure_id: Haus
property_id: cartesian filtered colimits
type: category
proof:
It is shown in <a href="https://math.stackexchange.com/questions/1255678">MSE/1255678</a> that $\IQ \times - : \Top \to \Top$ does not preserve sequential colimits (so that it cannot be a left adjoint). The same example also works in $\Haus$: Surely $\IQ$ is Hausdorff, $X_n$ is Hausdorff, as is their colimit $X$, and the colimit (taken in $\Top$) of the $X_n \times \IQ$ admits a bijective continuous map to a Hausdorff space, therefore is also Hausdorff, meaning it is also the colimit taken in $\Haus$.


structure_id: Haus
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
Recall the counterexample for sets: The unique maps $\IN_{\geq n} \to 1$ are surjective, but their limit $0 = \bigcap_{n \geq 0} \IN_{\geq n} \to 1$ is not. This also works in $\Haus$ by using discrete topologies. We could also apply a variant of (the dual of) Lemma 2 <a href="/content/subcategories">here</a> to the discrete topology functor $\Set \to \Haus$, which does not preserve all cofiltered limits, but does preserve intersections.


structure_id: Haus
property_id: filtered-colimit-stable monomorphisms
type: category
proof:
The proof is similar to <a href="/category/Met">$\Met$</a>. For $n \geq 1$ let $X_n$ be the pushout of
$$(-\infty, -1/n] \cup [1/n, \infty) \hookrightarrow \IR$$
with itself. That is, $X_n$ is the union of two lines $\IR \times \{1\}$ and $\IR \times \{2\}$ where we identify $(x,1) \equiv (x,2)$ when $|x| \geq 1/n$. Then $X_n$ is Hausdorff, and there is a canonical surjective continuous map $X_n \to X_{n+1}$. The colimit in $\Top$ is the union of two lines where we identify $(x,1) \equiv (x,2)$ when $|x| \geq 1/n$ for some $n$, i.e. when $x \neq 0$. This is the line with the double origin, which is not Hausdorff. Its Hausdorff reflection is the line $\IR$ where all points of both lines are identified, and it provides the colimit in $\Haus$. Now, the injective continuous maps $\{1,2\} \to X_n$, $i \mapsto (0,i)$ (where $\{1,2\}$ is discrete) become the constant map $0 : \{1,2\} \to \IR$ in the colimit, which is not a monomorphism.


structure_id: Haus
property_id: accessible
type: category
proof:
In fact, it does not have any small colimit-dense subcategory by <a href="https://math.stackexchange.com/questions/4097315/" target="_blank">MSE/4097315</a>.


structure_id: Haus
property_id: cogenerator
type: category
proof:
Assume that $Q$ is a cogenerator. Since $Q$ is Hausdorff, $Q$ is $T_1$. By a theorem of Herrlich (<i>Wann sind alle stetigen Abbildungen in Y konstant</i>. Math. Z. 90 (1965): 152-154. <a href="http://eudml.org/doc/170472" target="_blank">EUMDL</a>), there is a regular Hausdorff space $X$ with $\geq 2$ points such that every continuous map $X \to Q$ is constant. (The author only states that $X$ is regular, but actually, $X$ is regular and $T_1$, hence Hausdorff.) But since $Q$ is a cogenerator, this implies that all maps $1 \rightrightarrows X$ are equal, i.e. that $X$ has just one point. This is a contradiction.


structure_id: Haus
property_id: regular
type: category
proof:
The regular epimorphisms are precisely the surjective quotient maps of Hausdorff spaces (see below). In a regular category, for every regular epimorphism $X \to Y$ and every object $Z$, the induced morphism $X \times Z \to Y \times Z$ is again a regular epimorphism. This is not the case in $\Haus$ (or $\Top$, for that matter). The standard example is the quotient map $\IR \to \IR / \IZ^+$, for which the induced map $\IR \times \IQ \to \IR/\IZ^+ \times \IQ$ is not a quotient map (<a href="https://math.stackexchange.com/questions/1907972/">MSE/1907972</a>).


structure_id: Haus
property_id: coregular
type: category
proof:
Let $\Gamma$ be the <a href="https://en.wikipedia.org/wiki/Moore_plane" target="_blank">Moore plane</a>. Its underlying set is $\{(x,y) \in \IR^2 : y \geq 0 \}$. The open neighborhoods of points $(x,y)$ with $y > 0$ are those of $\IR^2$ (intersected with $\Gamma$), and the basic open neighborhoods of a point $(x,0)$ are open disks centered at $(x,\varepsilon)$ with radius $\varepsilon$ for some $\varepsilon > 0$. Then $\Gamma$ is Hausdorff, and the $x$-axis $A \coloneqq \{(x,0) : x \in \IR\}$ is a closed discrete subspace of $\Gamma$. In particular, by the classification of regular monomorphisms below, the inclusion map $i : A \to \Gamma$ is a regular monomorphism. Consider the two subsets $A_1 \coloneqq \{(x,0) : x \in \IQ \}$ and $A_2 \coloneqq \{(x,0) : x \in \IR \setminus \IQ \}$ of $A$. They are closed in $A$ (since $A$ is closed and discrete), disjoint, but cannot be separated by disjoint open neighborhoods in $\Gamma$; this is part of the proof of the well-known fact that $\Gamma$ is not normal (<a href="https://math.stackexchange.com/questions/2528435" target="_blank">MSE/2528435</a>). Let $C \coloneqq \{1,2\}$ be the discrete two-point space. The map $f : A \to C$ defined by $f(a)=1$ for $a \in A_1$ and $f(a)=2$ for $a \in A_2$ is continuous, since $A$ is discrete. The pushout $C \sqcup_A \Gamma$ in $\Haus$ is the Hausdorff reflection of the pushout $Q$ in $\Top$. Notice that $Q$ is the quotient space of $\Gamma$ in which $A_1$ and $A_2$ are each collapsed to a point, denoted by $[A_1]$ and $[A_2]$. The canonical map $C \to Q$ is given by $i \mapsto [A_i]$. Now, $[A_1]$ and $[A_2]$ cannot be separated by disjoint open neighborhoods in $Q$, since such neighborhoods would pull back to disjoint open neighborhoods of $A_1$ and $A_2$ in $\Gamma$. Thus, they are identified in the Hausdorff reflection. This shows that the canonical map $C \to C \sqcup_A \Gamma$ is not injective and hence not a regular monomorphism.


structure_id: J2
property_id: finitary algebraic
type: category
proof:
The structure of a Jónsson-Tarski algebra on a set $X$ is equivalent to one binary operation $\mu : X^2 \to X$ and two unary operations $\lambda, \rho : X \rightrightarrows X$ such that $\mu(\lambda(x),\rho(x)) = x$, $\lambda(\mu(x,y))=x$, and $\rho(\mu(x,y))=y$.


structure_id: J2
property_id: semi-strongly connected
type: category
proof:
There is a bijection $\alpha = (\lambda,\rho) : \IN \to \IN \times \IN$ such that $\lambda$ has a fixed point, but $\rho$ does not (see below). Then the isomorphism $\beta \coloneqq (\rho,\lambda)$ has the opposite property. There cannot be any morphism $(\IN,\alpha) \to (\IN,\beta)$, as it would map the fixed point of $\lambda$ to a fixed point of $\rho$, and likewise there is no morphism $(\IN,\beta) \to (\IN,\alpha)$.
To construct $\alpha$ or rather $\alpha^{-1} : \IN \times \IN \to \IN$, we can alter the standard bijection $(n,m) \mapsto 2^n (2m+1) - 1$ as follows: $$\alpha^{-1}(n,m) = \begin{cases} 2 & (n,m) = (0,0) \\ 0 & (n,m) = (0,1) \\ 2^n (2m+1) - 1 & \text{otherwise} \end{cases}$$ Then $\alpha(0)=(0,1)$, i.e. $\lambda(0)=0$. The function $\rho$ has no fixed point, i.e. $\alpha^{-1}(n,m) \neq m$ for all $n,m$. Namely, if $(n,m)=(0,0)$, then $\alpha^{-1}(n,m)=2 \neq m$. If $(n,m)=(0,1)$, then $\alpha^{-1}(n,m)=0 \neq m$. Otherwise, $$\alpha^{-1}(n,m) = 2^n (2m+1) - 1 \geq (2m+1)-1 = 2m \geq m,$$ and equality can only hold if $m=0$ and $n=0$, which we already excluded.


structure_id: LRS_R
property_id: locally small
type: category
proof:
For two ringed spaces $(X,\O_X), (Y,\O_Y)$, the collection of morphisms $X \to Y$ of ringed spaces is the collection $\prod_{f \in \Hom(X,Y)} \Hom(\O_Y,f_* \O_X)$, which is a set since $\Hom(X,Y)$ is a set and each $\Hom(\O_Y,f_* \O_X)$ is a set.


structure_id: LRS_R
property_id: complete
type: category
proof:
See <a href="https://arxiv.org/abs/1103.2139" target="_blank">Localization of ringed spaces</a> by W. Gillam. See also <a href="https://math.stackexchange.com/questions/1033675" target="_blank">MSE/1033675</a>.


structure_id: LRS_R
property_id: cocomplete
type: category
proof:
See Demazure-Gabriel's "Groupes algébriques", I. §1. 1.6. Specifically, the forgetful functor from locally ringed spaces to ringed spaces preserves colimits, and colimits of ringed spaces are built from colimits of topological spaces and limits of commutative rings, see <a href="https://math.stackexchange.com/questions/1646202" target="_blank">MSE/1646202</a>.


structure_id: LRS_R
property_id: infinitary extensive
type: category
proof:
[Sketch] Since <a href="/category/Top">$\Top$</a> is infinitary extensive, a morphism $f : Y \to \coprod_i X_i \eqqcolon X$ yields a decomposition $Y = \coprod_i Y_i$ (as topological spaces) with continuous maps $f_i : Y_i \to X_i$. Endow the open subset $Y_i \subseteq Y$ with the restricted sheaf. Then each $f_i$ becomes a morphism of locally ringed spaces, and $Y = \coprod_i Y_i$ holds as locally ringed spaces.


structure_id: LRS_R
property_id: balanced
type: category
proof:
Choose a residue field $K$ of $R$. The canonical morphism $$\Spec(K[X]/X \times K[X,X^{-1}]) \longrightarrow \Spec(K[X])$$ is a mono- and an epimorphism, but no isomorphism.


structure_id: LRS_R
property_id: semi-strongly connected
type: category
proof:
This is because already the full subcategory <a href="/category/Sch_R">$\Sch_R$</a> of $R$-schemes is not semi-strongly connected.


structure_id: LRS_R
property_id: Malcev
type: category
proof:
This is because already the full subcategory <a href="/category/Sch_R">$\Sch_R$</a> of $R$-schemes is not Malcev.


structure_id: LRS_R
property_id: co-Malcev
type: category
proof:
We can adjust the proof for <a href="/category/Top">$\Top$</a> (see <a href="https://mathoverflow.net/questions/509548" target="_blank">MO/509548</a>) as follows: Let $K$ be a residue field of $R$, let $X$ be a singleton and $Y = \{u,v\}$ be the Sierpinski space where $\{u\}$ is open, but $\{v\}$ is not. Endow both with the sheaf of locally constant functions to $K$. Thus, $\O_X(X) = K$, $\O_Y(Y) = \O_Y(\{u\}) = K$. There is a canonical morphism $p : X + X \to Y$. It is a coreflexive corelation that is not cosymmetric.


structure_id: LRS_R
property_id: generating set
type: category
proof:
Out of any small set $S$ of locally ringed spaces, there is only a small set of residue fields at their points. Therefore, if $K$ is a field over $R$ with a strictly larger cardinality than any of these residue fields, then the only possible morphism from an element of $S$ to $\Spec K(X,Y)$ is one with an empty domain. However, that makes it impossible for $S$ to distinguish the two canonical automorphisms of $\Spec K(X,Y)$.
Alternatively, using the usual adjunction between affine schemes and locally ringed spaces (<a href="https://en.wikipedia.org/wiki/%C3%89l%C3%A9ments_de_g%C3%A9om%C3%A9trie_alg%C3%A9brique" target="_blank">EGA I</a> (1971), Ch. 1, Prop. 1.6.3), a generating set in $\LRS_R$ would induce a generating set in the category of affine $R$-schemes, which contradicts the fact that <a href="/category/CAlg(R)">$\CAlg(R)$</a> does not have a cogenerating set.


structure_id: M-Set
property_id: semi-strongly connected
type: category
proof:
If this category is semi-strongly connected depends on the choice of $M$. For $M = 1$ it is, for $M = \IZ$ it is not. In general, if $G$ is a group, then $G{-}\Set$ is semi-strongly connected if and only if for all subgroups $H,K \subseteq G$, $H$ is subconjugated to $K$ or $K$ is subconjugated to $H$. If $G$ is abelian, this means that the poset of subgroups is linear, in which case $G$ is either isomorphic to $\IZ/p^n$ or to $\IZ/p^{\infty}$ for a prime $p$. See also <a href="https://math.stackexchange.com/questions/5129804" target="_blank">MSE/5129804</a>.


structure_id: Man
property_id: finite products
type: category
proof:
In short, this follows from the corresponding statement for topological spaces and $\IR^n \times \IR^m \cong \IR^{n+m}$.


structure_id: Man
property_id: generator
type: category
proof:
The $0$-dimensional one-point manifold is a generator since it represents the forgetful functor $\Top \to \Set$.


structure_id: Man
property_id: cogenerator
type: category
proof:
The manifold $\IR$ is a cogenerator, since for every smooth manifold $M$ and points $p \neq q$ in $M$ there is a smooth function $f : M \to \IR$ with $f(p) = 1$ and $f(q) = 0$ (John Lee, Introduction to Smooth Manifolds, Prop. 2.25).


structure_id: Man
property_id: essentially small
type: category
proof:
This is a consequence of the <a href="https://en.wikipedia.org/wiki/Whitney_embedding_theorem" target="_blank">Whitney embedding theorem</a>. But there is also a more direct proof: Since a manifold is second-countable, it is Lindelöf (<a href="https://topospaces.subwiki.org/wiki/Second-countable_implies_Lindelof" target="_blank">proof</a>). In particular, there is a countable atlas. It is then completely determined by countable many open subsets of Euclidean spaces and the transition maps.


structure_id: Man
property_id: extensive
type: category
proof:
[Sketch] Since <a href="/category/Top">$\Top$</a> is infinitary extensive, a continuous map $f : M \to \coprod_i N_i$ corresponds to a decomposition $M = \coprod_i M_i$ (as topological spaces) with continuous maps $f_i : M_i \to N_i$. Endow the open subset $M_i \subseteq M$ with the smooth structure inherited from $M$. Now remark that $f$ is smooth iff each $f_i$ is smooth.


structure_id: Man
property_id: countably distributive
type: category
proof:
To construct countable coproducts, take the usual disjoint union of spaces, which is clearly locally Euclidean and Hausdorff, and it is second countable since we are using only countable many spaces. (Without that condition, all coproducts would exist.) Now we need to check that the canonical smooth map $\coprod_i X \times Y_i \to X \times \coprod_i Y_i$ is a diffeomorphism (for countable families). It is a homeomorphism since <a href="/category/Top">$\Top$</a> is infinitary distributive. The inverse $X \times \coprod_i Y_i \to \coprod_i X \times Y_i$ is smooth since the domain is covered by the open subsets $X \times Y_i$ on which the map is clearly smooth.


structure_id: Man
property_id: Cauchy complete
type: category
proof:
See Theorem 2.1 at the <a href="https://ncatlab.org/nlab/show/smooth+manifold" target="_blank">nLab</a>.


structure_id: Man
property_id: coquotients of cocongruences
type: category
proof:
Let $p : X + X \twoheadrightarrow E$ be a cocongruence with coreflexivity morphism $r : E \to X$, so that $r \circ p : X + X \to X$ is the codiagonal. Since $p$ is an epimorphism, it has dense image (see below). We first claim that in fact $p$ also has closed image and therefore is surjective. Because $r \circ (p \circ i_1) : X \to X$ is the identity, the image of $p \circ i_1$ is the equalizer of $\id_E$ and $(p \circ i_1) \circ r$, hence closed. Likewise, the image of $p \circ i_2$ is closed. Thus, the image of $p$, which is the union of these images, is closed.
Now, since the pushforward maps of tangent spaces compose to the identity, we see that $p$ must be a local immersion and $r$ must be a submersion. Also, since the fibers of $r$ have one or two points each, we see that the dimension of $E$ must locally be the same as the dimension of $X$. This implies that in fact $p$ and $r$ are local diffeomorphisms. Therefore, the cardinality of the fiber of $r$ is locally constant. Thus, if $U$ is the subset of $X$ where $r$ has fiber of a single point, with the subspace topology, then $U$ is a clopen submanifold of $X$ which serves as the equalizer of $p \circ i_1$ and $p \circ i_2$.


structure_id: Man
property_id: effective cocongruences
type: category
proof:
From the proof that $\Man$ has coquotients of cocongruences, we know that for any cocongruence $X \rightrightarrows E$, there is a clopen submanifold $U$ of $X$ such that the fibers of $r : E \twoheadrightarrow X$ have one point on $U$, and two points on $X \setminus U$. Therefore, $E$ is the cokernel pair of the inclusion map $U \hookrightarrow X$.


structure_id: Man
property_id: balanced
type: category
proof:
The irrational winding of a torus $\IR \to S^1 \times S^1$, $t \mapsto (e^{i t}, \, e^{i \alpha t})$, where $\alpha \in \IR \setminus \IQ$, provides a counterexample. It is injective and has a dense image. Hence, it is a mono- and an epimorphism. But it is not surjective, hence no isomorphism.


structure_id: Man
property_id: countable powers
type: category
proof:
The power $\IR^{\IN}$ does not exist. More generally, let $(M_n)_{n \geq 0}$ be a sequence of smooth manifolds of positive dimension whose product $(\pi_n : P \to M_n)_{n \geq 0}$ exists. This product cone in $\Man$ yields a product cone in $\Set$ since the forgetful functor $\Man \to \Set$ is representable, hence preserves all limits. Choose points $x_n \in M_n$ with $T_{x_n}(M_n) \neq 0$. Choose the point $x \in P$ with $\pi_n(x) = x_n$. Consider the linear map $T_x(P) \to \prod_{n \geq 0} T_{x_n}(M_n)$ induced by the derivatives $d_x(\pi_n) : T_x(P) \to T_{x_n}(M_n)$. Since $T_x(P)$ is finite-dimensional and $\prod_{n \geq 0} T_{x_n}(M_n)$ is not, it cannot be surjective. But actually, it is: Choose tangent vectors $v_n \in T_{x_n}(M_n)$. Choose smooth curves $\gamma_n : \IR \to M_n$ with $\gamma_n(0)=x_n$ and ${\gamma_n}'(0) = v_n$. By the universal property there is a unique smooth curve $\gamma : \IR \to P$ with $\pi_n \gamma = \gamma_n$. In particular, $\gamma(0) = x$. The chain rule now implies that $\gamma'(0) \in T_x(P)$ is a preimage of $(v_n)$ – a contradiction.


structure_id: Man
property_id: pullbacks
type: category
proof:
See <a href="https://math.stackexchange.com/questions/5129579/" target="_blank">MSE/5129579</a> or <a href="https://math.stackexchange.com/questions/322485" target="_blank">MSE/322485</a>.


structure_id: Man
property_id: sequential colimits
type: category
proof:
If $\Man$ had sequential colimits, then by <a href="/content/special_sequential_colimits">this lemma</a> there would be a manifold $M$ that admits a split epimorphism $M \to \IR^n$ for every $n$. But then $M$ will have an infinite-dimensional tangent space, which is a contradiction.


structure_id: Man
property_id: ℵ₂-small copowers
type: category
proof:
A proof can be found at <a href="https://math.stackexchange.com/a/5083683/1650" target="_blank">MSE/5083641</a> (ignore the arguments regarding equal dimension).


structure_id: Man
property_id: ℵ₁-cofiltered limits
type: category
proof:
We continue the counterexample from <a href="/category/Set_c">$\Set_c$</a>. There, we started with an arbitrary uncountable set $R$ and constructed a diagram $X : P_{<\aleph_1}(R)^{\op} \to \Set$ by sending a countable subset $C \subseteq R$ to the pointed countable set $C \sqcup \{\ast\}$. Its limit in $\Set$ is uncountable. Now endow each set $C \sqcup \{\ast\}$ with the discrete topology. Each of them becomes a $0$-dimensional smooth manifold. Assume that there exists a limit cone $(\pi_C : M \to C \sqcup \{\ast\})$ in $\Man$. Since the forgetful functor $\Man \to \Set$ is representable, it preserves limits. In particular, the underlying set of $M$ is the limit of the underlying sets and is therefore uncountable. Since the codomain of $\pi_C$ is discrete, the map $\pi_C$ is locally constant. Thus, for every connected component $K \subseteq M$, the map $\pi_C$ is constant on $K$. Since the maps $\pi_C$ are jointly injective, it follows that every connected component of $M$ consists of a single point. Hence, $M$ is totally disconnected. Since $M$ is a smooth manifold, it follows that $M$ has dimension $0$. Since $M$ is also second-countable, it follows that the underlying set of $M$ is countable, contradicting our previous observation.


structure_id: Man
property_id: quotients of congruences
type: category
proof:
If $\Man$ had quotients of congruences, then by <a href="/content/pushouts-of-monos-via-congruence-quotients">this lemma</a>, it would have a pushout of $\IR \leftarrow \{ 0 \} \rightarrow \IR$.  This contradicts <a href="https://mathoverflow.net/questions/19116">MO/19916</a>.


structure_id: Meas
property_id: generator
type: category
proof:
The one-point measurable space (with the unique $\sigma$-algebra) is a generator since it represents the forgetful functor $\Meas \to \Set$.


structure_id: Meas
property_id: cogenerator
type: category
proof:
Take the two-element set $2$ endowed with the trivial $\sigma$-algebra (where only $\varnothing$ and $2$ are measurable), and use that $2$ is a cogenerator for $\Set$.


structure_id: Meas
property_id: complete
type: category
proof:
Take the limit of the underlying sets and take the smallest $\sigma$-algebra making all projections measurable.


structure_id: Meas
property_id: cocomplete
type: category
proof:
Take the colimit of the underlying sets and take the largest $\sigma$-algebra making all inclusions measurable. That is, a set is measurable iff its preimage under each inclusion is measurable.


structure_id: Meas
property_id: infinitary extensive
type: category
proof:
[Sketch] Since <a href="/category/Set">$\Set$</a> is infinitary extensive, a map $f : Y \to \coprod_i X_i \eqqcolon X$ corresponds to a decomposition $Y = \coprod_i Y_i$ (as sets) with maps $f_i : Y_i \to X_i$. Endow the measurable subset $Y_i \subseteq Y$ with the restricted $\sigma$-algebra. If $f$ is measurable, each $f_i$ is measurable, and $Y = \coprod_i Y_i$ holds as measurable spaces.


structure_id: Meas
property_id: filtered-colimit-stable monomorphisms
type: category
proof:
This follows from Lemma 2 <a href="/content/subcategories">here</a> applied to the forgetful functor to $\Set$.


structure_id: Meas
property_id: regular subobject classifier
type: category
proof:
The set $\{0,1\}$ with the trivial $\sigma$-algebra is a regular subobject classifier since measurable maps $X \to \{0,1\}$ correspond to subsets of $X$.


structure_id: Meas
property_id: balanced
type: category
proof:
Take a set $X$ with two different $\sigma$-algebras $\A \subset \B$ (for example, $\A = \{\varnothing,X\}$ and $\B = P(X)$ when $X$ has at least $2$ elements), then the identity map $(X,\B) \to (X,\A)$ provides a counterexample.


structure_id: Meas
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Meas$ which equips a set with the trivial $\sigma$-algebra.


structure_id: Meas
property_id: effective cocongruences
type: category
proof:
The proof is similar to the one for <a href="/category/Top">$\Top$</a>: Use the trivial $\sigma$-algebra on a two-point set.


structure_id: Met
property_id: generator
type: category
proof:
The one-point metric space is a generator since it represents the forgetful functor $\Met \to \Set$.


structure_id: Met
property_id: cogenerator
type: category
proof:
We claim that $\IR$ with the usual metric is a cogenerator. Let $a,b \in X$ be two points of a metric space such that $f(a)=f(b)$ for all non-expansive maps $f : X \to \IR$. This applies in particular to $f(x) \coloneqq d(a,x)$ and shows that $0=d(a,a)=d(a,b)$, so that $a=b$.


structure_id: Met
property_id: binary products
type: category
proof:
The product of two metric spaces $(X,d)$, $(Y,d)$ is $(X \times Y,d)$ with $d((x_1,y_1),(x_2,x_2)) \coloneqq \sup(d(x_1,x_2),d(y_1,y_2))$.


structure_id: Met
property_id: coequalizers
type: category
proof:
This is because the category of pseudo-metric spaces <a href="/category/PMet">$\PMet$</a> has coequalizers and $\Met \hookrightarrow \PMet$ has a left adjoint, mapping a pseudo-metric space $X$ to $X /{\sim}$ where $x \sim y \iff d(x,y)=0$. Concretely, we take the coequalizer in the category of pseudo-metric spaces and then identify points with distance zero.


structure_id: Met
property_id: filtered colimits
type: category
proof:
This is because the category of pseudo-metric spaces <a href="/category/PMet">$\PMet$</a> has filtered colimits and $\Met \hookrightarrow \PMet$ has a left adjoint, mapping a pseudo-metric space $X$ to $X /{\sim}$ where $x \sim y \iff d(x,y)=0$. Concretely, we take the filtered colimit in the category of pseudo-metric spaces and then identify points with distance zero.


structure_id: Met
property_id: cartesian filtered colimits
type: category
proof:
We already saw that filtered colimits and finite products exist. The canonical map $\colim_i (X \times Y_i) \to X \times \colim_i Y_i$ is an isomorphism for filtered diagrams $(Y_i)$: It is surjective by the concrete description of filtered colimits. It is isometric because of the elementary observation $$\textstyle\inf_i \max(r, s_i) = \max(r, \inf_i s_i)$$ for $r, s_i \in \IR$, where $i \leq j \implies s_i \geq s_j$.


structure_id: Met
property_id: ℵ₁-cofiltered limits
type: category
proof:
Let $D : \I \to \Met$ be an $\aleph_1$-cofiltered diagram. This diagram has a limit in the complete category <a href="/category/Met_oo">$\Met_\infty$</a>, obtained by equipping the limit of the underlying sets with the supremum metric $d(x,y) = \sup_i d(x_i,y_i)$. It remains to show that $d(x,y) < \infty$. Suppose otherwise. Then for every $n \in \IN$ there exists an object $i_n \in \I$ such that $d(x_{i_n},y_{i_n}) \geq n$. Since $\I$ is $\aleph_1$-cofiltered, there exists a cone $(j \to i_n)_{n \in \IN}$. Since the maps $D(j) \to D(i_n)$ are non-expansive, we have $$d(x_j,y_j) \geq d(x_{i_n},y_{i_n}) \geq n$$ for every $n \in \IN$. This is impossible since the metric on $D(j)$ takes only finite values.


structure_id: Met
property_id: well-copowered
type: category
proof:
If $f : X \to Y$ is an epimorphism, then $f(X)$ is dense in $Y$ (see below). Hence, there is an injective map $Y \to X^{\IN}$, which bounds the size of $Y$.


structure_id: Met
property_id: balanced
type: category
proof:
The inclusion $\IQ \hookrightarrow \IR$ is a counterexample; it is an epimorphism since $\IQ$ is dense in $\IR$. Alternatively, consider the identity map $(X,2d) \to (X,d)$ for any non-trivial metric space $(X,d)$.


structure_id: Met
property_id: countable powers
type: category
proof:
Assume that the power $P = \IR^{\IN}$ exists, where $\IR$ has the usual metric. Since the forgetful functor $\Met \to \Set$ is representable, it preserves limits, powers in particular. Thus, we may assume that $P$ is the set of sequences of numbers and that the projections $p_n : P \to \IR$ are given by $p_n(x) = x_n$. Now consider the sequences $x = (n)_n$ and $y = (0)_n$. Since each $p_n$ is non-expansive, we get $d(x,y) \geq d(p_n(x),p_n(y)) = d(n,0) = n$. But then $d(x,y) = \infty$, a contradiction.


structure_id: Met
property_id: binary copowers
type: category
proof:
The coproduct of two non-empty metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a>. For example, the copower $\IR \sqcup \IR$ does not exist. We only get coproducts when allowing $\infty$ as a distance, as in <a href="/category/Met_oo">$\Met_{\infty}$</a>.


structure_id: Met
property_id: cartesian closed
type: category
proof:
This is proven in <a href="https://math.stackexchange.com/questions/5131457" target="_blank">MSE/5131457</a>.


structure_id: Met
property_id: filtered-colimit-stable monomorphisms
type: category
proof:
The following example is taken from Remark 2.7 in <a href="https://arxiv.org/abs/2006.01399" target="_blank">Approximate injectivity and smallness in metric-enriched categories</a> by Adamek-Rosicky: For $n \geq 1$ let $X_n$ denote the metric space with underlying set $\{0,1\}$ in which $0,1$ have distance $1/n$. We have bijective non-expansive maps $X_n \to X_{n+1}$, $x \mapsto x$. The colimit of this sequence in <a href="/category/PMet">$\PMet$</a> is $\{0,1\}$ where $0,1$ have distance $0$, so the colimit in $\Met$ collapses to $\{0\}$. Therefore, the colimit of the monomorphisms $X_0 \to X_n$, $x \mapsto x$ is the non-injective map $X_0 \to \{0\}$.


structure_id: Met
property_id: natural numbers object
type: category
proof:
If $(N,z,s)$ is a natural numbers object in $\Met$, then $$1 \xrightarrow{z} N \xleftarrow{s} N$$ is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. Since there is a map $1 \to N$, we have $N \neq \varnothing$. However, the coproduct of two non-empty metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a>.


structure_id: Met
property_id: effective congruences
type: category
proof:
Any kernel pair of $h : X \to Z$ in $\Met$ corresponds to a closed subset of $X\times X$. However, there are plenty of non-closed congruences, such as $\Delta \cup (\IQ \times \IQ) \subseteq \IR \times \IR$ with the subspace metric.


structure_id: Met
property_id: effective cocongruences
type: category
proof:
We will define a cocongruence on the interval $(0,1) \subseteq \IR$ where $E \coloneqq (-1, 0) \cup (0, 1) \subseteq \IR$, and the two maps $(0, 1) \rightrightarrows E$ are the inclusion map and $x \mapsto -x$. Then for any metric space $X$, the induced relation on non-expansive maps $(0, 1) \to X$ is that $f \sim g$ if and only if $$d(f(x), g(y)) \le x+y$$ for each $x, y \in (0, 1)$. This is reflexive since $d(f(x), f(y)) \le |x-y| < x+y$, and it is clearly symmetric. For transitivity, suppose $f\sim g$ and $g\sim h$. Then for any $\varepsilon > 0$, we have $$d(f(x), h(y)) \le d(f(x), g(\varepsilon)) + d(g(\varepsilon), h(y)) \le (x + \varepsilon) + (y + \varepsilon).$$ Since this holds for every $\varepsilon > 0$, we conclude $d(f(x), h(y)) \le x+y$. On the other hand, if this cocongruence were effective, then by the dual of <a href="/content/effective-congruence-quotients">this result</a>, it would be the cokernel pair of the equalizer of the two inclusion maps. However, that equalizer is empty, so $E$ would have to be a binary copower of $(0,1)$, which does not exist in $\Met$.


structure_id: Met_c
property_id: countable products
type: category
proof:
For finite products, we take the cartesian product with, say, the sup-metric. The product of countably many metric spaces $(X_i,d_i)_{i \geq 0}$ is given by the cartesian product $\prod_{i \geq 0} X_i$ with the metric $d(x,y) \coloneqq \sum_{i \geq 0} d_i(x_i,y_i)/(1 + d_i(x_i,y_i))$. See Engelking's book <i>General Topology</i>.


structure_id: Met_c
property_id: generator
type: category
proof:
The one-point metric space is a generator since it represents the forgetful functor $\Met_c \to \Set$.


structure_id: Met_c
property_id: cogenerator
type: category
proof:
The same proof as for <a href="/category/Met">$\Met$</a> shows that $\IR$ with the usual metric is a cogenerator.


structure_id: Met_c
property_id: infinitary extensive
type: category
proof:
This follows from the existence of coproducts and finite products, and from the fact that <a href="/category/Top">$\Top$</a> is infinitary extensive.


structure_id: Met_c
property_id: effective cocongruences
type: category
proof:
Suppose we have a cocongruence $f, g : X \rightrightarrows E$ in $\Met_c$. Then the image in $\Haus$ is a coreflexive corelation (since epimorphisms in both categories are continuous maps with dense image). By <a href="https://mathoverflow.net/a/509582/2841" target="_blank">MO/509548</a>, that implies that image is of the form $X +_S X$ for a closed subset $S$ of $X$. Since $S$ is metrizable, and the functor $\Met_c \to \Haus$ is fully faithful <a href="https://ncatlab.org/nlab/show/reflected+limit#FullSubcategoryInclusionReflectCoLimits" target="_blank">and therefore reflects colimits</a>, we conclude that $E$ is effective in $\Met_c$.


structure_id: Met_c
property_id: ℵ₂-small powers
type: category
proof:
See <a href="https://math.stackexchange.com/questions/139168" target="_blank">MSE/139168</a> for a proof that uncountable powers do not exist.


structure_id: Met_c
property_id: regular subobject classifier
type: category
proof:
We recycle the proof from <a href="/category/Haus">$\Haus$</a>: Assume that there is a regular subobject classifier $\Omega$. By the classification of regular monomorphisms, we would have an isomorphism between $\Hom(X,\Omega)$ and the set of closed subsets of $X$ for any metric space $X$. If we take $X = 1$ we see that $\Omega$ has two points. Since $\Omega$ is Hausdorff, $\Omega \cong 1 + 1$ must be discrete. But then $\Hom(X,\Omega)$ is isomorphic to the set of all clopen subsets of $X$, of which there are usually far fewer than closed subsets (consider $X = [0,1]$).


structure_id: Met_c
property_id: sequential colimits
type: category
proof:
See <a href="https://mathoverflow.net/questions/510316" target="_blank">MO/510316</a> for a proof that the diagram $\IN \to \Met_c$, $n \mapsto \IR^n$ does not have a colimit.


structure_id: Met_c
property_id: ℵ₁-filtered colimits
type: category
proof:
See <a href="https://mathoverflow.net/questions/511433" target="_blank">MO/511433</a> for a proof that the diagram $\omega_1 \to \Met_c$, $\alpha \mapsto \IR^\alpha$ does not have a colimit.


structure_id: Met_c
property_id: quotients of congruences
type: category
proof:
If $\Met_c$ had quotients of congruences, then by <a href="/content/monic-sequential-colimits-via-congruence-quotients">this lemma</a> it would have sequential colimits of sequences of monomorphisms. This contradicts <a href="https://mathoverflow.net/questions/510316" target="_blank">MO/510316</a>.


structure_id: Met_oo
property_id: generator
type: category
proof:
The singleton metric space $1$ is a generator, since morphisms $1 \to X$ correspond to the elements of $X$.


structure_id: Met_oo
property_id: cogenerator
type: category
proof:
The proof is similar to $\Met$, a cogenerator is given by $\IR \cup \{\infty\}$ with the metric in which $d(a,\infty)=\infty$ for $a \in \IR$. Then one checks that the maps $d(a,-) : X \to \IR \cup \{\infty\}$ are non-expansive and finishes as for <a href="/category/Met">$\Met$</a>.


structure_id: Met_oo
property_id: locally ℵ₁-presentable
type: category
proof:
This is Example 4.5 in <a href="https://arxiv.org/abs/1504.02660" target="_blank">Metric abstract elementary classes as accessible categories</a>.


structure_id: Met_oo
property_id: cartesian filtered colimits
type: category
proof:
We can use the same proof as for <a href="/category/Met">$\Met$</a> since the equation $\inf_i \max(r, s_i) = \max(r, \inf_i s_i)$ also holds for for $r, s_i \in \IR \cup \{\infty\}$.


structure_id: Met_oo
property_id: infinitary extensive
type: category
proof:
[Sketch] Since <a href="/category/Set">$\Set$</a> is infinitary extensive, a map $f : Y \to \coprod_i X_i \eqqcolon X$ corresponds to a decomposition $Y = \coprod_i Y_i$ (as sets) with maps $f_i : Y_i \to X_i$. Endow $Y_i$ with the restricted metric. If $f$ is non-expansive, each $f_i$ is non-expansive, and for $x_i \in Y_i$ and $i \neq j$ we have $d_Y(x_i,x_j) \geq d_X(f(x_i),f(x_j)) = \infty$, so that $Y = \coprod_i Y_i$ holds as metric spaces.


structure_id: Met_oo
property_id: balanced
type: category
proof:
The inclusion $\IQ \hookrightarrow \IR$ provides a counterexample. Alternatively, consider the identity map $(X,2d) \to (X,d)$ for any non-trivial metric space $(X,d)$.


structure_id: Met_oo
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \Met_{\infty} \to \Set$ and the relation $R \subseteq U^2$ defined by $R(X) \coloneqq \{(a,b) \in U(X)^2 : d(x,y) \leq 1 \}$. Both are representable: $U$ by the singleton metric space and $R$ by the metric space $\{ 0,1 \}$ where $d(0,1) \coloneqq 1$. It is clear that $R$ is reflexive, but not transitive.


structure_id: Met_oo
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Met_{\infty}$ that equips a set with the discrete topology.


structure_id: Met_oo
property_id: effective cocongruences
type: category
proof:
The same counterexample as for <a href="/category/Met">$\Met$</a> works here. The difference in this case is that a binary copower of two copies of $(0,1)$ does exist in $\Met_\infty$. However, this would assign a distance of $\infty$ between points in $(-1,0)$ and points in $(0,1)$, which does not agree with the chosen subspace metric on $(-1,0) \cup (0,1)$.


structure_id: Mon
property_id: unital
type: category
proof:
If a submonoid of $X \times Y$ contains $X \times \{1\}$ and $\{1\} \times Y$, then for all $x \in X$ and $y \in Y$ it also contains $(x,1) \cdot (1,y) = (x,y)$.


structure_id: Mon
property_id: cogenerator
type: category
proof:
We apply <a href="/content/missing_cogenerator">this lemma</a> to the collection of simple groups: Any non-trivial homomorphism $G \to M$ from a simple group $G$ to a monoid $M$ must be injective (as it corestricts to a homomorphism of groups $G \to M^{\times}$), and for every infinite cardinal $\kappa$ there is a simple group of size $\geq \kappa$ (for example, the alternating group on $\kappa$ elements).


structure_id: Mon
property_id: counital
type: category
proof:
The canonical morphism $\IN \sqcup \IN \to \IN \times \IN$ is not a monomorphism since $\IN \sqcup \IN$ is not commutative.


structure_id: Mon
property_id: CSP
type: category
proof:
If $M \to N$ is an epimorphism in $\Mon$ and $M$ is infinite, then $\card(N) \leq \card(M)$ (see <a href="https://mathoverflow.net/questions/510431/" target="_blank">MO/510431</a>). This implies that in $\Mon$ the canonical homomorphism $\coprod_{n \geq 0} \IN \to \prod_{n \geq 0} \IN$ is not an epimorphism because its domain is countable and its codomain is uncountable.


structure_id: Mon
property_id: coregular
type: category
proof:
Consider the monoid $M \coloneqq \langle x_0, x_1, s : x_0 s = x_1 s = 1 \rangle$. Notice that every element in $M$ has a unique expression as $s^k \cdot u$ with $k \in \IN$ and $u \in \langle x_0,x_1 \rangle_M$. Moreover, the canonical homomorphism $\iota : \langle x_0, x_1 \rangle \to M$ (from the free monoid) is injective. We will prove that it is a regular monomorphism, which however is not pushout-stable. Consider $N \coloneqq \langle x_0, x_1, s_0, s_1 : x_i s_j = 1 \rangle$ and define $f_i : M \to N$ for $i=0,1$ by $f_i(x_j) = x_j$ and $f_i(s) = s_i$. Then $\iota$ is the equalizer of $f_0,f_1$. Now consider $g : \langle x_0,x_1 \rangle \to \langle y_0 \rangle$ defined by $g(x_0) = y_0$, $g(x_1) = 1$. The pushout of $\iota$ with $g$ is given by $\langle x_0, x_1, s, y_0 : x_0 s = x_1 s = 1 , \, x_0 = y_0, \, x_1 = 1 \rangle$, which simplifies to $\langle x_0, s : x_0 s = s = 1 \rangle$, which is trivial.


structure_id: Mon
property_id: regular subobject classifier
type: category
proof:
Assume that $\Omega$ is a regular subobject classifier. Since the trivial monoid is a zero object, every regular submonoid $U \subseteq M$ of any monoid $M$ would have the form $\{m \in M : h(m) = 1 \}$ for some homomorphism $M \to \Omega$. Now take any monoid $M$ with zero that has two different homomorphisms with zero $f,g : M \rightrightarrows N$ (for example, let $M = N = \{0\} \cup \{x^n : n \geq 0\}$ be the free monoid with zero on one generator, $f(x) = 0$,and $g(x) = x$). Take their equalizer $U \subseteq M$, and choose a homomorphism $h : M \to \Omega$ with $U = \{m \in M : h(m) = 1\}$. Since $0 \in U$, we have $h(0)=1$. But then for all $m \in M$ we have $h(m) = h(m) h(0) = h(m 0) = h(0) = 1$, i.e. $U = M$, which yields the contradiction $f = g$.


structure_id: Mon
property_id: regular quotient object classifier
type: category
proof:
We can just copy the proof for <a href="/category/CMon">$\CMon$</a>. Alternatively, we may use Lemma 1 <a href="/content/subcategories">here</a> (dualized) applied to the forgetful functor $\CMon \to \Mon$.


structure_id: Mon
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Grp">$\Grp$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the forgetful functor $\Grp \to \Mon$. It preserves epimorphisms since it has a right adjoint, the unit group functor.


structure_id: Mon
property_id: cocartesian cofiltered limits
type: category
proof:
We know that <a href="/category/Grp">$\Grp$</a> fails to satisfy this property. The same counterexample works here since the inclusion $\Grp \hookrightarrow \Mon$ preserves limits and colimits (it has a left and a right adjoint) and is conservative. A similar counterexample is given by the free monoids $N_n = \langle x_1,\dotsc,x_n \rangle$ and the Boolean monoid $M = \langle e : e^2=e \rangle$ with the maps $N_{n+1} \to N_n$, $x_{n+1} \mapsto 1$. Then the element $(x_1 e \cdots x_n e) \in \lim_n (M \sqcup N_n)$ does not come from $M \sqcup \lim_n N_n$ because its components have unbounded free product length.


structure_id: Mon
property_id: effective cocongruences
type: category
proof:
We adapt the counterexample from <a href="https://mathoverflow.net/a/510809" target="_blank">MO/510744</a> for $\Ring$. Namely, consider the monoids $$\begin{align*} X & \coloneqq \langle p \mid p^2 = p \rangle \cong (\{ 0, 1 \}, \cdot),\\ E & \coloneqq \langle p, q \mid p^2 = p,\, q^2 = q,\, pq = q,\, qp = p \rangle. \end{align*}$$ Then $X$ represents the functor sending a monoid $M$ to its idempotents, and $E$ represents the relation on idempotents $a, b$ of $M$ that $ab = b$, $ba = a$. The equations are equivalent to $aM = bM$, showing that the relation is indeed an equivalence relation. On the other hand, using the multiplicative map $$E \to M_{2\times 2}(\IZ), \quad p \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\quad q \mapsto \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix},$$ we can see that $p \ne q$ in $E$, so the equalizer of the two maps $X \rightrightarrows E$ is the trivial submonoid $\{ 1 \}$. Therefore, if $E$ were effective, it would be isomorphic to the coproduct $X \sqcup X$, whose underlying set consists of words in $p,q$ with $p,q$ strictly alternating. In particular, in this coproduct, $pq \ne q$.


structure_id: Mono
property_id: semi-strongly connected
type: category
proof:
This is immediate from the fact that $\Mono$ is equivalent to a full subcategory of <a href="/category/Set_arrow">$\Set^{\rightarrow}$</a>, and the latter is semi-strongly connected.


structure_id: Mono
property_id: generator
type: category
proof:
The object $(1, 0)$ is a generator. This is because it represents the functor taking an object $(X, X')$ to $X$, and taking a morphism $f : (X, X') \to (Y, Y')$ to the function $f : X \to Y$.


structure_id: Mono
property_id: cogenerator
type: category
proof:
Consider the forgetful functor $U : \Mono \to \Set$, $(X, X') \mapsto X$. This has right adjoint $R : \Set \to \Mono$, $X \mapsto (X, X)$. Therefore, by the dual of Lemma 9 <a href="/content/subcategories">here</a>, $R$ preserves cogenerators; and in particular, since $\Set$ has a cogenerator, so does $\Mono$.


structure_id: Mono
property_id: complete
type: category
proof:
The component-wise limit gives a pair of a set and a subset, and this pair forms the limit in $\Mono$.
More concretely, suppose we have a diagram $((X_i, X_i'))_{i \in \I}$, with limit cone $p_i : \lim_{i\in\I} X_i \to X_i$ in <a href="/category/Set">$\Set$</a>. Then the limit in $\Mono$ is $(\lim_{i\in\I} X_i, \bigcap_{i\in\I} p_i^{-1}(X_i'))$. (It is easy to see that in the subset component, we can also restrict $i$ to range over a <a href="https://ncatlab.org/nlab/show/weak+multilimit" target="_blank">weakly initial</a> set of objects of $\I$.)


structure_id: Mono
property_id: cocomplete
type: category
proof:
We have that $\Mono$ is a reflective subcategory of $\Set^{\rightarrow}$, with the reflector taking a function $f : X \to Y$ to the pair $(Y, \im(f))$. Therefore, the result follows from the fact that <a href="/category/Set_arrow">$\Set^{\rightarrow}$</a> is cocomplete.
More concretely, suppose we have a diagram $((X_i, X_i'))_{i \in \I}$, with colimit cone $c_i : X_i \to \colim_{i\in\I} X_i$ in <a href="/category/Set">$\Set$</a>. Then the colimit in $\Mono$ is $(\colim_{i\in\I} X_i, \bigcup_{i\in\I} c_i(X_i'))$. (It is easy to see that in the subset component, we can also restrict $i$ to range over a <a href="https://ncatlab.org/nlab/show/weak+multilimit" target="_blank">weakly terminal</a> set of objects of $\I$.)


structure_id: Mono
property_id: disjoint coproducts
type: category
proof:
This follows from the fact that coproducts in $\Mono$ can be computed component-wise, along with the fact that <a href="/category/Set">$\Set$</a> has disjoint coproducts.


structure_id: Mono
property_id: locally cartesian closed
type: category
proof:
For any object $(S, S')$ of $\Mono$, we can view objects of the slice $\Mono / (S, S')$ as indexed families $(\bigsqcup_{s\in S} X_s, \bigsqcup_{s\in S'} X'_s)$ where $X'_s \subseteq X_s$ for each $s\in S'$. Given any two such objects $X = (\bigsqcup_{s\in S} X_s, \bigsqcup_{s\in S'} X'_s)$ and $Y = (\bigsqcup_{s\in S} Y_s, \bigsqcup_{s\in S'} Y'_s)$, observe that $\Hom_{\Mono / (S, S')}(X, Y)$ is equivalent to the tuples of functions $f \in \prod_{s\in S} \Hom_{\Set}(X_s, Y_s)$ such that whenever $s \in S'$, then $f_s(X_s ') \subseteq Y_s '$. We claim that $\Mono$ has a relative exponential given by $$[X, Y]_S \coloneqq \left(\bigsqcup_{s\in S} [X_s, Y_s], \bigsqcup_{s\in S'} [X_s, Y_s]' \right)$$ where $$[X_s, Y_s]' \coloneqq \{ f \in [X_s, Y_s] : f(X_s ') \subseteq Y_s ' \}.$$ To check that this works, suppose we have a test object $U = (\bigsqcup_{s\in S} U_s, \bigsqcup_{s\in S'} U_s ')$. Then $U \times_S X = (\bigsqcup_{s\in S} (U_s \times X_s), \bigsqcup_{s\in S'} (U_s ' \times X_s '))$. Given a morphism $g : U \times_S X \to Y$ in $\Mono / (S, S')$, we define the corresponding morphism $h : U \to [X, Y]_S$ to have $s$-component given by $u \mapsto (x \mapsto g_s(u, x))$. To check this is a valid morphism in $\Mono / (S, S')$, suppose that $s \in S'$ and $u \in U_s '$; then for every $x \in X_s '$, we have $(u, x) \in U_s ' \times X_s '$, so $g_s(u, x) \in Y_s '$. This shows that $h_s(u) \in [X_s, Y_s]'$. Conversely, given a morphism $h : U \to [X, Y]_S$ in $\Mono / (S, S')$, we define the corresponding morphism $g : U \times_S X \to Y$ to have $s$-component given by $(u, x) \mapsto h_s(u)(x)$. To check this is a valid morphism in $\Mono / (S, S')$, suppose that $s \in S'$ and $(u, x) \in U_s ' \times X_s '$; then $u \in U_s '$ so $h_s(u) \in [X_s, Y_s]'$. Furthermore, $x \in X_s '$, so $h_s(u)(x) \in Y_s '$.
It is easy to check these give inverse operations which are natural in $U$ (in fact, the details are exactly the same as the details of an explicit proof that $\Set$ is locally cartesian closed), completing the proof.


structure_id: Mono
property_id: regular subobject classifier
type: category
proof:
The object $(\{ \top, \bot \}, \{ \top, \bot \})$ is a regular subobject classifier, with $\top : (1, 1) \to (\{ \top, \bot \}, \{ \top, \bot \})$ consisting of the function with image $\top$. This follows from the description of regular monomorphisms below.


structure_id: Mono
property_id: locally finitely presentable
type: category
proof:
It is equivalent to the category of models of the finite limit sketch with a single limit cone $$\begin{CD} X @> \id >> X \\ @V \id VV @VV f VV \\ X @>> f > Y. \end{CD}$$ Alternately, we already know $\Mono$ is cocomplete. We claim that both $(1, 0)$ and $(1, 1)$ are finitely presentable objects of $\Mono$. For the first, $(1, 0)$ represents the forgetful functor $(X, X') \mapsto X$; and this functor has a right adjoint $X \mapsto (X, X)$ so it preserves filtered colimits (in fact all colimits). For the second, $(1, 1)$ represents the functor $(X, X') \mapsto X'$. Now by the previous description of colimits, for a filtered diagram $((X_i, X_i'))_{i\in\I}$, first form the filtered colimit of $X_i$ in <a href="/category/Set">$\Set$</a> with cocone $c_i : X_i \to \colim_{i\in\I} X_i$. Then the filtered colimit in $\Mono$ is $(\colim_{i\in\I} X_i, \bigcup_{i\in\I} c_i(X_i'))$. This makes it easy to see that the canonical function $$\colim_{i\in\I} \Hom((1,1), (X_i,X_i')) \to \Hom((1,1), \colim_{i\in\I} (X_i,X_i'))$$ is surjective. To see it is injective, use the fact that the unique morphism $(1, 0) \to (1, 1)$ is an epimorphism (see below), and we have already seen that the corresponding function for $(1,0)$ is injective.
Therefore, any finite coproduct of $(1, 0)$ and $(1, 1)$ is finitely presentable; this means any object of the form $([n], [m])$ with $m, n \in \IN$ and $m \le n$ is finitely presentable. All such objects form a small set. Moreover, any object $(X,X')$ of $\Mono$ is a filtered colimit of the subobjects $(F, F \cap X')$ where $F$ is a finite subset of $X$, and each $(F, F \cap X')$ is isomorphic to some $([n], [m])$.


structure_id: Mono
property_id: co-Malcev
type: category
proof:
Suppose we have a coreflexive corelation $p : (X \sqcup X, X' \sqcup X') \twoheadrightarrow (E, E')$ with coreflexivity morphism $r : (E, E') \to (X, X')$. From the assumption that $p$ is an epimorphism, we have that $p : X \sqcup X \to E$ is a surjective function. Since <a href="/category/Set">$\Set$</a> is co-Malcev, it follows that $E \cong X \sqcup_Y X$ for some subset $Y \subseteq X$. It remains to show that $E' = i_1(X') \cup i_2(X') \subseteq X \sqcup_Y X$. Certainly, since we have a morphism $(X \sqcup_Y X, i_1(X') \cup i_2(X')) \to (E, E')$ induced by $p$, we must have $i_1(X') \cup i_2(X') \subseteq E'$. On the other hand, any element of $E'$ is an element of $E$ and hence is equal to either $i_1(x)$ or $i_2(x)$ for $x \in X$. In the first case, we must have $x = r(i_1(x)) \in X'$, so $i_1(x) \in i_1(X')$; and similarly for the second case.


structure_id: Mono
property_id: effective cocongruences
type: category
proof:
See the proof that $\Mono$ is co-Malcev: It shows that in fact any coreflexive corelation is equivalent to an effective cocongruence $X \sqcup X \twoheadrightarrow X \sqcup_Y X$.


structure_id: Mono
property_id: balanced
type: category
proof:
The unique morphism from $(1, 0)$ to $(1, 1)$ is both a monomorphism and an epimorphism, but not an isomorphism (see descriptions below).


structure_id: Mono
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Mono$ which sends a set $X$ to the pair $(X, X)$.


structure_id: N_oo
property_id: locally strongly finitely presentable
type: category
proof:
We already saw that coproducts, and therefore colimits exist. Every natural number is strongly finitely presentable, and $\infty$ is the colimit of all $n < \infty$.


structure_id: On
property_id: binary products
type: category
proof:
For ordinal numbers $\alpha,\beta$ their minimum exists, it is $\alpha$ when $\alpha \leq \beta$ and $\beta$ otherwise.


structure_id: On
property_id: well-powered
type: category
proof:
This is because for every ordinal $\alpha$ the collection of ordinals $\leq \alpha$ is a set (namely, $\alpha + 1$).


structure_id: On
property_id: semi-strongly connected
type: category
proof:
It is well-known that for ordinals $\alpha,\beta$ we have $\alpha \leq \beta$ or $\beta \leq \alpha$.


structure_id: PMet
property_id: generator
type: category
proof:
The one-point (pseudo-)metric space is a generator since it represents the forgetful functor $\PMet \to \Set$.


structure_id: PMet
property_id: cogenerator
type: category
proof:
The set $\{0,1\}$ equipped with the pseudo-metric $d(0,1)=0$ is a cogenerator since every map into is automatically non-expansive and since $\{0,1\}$ is a cogenerator in $\Set$.


structure_id: PMet
property_id: binary products
type: category
proof:
The product of two pseudo-metric spaces $(X,d)$, $(Y,d)$ is $(X \times Y,d)$ with $d((x_1,y_1),(x_2,x_2)) \coloneqq \sup(d(x_1,x_2),d(y_1,y_2))$.


structure_id: PMet
property_id: filtered colimits
type: category
proof:
Given a filtered diagram $(X_i)$ of pseudo-metric spaces, take the filtered colimit $X$ of the underlying sets with the following pseudo-metric: If $x,y \in X$, let $d(x,y)$ be infimum of all $d(x_i,y_i)$, where $x_i,y_i \in X_i$ are some preimages of $x,y$ in some $X_i$. The definition ensures that each $X_i \to X$ is non-expansive, and the universal property is easy to check.


structure_id: PMet
property_id: exact filtered colimits
type: category
proof:
We already saw that finite limits and filtered colimits exist. Now let $\I$ be a finite category and $\J$ be a small filtered category, w.l.o.g. a directed poset. Let $X : \I \times \J \to \PMet$ be a diagram. We need to show that the canonical map $\colim_{j \in \J} \lim_{i \in \I} X(i,j) \to \lim_{i \in \I} \colim_{j \in \J}  X(i,j)$ is an isomorphism. It is bijective since the forgetful functor to $\Set$ preserves finite limits and filtered colimits and since <a href="/category/Set">$\Set$</a> has exact filtered colimits. That the map is isometric can easily be reduced to the following lemma: If $d_{i,j} \in \IR_{\geq 0}$ are numbers for $i \in \I$, $j \in \J$ with $j \leq k \implies d_{i,k} \leq d_{i,j}$, then $\inf_j \sup_i d_{i,j} = \sup_i \inf_j d_{i,j}$. This can be proven directly. Alternatively, use that the thin category $(\IR_{\geq 0} \cup \{\infty\},\leq)$ is isomorphic to $([0,1],\leq)$, and we already know that <a href="/category/real_interval">it has exact filtered colimits</a>.


structure_id: PMet
property_id: balanced
type: category
proof:
Let $d : \IR \times \IR \to \IR_{\geq 0}$ be the usual Euclidean metric on $\IR$ and $0 : \IR \times \IR \to \IR_{\geq 0}$ be the zero pseudo-metric. Then the identity map $(\IR,d) \to (\IR,0)$ provides a counterexample.


structure_id: PMet
property_id: Malcev
type: category
proof:
Take any counterexample in <a href="/category/Set">$\Set$</a> and equip it with the zero pseudo-metric.


structure_id: PMet
property_id: countable powers
type: category
proof:
Assume that the power $P = \IR^{\IN}$ exists, where $\IR$ has the usual (pseudo-)metric. Since the forgetful functor $\PMet \to \Set$ is representable, it preserves limits, powers in particular. Thus, we may assume that $P$ is the set of sequences of numbers and that the projections $p_n : P \to \IR$ are given by $p_n(x) = x_n$. Now consider the sequences $x = (n)_n$ and $y = (0)_n$. Since each $p_n$ is non-expansive, we get $d(x,y) \geq d(p_n(x),p_n(y)) = d(n,0) = n$. But then $d(x,y) = \infty$, a contradiction.


structure_id: PMet
property_id: binary copowers
type: category
proof:
The coproduct of two non-empty pseudo-metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a> (the proof also works for pseudo-metric spaces). For example, the copower $\IR \sqcup \IR$ does not exist. We only get coproducts when allowing $\infty$ as a distance.


structure_id: PMet
property_id: natural numbers object
type: category
proof:
If $(N,z,s)$ is a natural numbers object in $\PMet$, then $$1 \xrightarrow{z} N \xleftarrow{s} N$$ is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. Since there is a map $1 \to N$, we have $N \neq \varnothing$. However, the coproduct of two non-empty pseudo-metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a>.


structure_id: PMet
property_id: effective cocongruences
type: category
proof:
The proof is similar to the one for <a href="/category/Top">$\Top$</a>: Equip a two-point set with the zero metric; this pseudo-metric space represents the functor taking a pseudo-metric space to the pairs of points with $d(x,y) = 0$. In this case, once you conclude $Z = \varnothing$, the map $h : Z \to 1$ does not have any cokernel pair, since that would have to be a coproduct $1+1$, which does not exist.


structure_id: PMet
property_id: regular
type: category
proof:
We can adapt Example 3.14 at the <a href="https://ncatlab.org/nlab/show/regular+category" target="_blank">nLab</a> (which disproves regularity for $\Pos$ and related categories) as follows: Consider the subspaces $X = \{0,1,2,3\}$ and $Y = \{0,1,2\}$ of $\IR$ with the usual metric. Define a surjective map $p : X \to Y$ by $p(0)=0$, $p(1)=p(2)=1$, and $p(3)=2$. Clearly, $p$ is non-expansive. Moreover, one can check that $p$ satisfies the universal property in $\PMet$ of a coequalizer of the two maps $1,2 : \{\ast\} \rightrightarrows X$. Thus, $p$ is a regular epimorphism. Now consider the subspace $Z = \{0,2\}$ of $Y$. As a set, the pullback $X \times_Y Z$ is $p^*(Z) = \{0,3\}$. Using the definition of the product metric, one can verify that $d(0,3) = 3$ in this pullback. The projection $X \times_Y Z \to Z$ identifies with the evident bijective and non-expansive map $\{0,3\} \to \{0,2\}$. It is a monomorphism and not an isomorphism (the distances do not match), hence cannot be a regular epimorphism.


structure_id: Pos
property_id: locally finitely presentable
type: category
proof:
See <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>, Example 1.10.


structure_id: Pos
property_id: cartesian closed
type: category
proof:
For posets $P,Q$ we endow $\Hom(P,Q)$ with the partial order in which $f \leq g$ holds iff $f(p) \leq g(p)$ for all $p \in P$. The universal evaluation map is $\Hom(P,Q) \times P \to Q$, $(f,p) \mapsto f(p)$, it is order-preserving, and it satisfies the universal property.


structure_id: Pos
property_id: generator
type: category
proof:
The singleton poset $1$ is a generator, since morphisms $1 \to P$ correspond to the elements of $P$.


structure_id: Pos
property_id: cogenerator
type: category
proof:
We prove that the poset $\{0 < 1\}$ is a cogenerator: Let $P$ be a poset and $a,b \in P$ be two elements such that $f(a) = f(b)$ for all order-preserving maps $f : P \to \{0 < 1 \}$. This means that $a$ and $b$ lie in the same upper sets. In particular, $b$ lies in the upper set generated by $a$, meaning $a \leq b$, and similarly we deduce $b \leq a$. Thus, $a = b$.


structure_id: Pos
property_id: infinitary extensive
type: category
proof:
[Sketch] Since <a href="/category/Set">$\Set$</a> is infinitary extensive, a map $f : P \to \coprod_i Q_i$ corresponds to a decomposition $P = \coprod_i P_i$ (as sets) with maps $f_i : P_i \to Q_i$. Endow $P_i$ with the induced order. If $f$ is order-preserving, the elements in different $P_i$ cannot be comparable (since their $f$-images are not comparable), so that $P = \coprod_i P_i$ as posets, and each $f_i$ is order-preserving.


structure_id: Pos
property_id: balanced
type: category
proof:
The inclusion $\{0,1\} \to \{0 < 1\}$ provides a counterexample (where in the domain there is no relation between $0$ and $1$).


structure_id: Pos
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \Pos \to \Set$ and the relation $R \subseteq U^2$ defined by $R(A) \coloneqq \{(a,b) \in U(A)^2 : a \leq b\}$. Both are representable: $U$ by the singleton poset and $R$ by $\{0 \leq 1 \}$. It is clear that $R$ is reflexive, but not symmetric.


structure_id: Pos
property_id: regular subobject classifier
type: category
proof:
Assume that there is a regular subobject classifier $\Omega$, so that (by the classification of regular monomorphisms) $\Hom(P,\Omega)$ is isomorphic to the set of subsets of $P$. By taking $P = \{0\}$ we see that $\Omega$ has $2$ elements, so that either $\Omega \cong \{0,1\}$ (discrete) or $\Omega \cong \{0 < 1\}$. By taking $P = \{0 < 1\}$ we see that $\Omega$ has four pairs $(x,y)$ with $x \leq y$. But $\{0,1\}$ has only two and $\{0 < 1\}$ has only three such pairs.


structure_id: Pos
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
Pick any poset $X$ which has a decreasing sequence of non-empty sets $X = X_0 \supseteq X_1 \supseteq \cdots$ with empty intersection, such as $X_n = \IN_{\geq n}$ with the natural ordering. The unique map $X_n \to 1$ is surjective, but their limit $\varnothing \to 1$ is not surjective.


structure_id: Pos
property_id: effective cocongruences
type: category
proof:
Let $X$ be $\IR$ with the standard (total) order, and let $E$ be the poset with underlying set $\IR \times \{ 0, 1 \}$ and partial order such that $(x, m) \le (y, n)$ if and only if $x < y$ or $(x, m) = (y, n)$. The two maps $\IR \rightrightarrows E$ will be $x \mapsto (x, 0)$ and $x \mapsto (x, 1)$ respectively. For any partial order $(\IP, \le)$, the induced equivalence relation on the set of order-preserving functions $\IR \to \IP$ is that $f \sim g$ if and only if $f(x) \le g(y)$ and $g(x) \le f(y)$ whenever $x < y$. This relation is clearly reflexive and symmetric; for transitivity, if $f \sim g$ and $g \sim h$, then whenever $x < y$, we have $f(x) \le g(\frac{x+y}{2}) \le h(y)$ and similarly $h(x) \le g(\frac{x+y}{2}) \le f(y)$, showing that $f \sim h$.
On the other hand, if this cocongruence on $\IR$ were effective, then by the dual of <a href="/content/effective-congruence-quotients">this result</a>, $E$ would be the cokernel pair of the equalizer of the two maps $\IR \rightrightarrows E$. However, that equalizer is the empty poset, so $E$ would have to be the coproduct poset $\IR + \IR$, giving a contradiction.


structure_id: Prost
property_id: locally finitely presentable
type: category
proof:
The same proof as for <a href="/category/Pos">$\Pos$</a> works, cf. <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>, Example 1.10.


structure_id: Prost
property_id: cartesian closed
type: category
proof:
For prosets $P,Q$ we endow $\Hom(P,Q)$ with the preorder in which $f \leq g$ holds iff $f(p) \leq g(p)$ for all $p \in P$. The universal evaluation map is $\Hom(P,Q) \times P \to Q$, $(f,p) \mapsto f(p)$, it is order-preserving, and it satisfies the universal property.


structure_id: Prost
property_id: generator
type: category
proof:
The singleton proset $1$ is a generator, since morphisms $1 \to P$ correspond to the elements of $P$.


structure_id: Prost
property_id: cogenerator
type: category
proof:
Endow the set $\{ 0,1 \}$ with the preorder $0 \leq 1$, $1 \leq 0$ (which is not a partial order). Then every map $P \to \{0,1\}$ is order-preserving. Now the claim follows since the set $\{ 0,1 \}$ is a cogenerator in $\Set$.


structure_id: Prost
property_id: infinitary extensive
type: category
proof:
[Sketch] Since <a href="/category/Set">$\Set$</a> is infinitary extensive, a map $f : P \to \coprod_i Q_i$ corresponds to a decomposition $P = \coprod_i P_i$ (as sets) with maps $f_i : P_i \to Q_i$. Endow $P_i$ with the induced order. If $f$ is order-preserving, the elements in different $P_i$ cannot be comparable (since their $f$-images are not comparable), so that $P = \coprod_i P_i$ as prosets, and each $f_i$ is order-preserving.


structure_id: Prost
property_id: regular subobject classifier
type: category
proof:
The set $\{0,1\}$ with the chaotic preorder $(0 \leq 1$, $1 \leq 0)$ is a regular subobject classifier since order-preserving maps $P \to \{0,1\}$ correspond to subsets of $P$.


structure_id: Prost
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \Prost \to \Set$ and the relation $R \subseteq U^2$ defined by $R(A) \coloneqq \{(a,b) \in U(A)^2 : a \leq b\}$. Both are representable: $U$ by the singleton preordered set and $R$ by $\{0 \leq 1 \}$. It is clear that $R$ is reflexive, but not symmetric.


structure_id: Prost
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We know that <a href="/category/Set">$\Set$</a> does not have this property. Now use the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> applied to the functor $\Set \to \Prost$ that equips a set with the chaotic preorder.


structure_id: Prost
property_id: effective cocongruences
type: category
proof:
Consider the proset $E \coloneqq \{ a, b \}$ with the chaotic preorder. This represents the functor which sends a proset to the pairs of elements $x,y$ with $x \le y$ and $y \le x$. Therefore, it defines a cocongruence $1 \rightrightarrows E$, where the maps are the two possible functions. However, this cannot be effective: for any map $h : Z \to 1$ which equalizes the two functions, $Z$ must be empty. But that means the cokernel pair of $h$ is the two-element proset with the trivial preorder.


structure_id: R-Mod
property_id: finitary algebraic
type: category
proof:
Take the algebraic theory of an $R$-module (given by the algebraic theory of an abelian group and for each $r \in R$ a unary operation).


structure_id: R-Mod
property_id: split abelian
type: category
proof:
By assumption, $R$ is not semisimple, so that there is a non-projective $R$-module, which yields a non-split sequence.


structure_id: R-Mod
property_id: CSP
type: category
proof:
The canonical homomorphism $\bigoplus_{n \geq 0} R \to \prod_{n \geq 0} R$ is not surjective, hence no epimorphism.


structure_id: R-Mod_div
property_id: split abelian
type: category
proof:
It is a standard fact that the category of $R$-modules is abelian for any ring $R$, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. VIII. If $R$ is a division ring, then by linear algebra every $R$-module has a basis, hence is projective, so that every short exact sequence splits.


structure_id: Rel
property_id: self-dual
type: category
proof:
There is an isomorphism $\Rel \to \Rel^{\op}$ that is the identity on objects (sets) and maps a relation $R \subseteq X \times Y$ to the opposite relation $R^{\op} \subseteq Y \times X$ defined by $R^{\op} \coloneqq \{(y,x) : (x,y) \in R \}$.


structure_id: Rel
property_id: coproducts
type: category
proof:
It is an easy exercise to deduce this from the corresponding fact for sets and that sets form a distributive category.


structure_id: Rel
property_id: biproducts
type: category
proof:
This is a consequence of the description of coproducts and products, both are disjoint unions (even for infinite families).


structure_id: Rel
property_id: well-powered
type: category
proof:
Any relation $R : A \to B$ induces an injective map $P(A) \to P(B)$, so that in particular there is an injective map $A \to P(B)$.


structure_id: Rel
property_id: kernels
type: category
proof:
Let $R : A \to B$ be a relation. Define $K = \bigl\{a \in A : \neg \exists b \in B ((a,b) \in R) \bigr\}$. We have an inclusion map $I : K \to A$, which can also be regarded as relation, and $R \circ I = \empty$ is the empty relation, i.e. the zero morphism. It is easy to check the universal property.


structure_id: Rel
property_id: normal
type: category
proof:
The construction of equalizers in $\Rel$ shows that they are injective functions, but <a href="https://math.stackexchange.com/questions/350716" target="_blank">MSE/350716</a> shows that monomorphisms in $\Rel$ don't have to be functions.


structure_id: Ring
property_id: disjoint finite products
type: category
proof:
To show that $A \sqcup_{A \times B} B$ is trivial, let $R$ be a ring which admits homomorphisms $f : A \to R$, $g : B \to R$ with $f(p_1(a,b))=g(p_2(a,b))$ for all $(a,b) \in A \times B$, i.e. $f(a)=g(b)$. Applying this to $a=0$, $b=1$ yields $1=0$ in $R$.


structure_id: Ring
property_id: semi-strongly connected
type: category
proof:
This is because already the full subcategory <a href="/category/CRing">$\CRing$</a> does not have this property.


structure_id: Ring
property_id: cogenerating set
type: category
proof:
We apply <a href="/content/missing_cogenerating_sets">this lemma</a> to the collection of fields: If $F$ is a field and $R$ is a non-trivial ring, any ring homomorphism $F \to R$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).


structure_id: Ring
property_id: codistributive
type: category
proof:
If $\sqcup$ denotes the coproduct of rings (see <a href="https://math.stackexchange.com/questions/625874" target="_blank">MSE/625874</a> for their description) and $R$ is a ring, the canonical morphism $R \sqcup \IZ^2 \to (R \sqcup \IZ)^2 = R^2$ is usually no isomorphism. For example, for $R = \IZ[X]$ the coproduct on the LHS is not commutative, it has the ring presentation $\langle X,E : E^2=E \rangle$.


structure_id: Ring
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \Ring \to \Set$ and the relation $R \subseteq U^2$ defined by $R(A) \coloneqq \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $\IZ[X]$ and $S$ by $\IZ \langle X,Y \rangle / \langle XY-X^2 \rangle$. It is clear that $R$ is reflexive, but not symmetric.


structure_id: Ring
property_id: coregular
type: category
proof:
Let $B \coloneqq M_2(\IQ)$ and $A \coloneqq \IQ^2$. Then $A \to B$, $(x,y) \mapsto \diag(x,y)$ is a regular monomorphism: A direct calculation shows that a matrix is diagonal iff it commutes with $M \coloneqq \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & 2 \end{smallmatrix}\bigr)$, so that $A \to B$ is the equalizer of the identity $B \to B$ and the conjugation $B \to B$, $X \mapsto M X M^{-1}$. Consider the homomorphism $A \to K$, $(a,b) \mapsto a$. We claim that $K \to K \sqcup_A B$ is not a monomorphism, because in fact, the pushout $K \sqcup_A B$ is zero: Since $A \to K$ is surjective with kernel $0 \times K$, the pushout is $B/\langle 0 \times K \rangle$, which is $0$ because $B$ is simple (<a href="https://math.stackexchange.com/questions/22629" target="_blank">proof</a>) or via a direct calculation with elementary matrices.


structure_id: Ring
property_id: regular quotient object classifier
type: category
proof:
We may copy the proof for <a href="/category/CRing">$\CRing$</a> (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\Ring$ would produce one in $\CRing$ by Lemma 1 <a href="/content/subcategories">here</a> (dualized).


structure_id: Ring
property_id: cocartesian cofiltered limits
type: category
proof:
Consider the ring $A = \IZ[X]$ and the sequence of rings $B_n = \IZ[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $\IZ[[Y]]$. Every element in the coproduct of rings $\IZ[X] \sqcup \IZ[[Y]]$ has a finite "free product" length. Now consider the elements $$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n.$$ Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.


structure_id: Ring
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We know that <a href="/category/CRing">$\CRing$</a> does not have this property. Now use the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> applied to the forgetful functor $\CRing \to \Ring$. It preserves epimorphisms by <a href="https://math.stackexchange.com/questions/5133488" target="_blank">MSE/5133488</a>.


structure_id: Rng
property_id: balanced
type: category
proof:
The inclusion $\IZ \hookrightarrow \IQ$ is a counterexample. (The proof can be reduced to the unital case.)


structure_id: Rng
property_id: cogenerator
type: category
proof:
We apply <a href="/content/missing_cogenerator">this lemma</a> to the collection of fields: Any non-zero rng homomorphism from a field to a rng must be injective, and for every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$.


structure_id: Rng
property_id: counital
type: category
proof:
If $\IZ\langle X_1,\dotsc,X_n \rangle_0$ denotes the free rng on $n$ generators (non-commutative polynomials without constant term), then the canonical homomorphism $$\IZ\langle X,Y \rangle_0 = \IZ\langle X \rangle_0 \sqcup \IZ\langle Y \rangle_0 \to \IZ\langle X \rangle_0 \times \IZ\langle Y \rangle_0$$ is not a monomorphism since $\IZ\langle X,Y \rangle_0$ is not commutative.


structure_id: Rng
property_id: CSP
type: category
proof:
Assume that $\coprod_n \IZ \to \prod_n \IZ$ is an epimorphism in $\Rng$. Then $((\coprod_n \IZ)^+)^{\ab} \to \prod_n \IZ$ would be an epimorphism in $\CRing$, where $(-)^+$ denotes the unitalization and $(-)^{\ab}$ the abelianization. But if $R \to S$ is an epimorphism of commutative rings, then $\card(S) \leq \card(R)$ by <a href="https://stacks.math.columbia.edu/tag/04W0" target="_blank">SP/04W0</a>. Since $((\coprod_n \IZ)^+)^{\ab}$ is countable and $\prod_n \IZ$ is not, we get a contradiction.


structure_id: Rng
property_id: coregular
type: category
proof:
We can copy the proof for <a href="/category/Ring">$\Ring$</a>. In short, the inclusion of diagonal matrices $\IQ^2 \hookrightarrow M_2(\IQ)$ is a regular monomorphism, but becomes zero after taking the pushout with $p_1 : \IQ^2 \twoheadrightarrow \IQ$ because $M_2(\IQ)$ is simple.


structure_id: Rng
property_id: regular quotient object classifier
type: category
proof:
Assume that $\Rng$ has a regular quotient object classifier $P$. Consider the functor $N : \Ab \to \Rng$ that equips an abelian group with zero multiplication. It is fully faithful and has a left adjoint mapping a rng $R$ to the abelian group $R/R^2$. If $R$ is a rng with zero multiplication and $R \to S$ is a surjective homomorphism, then $S$ has zero multiplication. Therefore, the assumptions of Lemma 1 <a href="/content/subcategories">here</a> (dualized) apply and we conclude that $P/P^2$ is a regular quotient object classifier of $\Ab$. But we already know that <a href="/category/Ab">$\Ab$</a> has no such object (in fact, the only additive categories with such an object are trivial by <a href="https://math.stackexchange.com/questions/4086192" target="_blank">MSE/4086192</a>).


structure_id: Rng
property_id: cocartesian cofiltered limits
type: category
proof:
Consider the ring $A = \IZ[X]$ and the sequence of rings $B_n = \IZ[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $\IZ[[Y]]$ (both in $\Ring$ and $\Rng$). Every element in the coproduct of rngs $\IZ[X] \sqcup \IZ[[Y]]$ has a finite "free product" length. Now consider the elements $$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) - 1 \in A \sqcup B_n.$$ Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.


structure_id: Rng
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We know that <a href="/category/Ring">$\Ring$</a> does not have this property. Now use the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> applied to the forgetful functor $\Ring \to \Rng$. We only need to verify that it preserves epimorphisms: Let $f : R \to S$ be an epimorphism in $\Ring$ and let $g,h : S \rightrightarrows T$ be two homomorphisms of rngs with $gf = hf$. The element $e = g(1) = h(1) \in T$ is idempotent, and $g,h$ become homomorphisms of rings $S \rightrightarrows eTe$. Hence, $g=h$.


structure_id: Rng
property_id: effective cocongruences
type: category
proof:
The counterexample is similar to the one at <a href="https://mathoverflow.net/a/510809" target="_blank">MO/510744</a> for $\Ring$: in this case, $$X \coloneqq \langle p \mid p^2 = p \rangle_{\Rng} \cong \IZ$$ and $$E \coloneqq \langle p, q \mid p^2 = p, q^2 = q, pq = q, qp = p \rangle_{\Rng} \cong \begin{pmatrix} \IZ & \IZ \\ 0 & 0 \end{pmatrix}$$ via $$p \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad q \mapsto \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}.$$ From here, the rest of the proof is similar to the one for <a href="/category/Ring">$\Ring$</a>.


structure_id: Sch_R
property_id: pullbacks
type: category
proof:
This is the well-known construction of the fiber product of schemes, see e.g. EGA I, Chap. I, Thm. 3.2.1.


structure_id: Sch_R
property_id: infinitary extensive
type: category
proof:
One uses the same proof as for locally ringed spaces, using that open subspaces of schemes are also schemes.


structure_id: Sch_R
property_id: countable powers
type: category
proof:
While all diagrams of affine schemes have a limit in the category of schemes, one can show that an infinite product of quasi-compact non-empty schemes only exists when almost all of them are affine, see <a href="https://mathoverflow.net/questions/65506" target="_blank">MO/65506</a>. Thus, for example the power $(\IP^1_R)^{\IN}$ does not exist.


structure_id: Sch_R
property_id: Malcev
type: category
proof:
Consider the subscheme $V(x-y) \cup V(y) \subseteq \IA^2_R$. It contains the diagonal but it is not symmetric.


structure_id: Sch_R
property_id: semi-strongly connected
type: category
proof:
This is because already the full subcategory of affine $R$-schemes is not semi-strongly connected, because its dual <a href="/category/CAlg(R)">$\CAlg(R)$</a> is not semi-strongly connected.


structure_id: Sch_R
property_id: generating set
type: category
proof:
If $S$ is a generating set of $R$-schemes, then the set of affine open subsets of the schemes in $S$ would also be a generating set. This is then also a generating set in the category of affine $R$-schemes, corresponding to a cogenerating set in <a href="/category/CAlg(R)">$\CAlg(R)$</a>, which we know does not exist.


structure_id: Sch_R
property_id: quotients of congruences
type: category
proof:
If $\Sch_R$ had quotients of congruences, then by <a href="/content/pushouts-of-monos-via-congruence-quotients">this lemma</a> it would also have pushouts of monomorphisms, contradicting the fact that the span $\IA^1_K \leftarrow \Spec(K(t)) \rightarrow \IA^1_K$ has no pushout where $K$ is a residue field of $R$; see <a href="https://mathoverflow.net/questions/9961" target="_blank">MO/9961</a>.


structure_id: Sch_R
property_id: sequential colimits
type: category
proof:
It is shown in <a href="https://mathoverflow.net/questions/511944" target="_blank">MO/511944</a> that the diagram $$\IA^1_R \hookrightarrow \IA^2_R \hookrightarrow \IA^3_R \hookrightarrow \cdots$$ has no colimit.


structure_id: SemiGrp
property_id: strict initial object
type: category
proof:
This is because the initial object is the empty semigroup, and a non-empty set has no map to an empty set.


structure_id: SemiGrp
property_id: cocartesian cofiltered limits
type: category
proof:
We need to prove that for two cofiltered diagram of semigroups $(B_i)$, $(C_i)$ the canonical map $$\textstyle \alpha : \lim_i B_i + \lim_i C_i \to \lim_i (B_i + C_i)$$ is an isomorphism, i.e. bijective. The underlying set of the coproduct $B_i + C_i$ is identified with the disjoint union of sets of the form $$B_i \times C_i \times B_i \times \cdots \times C_i$$ with any positive product length, starting and ending either with $B_i$ or $C_i$. Moreover, $\lim_i$ commutes with arbitrary coproducts in $\Set$, and of course also with products. This shows that $\alpha$ is bijective.


structure_id: SemiGrp
property_id: balanced
type: category
proof:
The inclusion of additive semigroups $\IN \hookrightarrow \IZ$ is a counterexample. Indeed, if $f,g : \IZ \to G$ are semigroup homomorphisms which agree on $\IN$, the element $e \coloneqq f(0) = g(0)$ provides a neutral element for $eGe$, which therefore becomes a monoid, and $f,g$ corestrict to monoid homomorphisms $f,g : \IZ \to eGe$ which agree on $\IN$. And we already know that $f = g$ in that case (inverses are uniquely determined).
Another example can be found in <a href="https://mathoverflow.net/a/510442/2841">MO/510431</a>.


structure_id: SemiGrp
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \SemiGrp \to \Set$ and the relation $R \subseteq U^2$ defined by $R(A) \coloneqq \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by the free semigroup on a single generator and $R$ by the free semigroup on two generators $x,y$ subject to the relation $xy=x^2$. It is clear that $R$ is reflexive, but not symmetric.


structure_id: SemiGrp
property_id: semi-strongly connected
type: category
proof:
Let us first remark that every non-empty finite semigroup $A$ has an idempotent element $e$, and then $B \to A$, $x \mapsto e$ does define a semigroup homomorphism for any $B$. Therefore, counterexamples need to be infinite and also without idempotent elements.
Let $A$ be the set of positive rational numbers of the form $m/2^n$ (with $m > 0$, $n \geq 0$), and let $B$ be the set of positive rational numbers of the form $m/3^n$ (with $m > 0$, $n \geq 0$). Both are semigroups under addition. The element $1 \in A$ is $2^\infty$-divisible, meaning that for every $n \geq 0$ there is some $a \in A$ with $1 = 2^n \cdot a$. But $B$ has no $2^\infty$-divisible element. Hence, there is no semigroup homomorphism $A \to B$. Likewise, there is no semigroup homomorphism $B \to A$.


structure_id: SemiGrp
property_id: cogenerating set
type: category
proof:
The proof is similar to the proof for <a href="/category/Grp">$\Grp$</a>. Assume that there is a cogenerating set $S$. There is an infinite simple group $G$ larger than all the semigroups in $S$ (such as an alternating group). Since $\id_G, 1 : G \rightrightarrows G$ are different, there is a semigroup $H \in S$ and a homomorphism of semigroups $f : G \to H$ with $f \neq f \circ 1$. Then $$N \coloneqq \{g \in G : f(g) = f(1)\}$$ is a normal subgroup of $G$. It is proper, and hence trivial. But then $f$ is injective, which is a contradiction.


structure_id: SemiGrp
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Set">$\Set$</a> does not have this property (by <a href="/category-implication/topos_no_stable_epis">this result</a>). Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \SemiGrp$ that equips a set with the multiplication $a \cdot b \coloneqq a$.


structure_id: SemiGrp
property_id: effective cocongruences
type: category
proof:
The proof is similar to <a href="/category/Mon">$\Mon$</a>, i.e. we adapt the counterexample from <a href="https://mathoverflow.net/a/510809" target="_blank">MO/510744</a>. Namely, consider the semigroups $$\begin{align*} X & \coloneqq \langle p \mid p^2 = p \rangle,\\ E & \coloneqq \langle p, q \mid p^2 = p,\, q^2 = q,\, pq = q,\, qp = p \rangle, \end{align*}$$ whose underlying sets are $\{p\}$ and $\{p,q\}$, respectively. Then $X$ represents the functor sending a semigroup $A$ to its idempotents, and $E$ represents the relation on idempotents $a, b$ of $A$ that $ab = b$, $ba = a$. It is easy to check that this defines an equivalence relation (see <a href="https://mathoverflow.net/a/510809" target="_blank">MO/510744</a> for details). Since $p \ne q$ in $E$, the equalizer of the two maps $X \rightrightarrows E$ is the empty semigroup. Therefore, if $E$ were effective, it would be isomorphic to the coproduct $X \sqcup X$, whose underlying set consists of non-empty words in $p,q$ with $p,q$ strictly alternating. In particular, in this coproduct, $pq \ne q$.


structure_id: SemiGrp
property_id: natural numbers object
type: category
proof:
Assume that a natural numbers object exists. Then by <a href="/content/nno_distributive_criterion">this result</a>, for every semigroup $A$ the natural homomorphism $$\textstyle\alpha : \coprod_{n \geq 0} A \to A \times \coprod_{n \geq 0} 1$$ is a split monomorphism. But this is not true: For each $n \geq 0$ let $A_n$ denote a copy of $A$. The elements of the coproduct in $\alpha$'s domain have a unique representation as $$x_{n_1} * \cdots * x_{n_s}$$ with $x_i \in A_i$ and $n_i \neq n_{i+1}$, and we have $$\alpha(x_{n_1} * \cdots * x_{n_s}) = (x_{n_1} \cdots x_{n_s}, n_1 * \cdots * n_s).$$ In particular, if $A$ has two non-equal commuting elements $x,y$ (for example, if $A$ is any non-trivial monoid), we have $$\alpha(y_0 x_0) = \alpha(x_0 y_0),$$ showing that $\alpha$ is not injective.


structure_id: SemiGrp
property_id: coregular
type: category
proof:
We will find a regular monomorphism $\iota : F \to M$ of semigroups and a homomorphism $F \to K$ such that $K \to K \sqcup_F M$ is not injective. It is similar to our example for <a href="/category/Mon">$\Mon$</a>. Consider these semigroups defined by generators and relations: $$\begin{align*} F & \coloneqq \langle a,b,c,d \rangle \\ M & \coloneqq \langle a,b,c,d,s : as = c, \, bs = d \rangle \cong \langle a,b,s \rangle \\ K & \coloneqq \langle x,c,d \rangle \\ N & \coloneqq \langle a,b,c,d,s_0,s_1 : as_i = c, \, bs_i = d \rangle \\ & \cong \langle a,b,s_0,s_1 : a s_0 = a s_1, \, b s_0 = b s_1 \rangle \end{align*}$$ There is a canonical homomorphism $\iota : F \to M$, which is the equalizer of the two canonical homomorphisms $M \rightrightarrows N$ defined by $s \mapsto s_i$. We define $F \to K$ by $a \mapsto x$, $b \mapsto x$, $c \mapsto c$, $d \mapsto d$. Then $$K \sqcup_F M \cong \langle x,c,d,s : x s = c,\, x s = d \rangle$$ shows that $c,d \in K$ have the same image in the pushout.


structure_id: SemiGrp
property_id: regular subobject classifier
type: category
proof:
Assume that a regular subobject classifier $\Omega$ exists in $\SemiGrp$. The universal regular monomorphism $\top : 1 \to \Omega$ corresponds to an idempotent element $e \in \Omega$. It follows that $e \Omega e$ is a monoid with neutral element $e$. We claim that it is a regular subobject classifier in <a href="/category/Mon">$\Mon$</a>, which we know does not exist. Indeed, let $\iota : A \to B$ be a regular monomorphism of monoids. Since the forgetful functor $\Mon \to \SemiGrp$ preserves limits, we can also see $\iota$ as a regular monomorphism of semigroups. Hence, there is a unique homomorphism of semigroups $f : B \to \Omega$ with $\iota(A) = \{b \in B : f(b) = e\}$. Since $1 \in \iota(A)$, we have $f(1) = e$. Then $f$ corresponds to a homomorphism of monoids $f : B \to e \Omega e$ with kernel $\iota$, which proves our claim.


structure_id: Set_arrow
property_id: semi-strongly connected
type: category
proof:
Consider two objects $(X, Y, f)$ and $(X', Y', f')$. If $X'$ is non-empty with element $x'$, we can construct a morphism $(\ell, r) : (X, Y, f) \to (X', Y', f')$ where $\ell$ is the constant function with value $x'$, and $r$ is the constant function with value $f'(x')$; and similarly if $X$ is non-empty we can construct a morphism $(X', Y', f') \to (X, Y, f)$. Otherwise, if $X$ and $X'$ are both empty, we can reduce to the fact that $\Set$ is semi-strongly connected to find $r$ in one direction, and fill in $\ell$ as the unique function $\varnothing \to \varnothing$.


structure_id: Set_arrow
property_id: Grothendieck topos
type: category
proof:
It is equivalent to the category of presheaves on <a href="/category/walking_morphism">the walking morphism</a>.


structure_id: Set_arrow
property_id: locally strongly finitely presentable
type: category
proof:
It is equivalent to the category of models of the algebraic theory with two sorts, and a single unary operation from one sort to the other.


structure_id: Set_arrow
property_id: generator
type: category
proof:
Suppose $\Set^{\rightarrow}$ had a generator. Then by <a href="/content/topos-with-generator">this result</a>, every subterminal object would be either initial or terminal. However, in fact, $0 \to 1$ is a subterminal object which is isomorphic to neither the initial object $0 \to 0$ nor the terminal object $1 \to 1$.


structure_id: Set_c
property_id: locally small
type: category
proof:
The collection of maps between two (countable) sets $X,Y$ is a subset of $X \times Y$ and therefore a set.


structure_id: Set_c
property_id: finitely complete
type: category
proof:
The embedding $\Set_\c \hookrightarrow \Set$ is closed under finite products and equalizers, hence under finite limits.


structure_id: Set_c
property_id: finitely cocomplete
type: category
proof:
The embedding $\Set_\c \hookrightarrow \Set$ is closed under finite coproducts and coequalizers, hence under finite colimits.


structure_id: Set_c
property_id: subobject classifier
type: category
proof:
This is because $\{0,1\}$ is a subobject classifier in <a href="/category/Set">$\Set$</a>, which is countable, and the monomorphisms coincide.


structure_id: Set_c
property_id: extensive
type: category
proof:
The same proof as for <a href="/category/Set">$\Set$</a> applies. Actually, the category is "countably extensive".


structure_id: Set_c
property_id: countably distributive
type: category
proof:
By elementary set theory, a countable (disjoint) union of countable sets is again countable. Hence, countable coproducts exist in $\Set_\c$, and we already saw that finite products exist. The distributivity morphism is an isomorphism since this is the case in <a href="/category/Set">$\Set$</a> and the forgetful functor $\Set_\c \to \Set$ preserves finite products and countable coproducts.


structure_id: Set_c
property_id: effective congruences
type: category
proof:
Let $f, g : E \rightrightarrows X$ be a congruence in $\Set_\c$. Then using $1$ as a test object, we see that this induces an equivalence relation on $X$. We already know that <a href="/category/Set">$\Set$</a> has effective congruences (as does every topos). Using <a href="/content/effective-congruence-quotients">this result</a>, we see that $E$ is the kernel pair of $X \to (X/E)_{\Set}$ in $\Set$. Also, the quotient $(X/E)_{\Set}$ is countable; and the forgetful functor $\Set_\c \to \Set$ is fully faithful <a href="https://ncatlab.org/nlab/show/reflected+limit#FullSubcategoryInclusionReflectCoLimits" target="_blank">and therefore reflects limits</a>. Thus, we conclude that $E$ is the kernel pair of $X \to (X/E)_{\Set}$ in $\Set_\c$ as well.


structure_id: Set_c
property_id: regular
type: category
proof:
From the other properties we know that the category is finitely complete and that it has coequalizers. The regular epimorphisms are stable under pullback since this holds in <a href="/category/Set">$\Set$</a> and both regular epimorphisms (they are surjective maps) and pullbacks coincide.


structure_id: Set_c
property_id: coregular
type: category
proof:
From the other properties we know that the category is finitely cocomplete and that it has equalizers. The regular monomorphisms are stable under pushout since this holds in <a href="/category/Set">$\Set$</a> and both regular monomorphisms (they are injective maps) and pushouts coincide.


structure_id: Set_c
property_id: countable powers
type: category
proof:
Since the forgetful functor $\Set_\c \to \Set$ is representable, it preserves (countable) products. Therefore, if the power $\{0,1\}^{\IN}$ exists in $\Set_\c$, it must be the ordinary cartesian product, which however is uncountable.


structure_id: Set_c
property_id: ℵ₂-small copowers
type: category
proof:
Assume that the copower $X \coloneqq I \otimes 1$ exists in $\Set_\c$. We will prove that $I$ is countable; in particular, $\aleph_1 \otimes 1$ does not exist. The copower is a countable set $X$ with a universal family of elements $(x_i)_{i \in I}$. For every $i \in I$ there is a map $X \to \{0,1\}$ defined by $x_i \mapsto 1$ and $x_j \mapsto 0$ for $j \neq i$. In particular, $x_i \neq x_j$ whenever $i \neq j$. Thus, $x : I \to X$ is an injective map, which proves that $I$ is countable.


structure_id: Set_c
property_id: ℵ₁-cofiltered limits
type: category
proof:
Since the forgetful functor $\Set_\c \to \Set$ is representable, it preserves limits. So it suffices to construct an $\aleph_1$-cofiltered diagram of countable sets whose $\Set$-limit is not countable. Let $R$ be an uncountable set (such as $\IR$) and consider the poset $P_{<\aleph_1}(R)$ of countable subsets of $R$. It is $\aleph_1$-filtered, so its opposite is $\aleph_1$-cofiltered. Define a diagram $X : P_{<\aleph_1}(R)^{\op} \to \Set_\c$ by mapping a countable subset $C \subseteq R$ to the pointed countable set $C \sqcup \{\ast\}$ and an inclusion $C \subseteq D$ to the map $D \sqcup \{\ast\} \to C \sqcup \{\ast\}$ which restricts to the identity on $C$ and sends everything else to the base point. For $r \in R$ we define $\varepsilon(r) \in \lim(X)$ by $$\varepsilon(r)_C = \begin{cases} r & r \in C \\ \ast & r \notin C \end{cases}$$ It is easily checked that this indeed lies in the limit. Moreover, $\varepsilon(r) = \varepsilon(r')$ implies $r=r'$, since evaluating at $C \coloneqq \{r\}$ yields $r = r'$. Hence, $\lim(X)$ is uncountable.


structure_id: Set_f
property_id: generator
type: category
proof:
The singleton set (which is <i>not</i> terminal) is a generator as it represents the forgetful functor $\Set_\f \to \Set$.


structure_id: Set_f
property_id: cogenerator
type: category
proof:
The set $\{0,1\}$ is a cogenerator in $\Set_\f$: Assume that $f,g : X \rightrightarrows Y$ are two finite-to-one maps such that $h \circ f = h \circ g$ for all finite-to-one maps $h : Y \to \{0,1\}$. This exactly means $f^*(A)=g^*(A)$ for all finite subsets $A \subseteq Y$. Applying this to $A = \{f(x)\}$ for $x \in X$ we get $x \in f^*(\{f(x)\}) = g^*(\{f(x)\})$, so that $g(x) = f(x)$.


structure_id: Set_f
property_id: semi-strongly connected
type: category
proof:
From set theory it is known that for all sets $X,Y$ there is an injective map $X \to Y$ or an injective map $Y \to X$, and injective maps are finite-to-one.


structure_id: Set_f
property_id: extensive
type: category
proof:
We first show that finite coproducts exist. The empty set is clearly initial. The disjoint union $X+Y$ of two sets $X,Y$ with the inclusion maps $X \rightarrow X+Y \leftarrow Y$ is a coproduct: The inclusions are injective, hence finite-to-one. If $f : X \to T$, $g : Y \to T$ are finite-to-one maps, the induced map $(f;g) : X + Y \to T$ is finite-to-one since the fiber of $t \in T$ is $f^*(\{t\}) + g^*(\{t\})$, which is finite. Hence, finite coproducts exist. A map $A \to X + Y$ yields a decomposition $A = A_X + A_Y$ with maps $A_X \to X$, $A_Y \to Y$ (since <a href="/category/Set">$\Set$</a> is extensive). Here, $A \to X + Y$ is finite-to-one iff $A_X \to X$ and $A_Y \to Y$ are finite-to-one.


structure_id: Set_f
property_id: equalizers
type: category
proof:
Equalizers can be constructed as in <a href="/category/Set">$\Set$</a> because of the following trivial observation: if $f : X \to Y$ is a finite-to-one map and $E \subseteq Y$ is a subset with $f(X) \subseteq E$, then the induced map $f^E : X \to E$ is also finite-to-one.


structure_id: Set_f
property_id: epi-regular
type: category
proof:
If $f : X \to Y$ is an epimorphism in $\Set_\f$, i.e. a surjective finite-to-one map, it is a coequalizer of the two maps $p_1, p_2 : X \times_Y Y \rightrightarrows Y$ in $\Set$. These maps are finite-to-one since $p_i^*(\{y\}) \cong f^*(\{y\})$ for $i=1,2$, and their coequalizer is also $f$ in $\Set_\f$: It suffices to observe that if $h : Y \to T$ is a map such that $h \circ f$ is finite-to-one, then $h$ is finite-to-one as well. In fact, surjectivity of $f$ implies $h^*(\{t\}) = f_*((h \circ f)^*(\{t\}))$ for $t \in T$.


structure_id: Set_f
property_id: quotients of congruences
type: category
proof:
A congruence on a set $X$ in $\Set_\f$ is the same as an equivalence relation $R$ on $X$ whose equivalence classes are finite. In that case, the usual quotient map $p : X \to X/R$ is finite-to-one. Moreover, if $h : X/R \to Y$ is a map such that $h \circ p : X \to Y$ is finite-to-one, then $h$ is finite-to-one as well because $h^*(\{y\}) \subseteq p^*((h \circ p)^*(\{y\}))$ for all $y \in Y$. Therefore, $p$ is also the quotient in $\Set_\f$.


structure_id: Set_f
property_id: effective congruences
type: category
proof:
Let $f, g : E \rightrightarrows X$ be a congruence in $\Set_\f$. From the proof on quotients of congruences in $\Set_\f$, we have a quotient map $p : X \to X/E$ in $\Set_\f$, and $E$ is the kernel pair of $p$ in $\Set$. It remains to see that $E$ is also the kernel pair of $p$ in $\Set_f$. Thus, suppose we have $x_1, x_2 : T \rightrightarrows X$ with $p \circ x_1 = p \circ x_2$. Then there is a unique $e : T \to E$ in $\Set$ with $x_1 = f\circ e$ and $x_2 = g\circ e$. Since $f\circ e$ is finite-to-one, we must have $e$ is finite-to-one as well.


structure_id: Set_f
property_id: effective cocongruences
type: category
proof:
Suppose we have a cocongruence $f, g : X \rightrightarrows E$ in $\Set_f$. Then it is a coreflexive corelation in $\Set$. Since <a href="/category/Set">$\Set$</a> is co-Malcev and has effective cocongruences, that implies $E$ is the cokernel pair of some function $h : Z \to X$ in $\Set$. By the dual of <a href="/content/effective-congruence-quotients">this result</a>, if $\inc_Y : Y \hookrightarrow X$ is the equalizer of $f$ and $g$, then $E$ is also the cokernel pair of $\inc_Y$ in $\Set$. It remains to see that $E$ is the cokernel pair of $\inc_Y$ in $\Set_\f$ as well. Thus, suppose $a, b : X \rightrightarrows T$ are such that $a |_Y = b |_Y$. Then there is a unique $c : E\to T$ in $\Set$ with $a = c\circ f$ and $b = c\circ g$. Since $(f;g) : X + X \to E$ is surjective and $c \circ (f;g) = (a;b)$ is finite-to-one, we see $c$ is finite-to-one as well.


structure_id: Set_f
property_id: locally cartesian closed
type: category
proof:
If $X$ is a set, the equivalence $\Set/X \simeq \Set^X$, $f \mapsto (f^*(\{x\}))_{x \in X}$ restricts to an equivalence $\Set_\f / X \simeq \FinSet^X$. This category is cartesian closed since <a href="/category/FinSet">$\FinSet$</a> is cartesian closed and products of cartesian closed categories are cartesian closed.


structure_id: Set_f
property_id: ℵ₁-accessible
type: category
proof:
We first show that $\aleph_1$-directed colimits exist and are preserved by the forgetful functor to $\Set$. Let $(X_i)_{i \in I}$ be a diagram in $\Set_\f$ indexed by a $\aleph_1$-directed poset $I$. Let $(u_i : X_i \to X)$ be the colimit in $\Set$. Each map $u_i$ is finite-to-one: Otherwise, some fiber $u_i^*(\{x\}) \subseteq X_i$ contains infinitely many elements $a_1,a_2,\dotsc$. For every $n \geq 1$ we find $i_n \geq i$ such that $a_1,a_n$ have the same image in $X_{i_n}$. Let $i_\infty \in I$ be an upper bound of all $i_n$. Then all $a_n$ have the same image in $X_{i_\infty}$. This is a contradiction since $X_i \to X_{i_\infty}$ is finite-to-one. Now if $f : X \to Y$ is a map such that each $f \circ u_i$ is finite-to-one, then $f$ is finite-to-one as well: If a fiber $f^*(\{y\})$ contains infinitely many distinct elements $x_1,x_2,\dotsc$, there is some index $i$ such that all have a preimage in $X_i$, but then $(f \circ u_i)^*(\{y\})$ would be infinite. This concludes the proof that $\aleph_1$-directed colimits exist. In $\Set$, every set $X$ is the $\aleph_1$-directed colimit of its countable subsets. This remains true in $\Set_\f$ because a map $X \to Y$ is finite-to-one as long as each restriction to a countable subset of $X$ is finite-to-one. Moreover, every countable set is $\aleph_1$-presentable in $\Set$, but also in $\Set_\f$.


structure_id: Set_f
property_id: ℵ₁-cofiltered limits
type: category
proof:
We derive this from the fact that <a href="/category/FinSet">$\FinSet$</a> is closed under $\aleph_1$-cofiltered limits in $\Set$. Let $D : \I \to \Set_\f$ be an $\aleph_1$-cofiltered diagram. Let $(p_i : L \to D(i))_{i \in \I}$ be the limit cone in $\Set$. Each $p_i$ is finite-to-one, because the fiber over $d \in D(i)$ is the limit of the $\aleph_1$-cofiltered diagram $\tilde{D} : I/i \to \FinSet$ which maps $j \to i$ to $D(j \to i)^*(\{d\})$. To verify the universal property, let $(f_i : T \to D(i))_{i \in \I}$ be a cone in $\Set_\f$. There is a unique map $f : T \to L$ such that $p_i f = f_i$ for every $i \in \I$, and it remains to prove that $f$ is finite-to-one. Choose any $i \in \I$. Then for every $x \in L$ we have $f^*(\{x\}) \subseteq f_i^*(\{p_i(x)\})$, and the latter set is finite.


structure_id: Set_f
property_id: locally finite
type: category
proof:
If $X$ is an infinite set, there are infinitely many maps $1 \to X$. They are injective and hence finite-to-one.


structure_id: Set_f
property_id: binary powers
type: category
proof:
More generally, if $X,Y$ are two non-empty sets such that $X \times Y$ exists in $\Set_\f$, then both $X$ and $Y$ must be finite. In fact, the forgetful functor to $\Set$ is representable, so it must preserve products. This means we can assume $X \times Y$ is the usual cartesian product with the usual projections. Since $p_1 : X \times Y \to X$ is finite-to-one and $X$ is non-empty, $Y$ is finite. By symmetry, also $X$ is finite. (Conversely, if $X$ and $Y$ are finite, or one of them is empty, then indeed $X \times Y$ exists.)


structure_id: Set_f
property_id: countable copowers
type: category
proof:
Assume that the copower $X \coloneqq \IN \otimes 1$ exists, where $1$ is the singleton set. This is a set with a map $i : \IN \to X$ (not necessarily finite-to-one) such that for every other such map $j : \IN \to Y$ there is a unique finite-to-one map $f : X \to Y$ with $f \circ i = j$. Applying this to $j : \IN \to 1$, we see that $X$ is finite. Applying the universal property to maps $j : \IN \to \{0,1\}$, we see that for every subset $E \subseteq \IN$ there is a unique finite subset $F \subseteq X$ with $i^*(F) = E$. But finiteness of $F$ is automatic, so $i^* : P(X) \to P(\IN)$ is bijective. But then $P(\IN)$ is finite, which is absurd.


structure_id: Set_f
property_id: filtered
type: category
proof:
Consider the maps $f,g : \IN \rightrightarrows \IN$ defined by $f(x)=x$ and $g(x)=2x$. They are injective, hence finite-to-one. If a map $h : \IN \to X$ coequalizes them, we have $h(x)=h(2x)$, in particular $h(1)=h(2^n)$ for all $n \in \IN$. Thus, $h$ is not finite-to-one.


structure_id: Set_f
property_id: sequential limits
type: category
proof:
Consider the set $[n] \coloneqq \{0,\dotsc,n\}$ for $n \in \IN$. The forgetful functor to $\Set$ is representable (by the singleton set), hence preserves all limits. Thus, if the diagram of truncation maps $\cdots \twoheadrightarrow [2] \twoheadrightarrow [1] \twoheadrightarrow [0]$ has a limit in $\Set_\f$, its underlying set is isomorphic to the limit taken in $\Set$, which is $\IN \cup \{\infty\}$. But there is no finite-to-one map $\IN \cup \{\infty\} \to [0]$.


structure_id: Set_f
property_id: coaccessible
type: category
proof:
Assume that $\Set_\f$ is $\lambda$-coaccessible for some regular cardinal. Then there is an infinite cardinal $\kappa$ such that every $X \in \Set_\f$ is a $\lambda$-cofiltered limit of sets of cardinality $\leq \kappa$. In particular, since every $\lambda$-cofiltered category is non-empty, there is some set $Y$ of cardinality $\leq \kappa$ that admits a morphism $f : X \to Y$ in $\Set_f$. Since $f$ has finite fibers, this implies $\card(X) \leq \card(Y) \cdot \aleph_0 \leq \kappa$. Since $X$ was an arbitrary set, this is a contradiction.


structure_id: Set_*
property_id: coregular
type: category
proof:
From the other properties we know that (co-)limits exist and that monomorphisms coincide with injective pointed maps. So it suffices to prove that these maps are stable under pushouts. This follows from the corresponding fact for <a href="/category/Set">$\Set$</a> and the observation that the forgetful functor $\Set_* \to \Set$ preserves pushouts.


structure_id: Set_*
property_id: co-Malcev
type: category
proof:
Malcev categories are closed under slice categories by Prop. 2.2.14 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>. It follows that co-Malcev categories are closed under coslice categories, and $\Set_*$ is a coslice category of <a href="/category/Set">$\Set$</a>, which is co-Malcev since every elementary topos is co-Malcev.


structure_id: Set_*
property_id: cocartesian cofiltered limits
type: category
proof:
Let $X$ be a pointed set and $(Y_i)$ be a filtered diagram of pointed sets. Base points will be denoted by $0$. The canonical map $X \vee \lim_i Y_i \to \lim_i (X \vee Y_i)$ is injective since the wedge sum naturally embeds into the product and the natural map $X \vee \prod_i Y_i \to \prod_i (X \times Y_i)$ is injective. Now let $z = (z_i) \in \lim_i (X \vee Y_i)$.
Case 1: There is some index $i$ with $z_i \in X \setminus \{0\}$. We claim $z_j \in X$ for any index $j$ and $z_j = z_i$ in $X$, so that $z$ has a preimage in $X$. To see this, choose an index $k \geq i,j$. Since $X \vee Y_i \to X \vee Y_k$ maps $z_i \mapsto z_k$ and is the identity on $X$, we see that $z_k \in X$ and $z_k = z_i$ in $X$. Since $X \vee Y_j \to X \vee Y_k$ maps $z_j \mapsto z_k$, we see that $z_j \notin Y_j$, since otherwise $z_k \in Y_k \cap X = \{0\}$. Hence, $z_j \in X \setminus \{0\}$, and then $z_j = z_k = z_i$.
Case 2: We have $z_i \in Y_i$ for all $i$. Then clearly $(z_i) \in \lim_i Y_i$ is a preimage.


structure_id: Set_*
property_id: CIP
type: category
proof:
The coproduct (wedge sum) of a family of pointed sets $(X_i)_{i \in I}$ can be realized as the subset of $\prod_{i \in I} X_i$ consisting of those tuples $x$ such that $x_i = 0$ for all but (at most) one index.


structure_id: Set_*
property_id: Barr-coexact
type: category
proof:
We have that $\Set_*^{\op}$ is a slice category of $\Set^{\op}$, which in turn is monadic over $\Set$. Therefore, by combining results from <a href="http://www.springer.com/us/book/9781402019616" target="_blank">Borceux and Bourn</a> Appendix A and <a href="https://ncatlab.org/nlab/show/colimits+in+categories+of+algebras#exact" target="_blank">nLab</a>, $\Set_*^{\op}$ is Barr-exact.


structure_id: Set_*
property_id: unital
type: category
proof:
The joint image of $X \to X \times Y \leftarrow Y$ is just $\{(x,0) : x \in X\} \cup \{(0,y) : y \in Y\}$ (where $0$ denotes the base point), which is clearly a proper subset of $X \times Y$ when both $X,Y$ are non-trivial.


structure_id: Set_*
property_id: CSP
type: category
proof:
The image of $X \vee Y$ in $X \times Y$ is just $\{(x,0) : x \in X\} \cup \{(0,y) : y \in Y\}$ (where $0$ denotes the base point), which is clearly a proper subset when both $X,Y$ are non-trivial.


structure_id: Set_*
property_id: conormal
type: category
proof:
Every cokernel is "injective away from the base point". Formally, if $p : A \to B$ is a cokernel in $\Set_*$, it has the property that $p(x)=p(y) \neq 0$ implies $x=y$ (where $0$ denotes the base point). Clearly this is not satisfied for every surjective pointed map, consider $(\IN,0) \to (\{0,1\},0)$ defined by $0 \mapsto 0$ and $x \mapsto 1$ for $x > 0$.


structure_id: Set_*
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Set_*$ that freely adds a base point.


structure_id: Setne
property_id: products
type: category
proof:
Take the product of non-empty sets inside of <a href="/category/Set">$\Set$</a> and observe that it is non-empty by the axiom of choice.


structure_id: Setne
property_id: cartesian closed
type: category
proof:
This follows as for <a href="/category/Set">$\Set$</a>, since for non-empty sets $X,Y$ there is at least one function $X \to Y$.


structure_id: Setne
property_id: finitely accessible
type: category
proof:
Since the inclusion $\Setne \hookrightarrow \Set$ is closed under non-empty colimits, it is also closed under filtered colimits. Therefore, non-empty finite sets are still finitely presentable in $\Setne$, and every non-empty set is written as a filtered colimit of them.


structure_id: Setne
property_id: generalized variety
type: category
proof:
Since the inclusion $\Setne \hookrightarrow \Set$ is closed under non-empty colimits, it is also closed under sifted colimits. Therefore, non-empty finite sets are still strongly finitely presentable in $\Setne$, and every non-empty set is written as a sifted colimit of them.


structure_id: Setne
property_id: natural numbers object
type: category
proof:
Any natural numbers object in $\Set$, such as $(\IN,0,n \mapsto n+1)$, is clearly also one in $\Setne$.


structure_id: Setne
property_id: multi-complete
type: category
proof:
Let $D$ be a diagram in $\Setne$, and let $L$ be a limit of $D$ in $\Set$. If $L$ is non-empty, it gives a limit in $\Setne$ as well. If $L$ is the empty set, there is no cone over $D$ in $\Setne$; hence the empty set of cones gives a multi-limit of $D$ in $\Setne$.


structure_id: Setne
property_id: effective congruences
type: category
proof:
If a congruence $E \rightrightarrows X$ is the kernel pair of $h : X \to Z$, with both $E$ and $X$ non-empty, then certainly $Z$ must also be non-empty.


structure_id: Setne
property_id: cofiltered
type: category
proof:
The two maps $\{0\} \rightrightarrows \{0,1\}$ are equalized by no map $X \to \{0\}$ in this category.


structure_id: Setne
property_id: sequential limits
type: category
proof:
Assume that the sequence of inclusions $\cdots \to \IN_{\geq 2} \to \IN_{\geq 1} \to \IN_{\geq 0} = \IN$ as a limit $X$, consisting of maps $X \to \IN_{\geq n}$. Since $X$ is non-empty, there is a map $1 \to X$. This corresponds to a family of compatible maps $ 1 \to \IN_{\geq n}$, i.e. to compatible elements in $\IN_{\geq n}$. But the set $\bigcap_{n \geq 0} \IN_{\geq n}$ is empty.


structure_id: Setne
property_id: ℵ₁-cofiltered limits
type: category
proof:
Since the forgetful functor $\Setne \to \Set$ is representable, it preserves limits. Thus it suffices to find an $\aleph_1$-cofiltered diagram of non-empty sets whose limit in $\Set$ is empty. Let $R$ be an uncountable set (such as $\IR$) and consider the poset $P$ of subsets $A \subseteq R$ whose complement $R \setminus A$ is countable; in particular, every such $A$ is non-empty. Then $P$ is $\aleph_1$-cofiltered, since a countable union of countable sets is countable. The limit of the diagram $P \to \Setne$, $A \mapsto A$, in $\Set$ is the intersection $\bigcap_{A \in P} A$ in $R$, which is empty, because for every $r \in R$ we have $r \notin R \setminus \{r\} \in P$.


structure_id: Setne
property_id: coquotients of cocongruences
type: category
proof:
The two maps $\{0\} \rightrightarrows \{0,1\}$ form a cocongruence on $\{0\}$ &mdash; namely the cofull cocongruence on $\{0\}$ &mdash; but they do not have an equalizer.


structure_id: Setne
property_id: effective cocongruences
type: category
proof:
The two maps $\{0\} \rightrightarrows \{0,1\}$ form a cocongruence on $\{0\}$ &mdash; namely the cofull cocongruence on $\{0\}$ &mdash; but there is no map $Z \to \{0\}$ making the required commutative diagram, much less a cocartesian square.


structure_id: Setne
property_id: coaccessible
type: category
proof:
If $\Setne$ is coaccessible, then by the dual of Cor. 2.44 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a> also the coslice category $\{\ast\} / \Setne$ would be coaccessible. But this category is isomorphic to <a href="/category/Set_*">$\Set_*$</a>, from which we know that it is not coaccessible (namely, because of Thm. 1.64 in loc. cit.).


structure_id: SetxSet
property_id: generator
type: category
proof:
Assume that $(A,B)$ is a generator. Of course, $A$ and $B$ cannot be both empty. Assume w.l.o.g. that $A$ is non-empty. Then there is no morphism $(A,B) \to (0,1)$, but there are two different morphisms $(0,1) \rightrightarrows (0,2)$.
Alternatively, the claim follows from <a href="/content/topos-with-generator">this result</a>.


structure_id: Sh(X)
property_id: generator
type: category
proof:
Suppose $\Sh(X)$ had a generator. Then by <a href="/content/topos-with-generator">this result</a>, every subterminal object would be either initial or terminal. On the other hand, the subterminal objects of $\Sh(X)$ are of the form $y_U \coloneqq \Hom({-}, U)$ for $U$ open. (Alternative descriptions of $y_U$ include: $$y_U(V) = \begin{cases} 1, & V \subseteq U; \\ \varnothing & \mathrm{otherwise} \end{cases}$$ with the unique restriction maps; and $y_U$ represents the functor $\Gamma(U, {-})$ of sections over $U$.) Therefore, the lemma implies that every open set of $X$ is either $\varnothing$ or $X$, contradicting our assumption that $X$ does not have the indiscrete topology.


structure_id: Sh(X,Ab)
property_id: Grothendieck abelian
type: category
proof:
This is standard, see for example Theorem 18.1.6. in <a href="https://ncatlab.org/nlab/show/Categories+and+Sheaves" target="_blank">Kashiwara-Schapira</a>.


structure_id: Sh(X,Ab)
property_id: split abelian
type: category
proof:
Choose a point $x \in X$. The functor $x_* : \Ab \to \Sh(X,\Ab)$ (skyscraper sheaf) is exact, and its left adjoint $x^* : \Sh(X,\Ab) \to \Ab$ (stalk) satisfies $x^* x_* \cong \id_{\Ab}$. Now, since <a href="/category/Ab">$\Ab$</a> is not split abelian, there is a short exact sequence of abelian groups $0 \to A \to B \to C \to 0$ that does not split. Then $0 \to x_* A \to x_* B \to x_* C \to 0$ is also exact, but it does not split: Otherwise it would also be split after applying $x^*$, which however gives the original sequence in $\Ab$.


structure_id: Sp
property_id: essentially small
type: category
proof:
This holds because <a href="/category/FinSet">$\FinSet$</a> and <a href="/category/B">$\IB$</a> are essentially small.


structure_id: Sp
property_id: elementary topos
type: category
proof:
The category is equivalent to $\prod_{n \geq 0} \Sigma_n{-}\FinSet$ (where $\Sigma_n$ denotes the symmetric group of order $n$), and each $\Sigma_n{-}\FinSet$ is an elementary topos since is <a href="/category/FinSet">$\FinSet$</a> an elementary topos and $\Sigma_n$ is a finite group, cf. <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part B, Corollary 2.3.18.


structure_id: Sp
property_id: cogenerator
type: category
proof:
This follows from $\Sp \simeq \prod_{n \geq 0} \Sigma_n{-}\FinSet$, <a href="/content/cogenerators_in_product_categories">this lemma</a>, and the fact that if $G$ is a (finite) group, the power set $P(G)$ with the evident $G$-action is a weakly terminal cogenerator in $G{-}\Set$ (resp. $G{-}\FinSet$). For the proof, notice that $\varnothing,G \in P(G)$ are fixed points, yielding two $G$-maps $1 \rightrightarrows P(G)$. In particular, $P(G)$ is weakly terminal. If $X$ is a $G$-set with distinct points $x,y$, we construct a $G$-map $f : X \to P(G)$ that separates $x,y$: First, $X$ is a coproduct of orbits. If $x,y$ lie in different orbits, let $f|_{Gx}$ be constant $\varnothing$, $f|_{Gy}$ be constant $G$, and, say, $f$ be constant $\varnothing$ on all other orbits. If $x,y$ lie in the same orbit, say $y = g_0 x$, define $f|_{Gx} : Gx \to P(G)$ by $f(x) = G_x$ (stabilizer), which is well-defined, and choose $f$ to be $\varnothing$ on all other orbits. Then $f(y) = g_0 G_x \neq G_x = f(x)$.


structure_id: Sp
property_id: ℵ₁-accessible
type: category
proof:
We know that <a href="/category/FinSet">$\FinSet$</a> has $\aleph_1$-filtered colimits and that every object is $\aleph_1$-presentable. It follows that for every $n \in \IN$ also $\Sigma_n{-}\FinSet$ has $\aleph_1$-filtered colimits and that every object is $\aleph_1$-presentable. From this it follows formally that $\Sp \simeq \prod_{n \geq 0} \Sigma_n{-}\FinSet$ also has these properties. In particular, $\Sp$ is $\aleph_1$-accessible.


structure_id: Sp
property_id: ℵ₁-cofiltered limits
type: category
proof:
Since <a href="/category/FinSet">$\FinSet$</a> has $\aleph_1$-cofiltered limits, also $\Sigma_n{-}\FinSet$ has $\aleph_1$-cofiltered limits for every $n \geq 0$. Hence, $\Sp \simeq \prod_{n \geq 0} \Sigma_n{-}\FinSet$ has $\aleph_1$-cofiltered limits.


structure_id: Sp
property_id: locally finite
type: category
proof:
If $1$ denotes the terminal species, there are infinitely many morphisms $1 \to 1 \sqcup 1$ since they correspond to functions $\IN \to \{1,2\}$.


structure_id: Sp
property_id: locally small
type: category
proof:
Disclaimer: This result and its proof are not relevant for category theory and are also depending on the concrete model of set theory. That this category is locally essentially small is only what matters. Now, consider the terminal species $F=G=1$. Then $\Hom(F,G)$ has just a single element, namely the natural transformation $\alpha$ that sends every finite set $X$ to the unique map $\alpha_X : 1 \to 1$. Formally, $\alpha$ is a map, modelled as a set of ordered pairs $(X,\id_1)$, where $X$ is a finite set. Hence, $\alpha$ is not a set (since finite sets do not form a set), and therefore $\Hom(F,G) = \{\alpha\}$ is also not a set.


structure_id: Sp
property_id: essentially countable
type: category
proof:
Any function $f : \IN \to \IN$ can be regarded as a combinatorial species with trivial actions, and distinct functions yield non-isomorphic species.


structure_id: Sp
property_id: semi-strongly connected
type: category
proof:
For $n \geq 0$ let $E_n$ be the combinatorial species of $n$-sets: $E_n(A) = \{A\}$ when $A$ has cardinality $n$, otherwise $E_n(A) = \varnothing$. Then there is no morphism between $E_n$ and $E_m$ unless $n = m$.


structure_id: Sp
property_id: generator
type: category
proof:
Assume that a generator $G$ exists. For $n \geq 0$ let $F_n$ be the combinatorial species of sets of cardinality $\neq n$: $F_n(A) = \varnothing$ when $A$ has cardinality $n$, otherwise $F_n(A) = \{A\}$. There are two different morphisms $F_n \rightrightarrows F_n \sqcup F_n$. Hence, there must be at least one morphism $G \to F_n$. If $A$ has cardinality $n$, this implies $G(A) = \varnothing$. Since this holds for all $n$, $G$ is the initial object. But this is clearly no generator (it would mean that the category is thin).
Alternatively, the claim follows from <a href="/content/topos-with-generator">this result</a>.


structure_id: Sp
property_id: natural numbers object
type: category
proof:
If $(N,z,s)$ is a natural numbers object, then $$1 \xrightarrow{z} N \xleftarrow{s} N$$ is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. But there is no combinatorial species $N$ with $N \cong 1 + N$, since evaluating this at, say, $\varnothing$, would yield a finite set $N$ with this property.


structure_id: Top
property_id: complete
type: category
proof:
Take the limit of the underlying sets and endow it with the coarsest topology making all projections continuous.


structure_id: Top
property_id: cocomplete
type: category
proof:
Take the colimit of the underlying sets and endow it with the finest topology making all inclusions continuous.


structure_id: Top
property_id: infinitary extensive
type: category
proof:
[Sketch] Since <a href="/category/Set">$\Set$</a> is infinitary extensive, a map $f : Y \to \coprod_i X_i$ corresponds to a decomposition $Y = \coprod_i Y_i$ (as sets) with maps $f_i : Y_i \to X_i$. Endow $Y_i$ with the subspace topology. If $f$ is continuous, each $Y_i = f^{-1}(X_i)$ is open in $Y$, so that $Y = \coprod_i Y_i$ holds as topological spaces, and each $f_i$ is continuous.


structure_id: Top
property_id: regular subobject classifier
type: category
proof:
The indiscrete two-point space $\{0,1\}$ is a regular subobject classifier since continuous maps $X \to \{0,1\}$ correspond to subsets of $X$.


structure_id: Top
property_id: coregular
type: category
proof:
The category has all limits and colimits, and the regular monomorphisms are the subspace inclusions. Thus, it suffices to prove that subspace inclusions are stable under pushouts. For a proof see e.g. Lemma 3.6 at the <a href="https://ncatlab.org/nlab/show/subspace+topology#pushout" target="_blank">nLab</a>.


structure_id: Top
property_id: balanced
type: category
proof:
If $X$ is a set, consider the discrete space $X_d$ on $X$ and the indiscrete space $X_i$ on $X$. The identity map $X \to X$ lifts to a continuous map $X_d \to X_i$, which is bijective and therefore both a mono- and an epimorphism, but it is not an isomorphism unless $X$ has at most one element.


structure_id: Top
property_id: cartesian filtered colimits
type: category
proof:
The functor $\IQ \times - : \Top \to \Top$ does not preserve sequential colimits, see <a href="https://math.stackexchange.com/questions/1255678" target="_blank">MSE/1255678</a>.


structure_id: Top
property_id: accessible
type: category
proof:
In fact, it does not have any small colimit-dense subcategory by <a href="https://math.stackexchange.com/questions/4097315/" target="_blank">MSE/4097315</a>. For a related result, see <a href="https://mathoverflow.net/questions/288648" target="_blank">MO/288648</a>.


structure_id: Top
property_id: coaccessible
type: category
proof:
Assume $\Top$ is coaccessible. Let $p : S \to I$ be the identity map from the Sierpinski space to the two-element indiscrete space. Then, a topological space is discrete if and only if it is projective to the morphism $p$. This implies that the full subcategory spanned by all discrete spaces, which is equivalent to $\Set$, is coaccessible by Prop. 4.7 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. However, since <a href="/category/Set">$\Set$</a> is not coaccessible, this is a contradiction.


structure_id: Top
property_id: co-Malcev
type: category
proof:
See <a href="https://mathoverflow.net/questions/509548" target="_blank">MO/509548</a>. We can also phrase the proof as follows: Consider the forgetful functor $U : \Top \to \Set$ and the relation $R \subseteq U^2$ defined by $R(X) \coloneqq \{(x,y) \in U(X)^2 : x \in \overline{\{y\}} \}$. Both are representable: $U$ by the singleton and $R$ by the Sierpinski space. It is clear that $R$ is reflexive, but not symmetric.


structure_id: Top
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Top$ which equips a set with the indiscrete topology.


structure_id: Top
property_id: effective cocongruences
type: category
proof:
Consider the indiscrete topological space $I$ on two points. This represents the functor which takes a topological space $X$ to the pairs of indistinguishable points of $X$. Therefore, we get a cocongruence $1 \rightrightarrows I$, where the maps are the two possible functions. However, this cannot be effective: if we have $h : Z\to 1$ which equalizes the two maps, then $Z$ must be empty. But that means the cokernel pair of $h$ is the discrete space on two points.


structure_id: Top_*
property_id: complete
type: category
proof:
This follows from $\Top_* \cong 1 / \Top$ and the fact that <a href="/category/Top">$\Top$</a> is complete. Concretely, the limit of pointed spaces $(X_i,x_i)$ is the limit of the underlying spaces $X_i$ equipped with the base point that projects down to each $x_i$.


structure_id: Top_*
property_id: coproducts
type: category
proof:
This follows from $\Top_* \cong 1 / \Top$ and the fact that <a href="/category/Top">$\Top$</a> has wide pushouts.


structure_id: Top_*
property_id: generator
type: category
proof:
The discrete space $\{0,1\}$ with base point $0$ is a generator since it represents the forgetful functor $\Top_* \to \Set$.


structure_id: Top_*
property_id: cogenerator
type: category
proof:
It is easily checked that the indiscrete two-point space $\{0,1\}$ with base point $1$ is a cogenerator.


structure_id: Top_*
property_id: regular subobject classifier
type: category
proof:
The indiscrete two-point space $\{0,1\}$ with base point $1$ is a regular subobject classifier since pointed continuous maps $X \to \{0,1\}$ correspond to pointed subsets of $X$ (by taking the fiber of $1$ as usual).


structure_id: Top_*
property_id: counital
type: category
proof:
Since embeddings are regular monomorphisms in this category (see below) and hence strong monomorphisms, it suffices to prove that the canonical morphism $X \vee Y \hookrightarrow X \times Y$ is an embedding. For a proof, see <a href="https://math.stackexchange.com/questions/4055988" target="_blank">MSE/4055988</a>.


structure_id: Top_*
property_id: CIP
type: category
proof:
This follows since <a href="/category/Set_*">$\Set_*$</a> has this property and the forgetful functor preserves products and coproducts.


structure_id: Top_*
property_id: cocartesian cofiltered limits
type: category
proof:
We continue the proof for <a href="/category/Set_*">$\Set_*$</a> by showing that the natural bijective map $$\textstyle \alpha : X \vee \lim_i Y_i \to \lim_i (X \vee Y_i)$$ is open. It suffices to consider open sets of two types: (1) If $U \subseteq X$ is open, the $\alpha$-image of $U \vee \lim_i Y_i$ is $p_{i_0}^{-1}(U \vee Y_{i_0})$ for any chosen index $i_0$, hence open. (2) If $i$ is an index and $V_i \subseteq Y_i$ is open, then the $\alpha$-image of $X \vee (p_i^{-1}(V_i) \cap \lim_i Y_i)$ is $p_i^{-1}(X \vee V_i)$, hence open.


structure_id: Top_*
property_id: coregular
type: category
proof:
Regular monomorphisms coincide with the embeddings (see below). Since <a href="/category/Top">$\Top$</a> is coregular, they are stable under pushouts, and pushouts in $\Top_*$ are the same.


structure_id: Top_*
property_id: balanced
type: category
proof:
If $X$ is a set with a base point $x_0$, consider the discrete space $X_d$ on $X$ and the indiscrete space $X_i$ on $X$. The identity map $X \to X$ lifts to a continuous map $X_d \to X_i$ preserving $x_0$, which is bijective and therefore both a mono- and an epimorphism, but it is not an isomorphism unless $X = \{x_0\}$.


structure_id: Top_*
property_id: regular
type: category
proof:
See Example 3.14 at the <a href="https://ncatlab.org/nlab/show/regular+category" target="_blank">nLab</a>. The proof also works for pointed spaces (resp. posets) by using the base points $a$ and $0$.


structure_id: Top_*
property_id: accessible
type: category
proof:
In fact, it does not have any small colimit-dense subcategory by <a href="https://math.stackexchange.com/questions/4097315/" target="_blank">MSE/4097315</a>. The proof easily adapts to pointed spaces.


structure_id: Top_*
property_id: cartesian filtered colimits
type: category
proof:
The functor $\IQ \times - : \Top_* \to \Top_*$ does not preserve colimits, see <a href="https://math.stackexchange.com/questions/2969372" target="_blank">MSE/2969372</a>. The counterexample also works for pointed spaces.


structure_id: Top_*
property_id: co-Malcev
type: category
proof:
We can adjust the proof for <a href="/category/Top">$\Top$</a> as follows: Consider the forgetful functor $U : \Top_* \to \Set$ and the relation $R \subseteq U^2$ defined by $R(X) \coloneqq \{(x,y) \in U(X)^2 : x \in \overline{\{y\}} \}$. Both are representable: $U$ by the discrete space $\{0,1\}$ with base point $0$ and $R$ by the Sierpinski space with an isolated base point added. It is clear that $R$ is reflexive, but not symmetric.


structure_id: Top_*
property_id: regular quotient object classifier
type: category
proof:
We can recycle the proof for <a href="/category/Set_*">$\Set_*$</a> using discrete topological spaces.


structure_id: Top_*
property_id: coaccessible
type: category
proof:
We can adjust the proof for <a href="/category/Top">$\Top$</a> as follows: Assume $\Top_*$ is coaccessible. Let $S_0=\{x,*\}$ be the pointed topological space such that $\{*\}$ is the only non-trivial open set, and let $S_1=\{x,*\}$ be the pointed space such that $\{x\}$ is the only non-trivial open set. Let $p_i : S_i \to \{x,*\}$ be the identity function to the two-element indiscrete pointed space. Then, a pointed topological space is discrete if and only if it is projective to the morphisms $p_0$ and $p_1$. This implies that the full subcategory spanned by all discrete pointed spaces, which is equivalent to $\Set_*$, is coaccessible by Prop. 4.7 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>. However, since <a href="/category/Set_*">$\Set_*$</a> is not coaccessible, this is a contradiction.


structure_id: Top_*
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Set_*">$\Set_*$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set_* \to \Top_*$ that equips a pointed set with the indiscrete topology.


structure_id: Top_*
property_id: effective congruences
type: category
proof:
Suppose that $\Top_*$ had effective congruences. Then by <a href="/content/coslice-effective-congruences">this result</a>, <a href="/category/Top">$\Top$</a> would also have effective congruences, which we know is not the case.


structure_id: Top_*
property_id: effective cocongruences
type: category
proof:
This counterexample is adapted from the counterexample for <a href="/category/Top">$\Top$</a>. Consider the pointed topological space $I \coloneqq \{ *, a, b \}$ with topology $\{ \varnothing, \{ * \}, \{ a, b \}, \{ *, a, b \} \}$. This represents the functor which sends a pointed topological space $X$ to the pairs of indistinguishable points of $X$. Therefore, we get a cocongruence $\{ *, a \} \rightrightarrows I$ on the discrete space $\{ *, a \}$, where the maps are $*\mapsto *, a\mapsto a$ and $*\mapsto *, a\mapsto b$ respectively. However, this cannot be effective: if we have $h : Z \to \{ *, a \}$ which equalizes the cocongruence, then $h$ must be the constant function with value $*$. But that means the cokernel pair of $h$ is the discrete space on $\{ *, a, b \}$.


structure_id: TorsAb
property_id: cocomplete
type: category
proof:
The embedding $\TorsAb \hookrightarrow \Ab$ is closed under colimits and <a href="/category/Ab">$\Ab$</a> is cocomplete.


structure_id: TorsAb
property_id: complete
type: category
proof:
The embedding $\TorsAb \hookrightarrow \Ab$ has a right adjoint, sending an abelian group $A$ to its torsion subgroup $T(A)$. Since <a href="/category/Ab">$\Ab$</a> is complete, $\TorsAb$ is complete as well. The limit of a diagram of torsion abelian groups is the torsion subgroup of the limit of the underlying abelian groups. Notice that the torsion subgroup is not required in the case of equalizers, since a subgroup of a torsion abelian group is already torsion. Also, a <i>finite</i> product of torsion abelian groups is already torsion.


structure_id: TorsAb
property_id: normal
type: category
proof:
If $f : A \to B$ is a monomorphism, it is injective (see below). In <a href="/category/Ab">$\Ab$</a> it is then the kernel of $B \to B/f(A)$. Since $B/f(A)$ is torsion, it is also the kernel in $\TorsAb$.


structure_id: TorsAb
property_id: conormal
type: category
proof:
If $f : A \to B$ is an epimorphism, it is surjective (see below). In <a href="/category/Ab">$\Ab$</a> it is then the cokernel of its kernel $K \hookrightarrow A$. Since $K$ is torsion, it is also the cokernel in $\TorsAb$.


structure_id: TorsAb
property_id: finitely accessible
type: category
proof:
We already know that (filtered) colimits exist and are preserved by the forgetful functor to $\Ab$. Every torsion abelian group is the filtered colimit of its finitely generated subgroups (which are finite). These are finitely presentable in $\Ab$, hence also in $\TorsAb$.


structure_id: TorsAb
property_id: CSP
type: category
proof:
The canonical homomorphism $\bigoplus_{n \geq 0} \IZ/2 \to T(\prod_{n \geq 0} \IZ/2)$ is not surjective, since the torsion element $(1,1,\dotsc)$ does not lie in the image. Hence, it is no epimorphism.


structure_id: TorsAb
property_id: locally strongly finitely presentable
type: category
proof:
Assume that it is locally strongly finitely presentable, and therefore the category of algebras of a multi-sorted finitary algebraic theory. For each sort $s$ let $F_s$ be the free algebra on one generator of sort $s$. One of them must be non-trivial, since otherwise the category would be thin. So assume $F_s \neq 0$. It is finitely presentable, hence a finite abelian group, and regular projective. Any direct summand will have the same properties. So we find that some $\IZ/p^k$ (with $k \geq 1$ and $p$ prime) is regular projective. However, the exact sequence in $\TorsAb$ $$0 \to \IZ/p \to \IZ/p^{k+1} \to \IZ/p^k \to 0$$ does not split.


structure_id: TorsFreeAb
property_id: complete
type: category
proof:
The embedding $\TorsFreeAb \hookrightarrow \Ab$ is closed under limits and <a href="/category/Ab">$\Ab$</a> is complete.


structure_id: TorsFreeAb
property_id: cocomplete
type: category
proof:
The embedding $\TorsFreeAb \hookrightarrow \Ab$ has a left adjoint, sending an abelian group $A$ to its torsion-free reflection $A/T(A)$, where $T(A)$ is the torsion subgroup of $A$. Since <a href="/category/Ab">$\Ab$</a> is cocomplete, $\TorsFreeAb$ is cocomplete as well. The colimit of a diagram of torsion-free abelian groups is the torsion-free reflection of the colimit of the underlying abelian groups. Notice that the reflection is not required in the case of coproducts: the direct sum of torsion-free abelian groups is again torsion-free. It is also not required for filtered colimits.


structure_id: TorsFreeAb
property_id: finitely accessible
type: category
proof:
We already saw that filtered colimits exist and are preserved by the forgetful functor to $\Ab$. Every torsion-free abelian group is the filtered colimit of its finitely generated subgroups, which are in fact free. Finitely generated free abelian groups are finitely presentable in $\Ab$ and therefore also in $\TorsFreeAb$.


structure_id: TorsFreeAb
property_id: cogenerator
type: category
proof:
The additive group $\IQ$ is a cogenerator since every torsion-free abelian group $A$ embeds into $A \otimes \IQ$, which is a vector space over $\IQ$, and by linear algebra $K$ is a cogenerator in the category of vector spaces over $K$.


structure_id: TorsFreeAb
property_id: regular
type: category
proof:
The regular epimorphisms are exactly the surjective homomorphisms (see below), and these are clearly stable under pullbacks.


structure_id: TorsFreeAb
property_id: coregular
type: category
proof:
It suffices to prove that regular monomorphisms (which are classified below) are stable under pushouts. Let $i : A \to B$ be a regular monomorphism in $\TorsFreeAb$, i.e. $i$ is injective and its $\Ab$-cokernel $B/i(A)$ is torsion-free, and let $f : B \to C$ be any morphism in $\TorsFreeAb$. Their $\Ab$-pushout is $$P = (B \times C)/\{(i(a),-f(a)): a \in A\}.$$ It is torsion-free: If $n \in \IZ \setminus \{0\}$ and $n (b,c) = (i(a),-f(a))$, there is some $a' \in A$ with $b = i(a')$ since $B/i(A)$ is torsion-free. It follows $n a' = a$, and then $c = -f(a')$ since $C$ is torsion-free. Thus, $(b,c) = (i(a'),-f(a'))$, which proves our claim. Therefore, $P$ is also the pushout in $\TorsFreeAb$. The homomorphism $j : C \to P$, $j(c) = [0,c]$ is injective (since <a href="/category/Ab">$\Ab$</a> is coregular, but a direct proof is also easy), and by the universal property of $P$ its $\Ab$-cokernel is isomorphic to the $\Ab$-cokernel of $i$, which is torsion-free.


structure_id: TorsFreeAb
property_id: balanced
type: category
proof:
It can be checked directly that $2 : \IZ \to \IZ$ is both a monomorphism and an epimorphism, but no isomorphism. This also follows from the general classification of mono- and epimorphisms below.


structure_id: TorsFreeAb
property_id: CSP
type: category
proof:
The canonical homomorphism $\bigoplus_{n \geq 0} \IZ \to \prod_{n \geq 0} \IZ$ is injective, but not an epimorphism, since the quotient $\prod_{n \geq 0} \IZ / \bigoplus_{n \geq 0} \IZ$ is not torsion. In fact, it is torsion-free and non-zero.


structure_id: Vect
property_id: split abelian
type: category
proof:
That $\Vect$ is abelian is a standard fact, see <a href="https://ncatlab.org/nlab/show/Categories+for+the+Working+Mathematician" target="_blank">Mac Lane</a>, Ch. VIII. Furthermore, it is a fact from linear algebra that every subspace has a complement, which is why every short exact sequence splits.


structure_id: Vect
property_id: CSP
type: category
proof:
The canonical homomorphism $\bigoplus_{n \geq 0} K \to \prod_{n \geq 0} K$ is not surjective, hence no epimorphism.


structure_id: Z
property_id: effective congruences
type: category
proof:
If we have a congruence $E \rightrightarrows X$ in $[\CRing, \Set]$, then evaluating at any commutative ring gives a congruence in $\Set$. Defining $Y$ pointwise to be the quotient of this congruence, we get a morphism of functors $h : X \to Y$, and by <a href="/content/effective-congruence-quotients">this result</a> applied pointwise, the kernel pair of $h$ is $E$.


structure_id: Z
property_id: effective cocongruences
type: category
proof:
If we have a cocongruence $X\rightrightarrows E$ in $[\CRing, \Set]$, then evaluating at any commutative gives a cocongruence in $\Set$. Defining $Y$ pointwise to be the equalizer of the pair, we get a morphism of functors $h : Y \to X$, and by the dual of <a href="/content/effective-congruence-quotients">this result</a> applied pointwise, the cokernel pair of $h$ is $E$.


structure_id: Z
property_id: Malcev
type: category
proof:
Any counterexample for <a href="/category/Set">$\Set$</a> (i.e., any non-symmetric reflexive relation) yields one for this category by taking constant functors.


structure_id: Z
property_id: semi-strongly connected
type: category
proof:
This is because already the full subcategory of representable functors is not semi-strongly connected, because <a href="/category/CRing">$\CRing$</a> is not semi-strongly connected. Specifically, there is no morphism between $\Hom(\IF_2,-)$ and $\Hom(\IF_3,-)$.


structure_id: Z
property_id: cartesian closed
type: category
proof:
There are functors $F,G : \CRing \rightrightarrows \Set$ such that $\Hom(F,G)$ is not essentially small, see <a href="https://mathoverflow.net/questions/390611" target="_blank">MO/390611</a> for example. Now if the exponential $[F,G] : \CRing \to \Set$ exists, we get $[F,G](\IZ) \cong \Hom(\Hom(\IZ,-),[F,G])$ by Yoneda, which simplifies to $\Hom(1,[F,G]) \cong \Hom(1 \times F,G) \cong \Hom(F,G)$, a contradiction.


structure_id: Z
property_id: well-powered
type: category
proof:
Consider the functor $F$ from <a href="https://mathoverflow.net/questions/390611" target="_blank">MO/390611</a> for example. The collection of subobjects of $F$ is not isomorphic to a set: for each infinite cardinal $\kappa$, simply cut off the construction of $F$ at $\kappa$. This yields a different subobject for each $\kappa$.


structure_id: Z
property_id: cofiltered-limit-stable epimorphisms
type: category
proof:
We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to [\CRing, \Set]$ that maps a set to its constant functor.


structure_id: Z_div
property_id: distributive
type: category
proof:
We need to prove $\lcm_i \gcd(a, b_i) \cong \gcd(a, \lcm_i b_i)$ for finite families. If $x$ denotes the LHS and $y$ denotes the RHS, the relation $x \mid y$ is formal. If $v_p(-) : \IZ \to \IN_{\infty}$  denotes the multiplicity of a prime $p$, then $v_p(x)$ equals $\max_i \min(v_p(a),v_p(b_i))$, and $v_p(y)$ equals $\min(v_p(a), \max_i v_p(b_i))$. Since our family is finite, there is some $i_0$ with $\max_i v_p(b_i) = v_p(b_{i_0})$. Then $v_p(x) \geq \min(v_p(a),v_p(b_{i_0})) = v_p(y)$. This proves $y \mid x$.


structure_id: Z_div
property_id: self-dual
type: category
proof:
The only integer with infinitely many divisors (up to isomorphism) is $0$. But many integers have infinitely many multiples (up to isomorphism).


structure_id: Z_div
property_id: countably distributive
type: category
proof:
We have $2 \times \coprod_n 3^n = \gcd(2,\lcm_n(3^n)) = \gcd(2,0) = 2$, but $\coprod_n (2 \times 3^n) = \lcm_n \gcd(2,3^n) = \lcm_n 1  = 1$.


structure_id: Z_div
property_id: countably codistributive
type: category
proof:
If $p$ runs through all odd primes, we have $2 \sqcup \prod_p p = \lcm(2,\gcd_p p) = \lcm(2,0) = 0$, but $\prod_p (2 \sqcup p) = \gcd_p (\lcm(2,p)) = \gcd_p (2 \cdot p) = 2$.


structure_id: real_interval
property_id: locally ℵ₁-presentable
type: category
proof:
See <a href="https://math.stackexchange.com/questions/4481902/locally-presentable-vs-compactly-presentable-categories#comment9399784_4482185" target="_blank">MSE/4481902</a>.


structure_id: real_interval
property_id: locally finitely presentable
type: category
proof:
It suffices to prove that $0$ (the initial object) is the only finitely presentable object. If $s > 0$, then $s = \sup_{n \in \IN, \, s \geq 1/n } (s - 1/n)$, but there is no $n$ with $s \leq s - 1/n$.


structure_id: sSet
property_id: locally small
type: category
proof:
This follows from the general fact that $[\C,\D]$ is locally small when $\C$ is small and $\D$ is locally small, here applied to $\C = \Delta^{\op}$ and $\D = \Set$.


structure_id: sSet
property_id: semi-strongly connected
type: category
proof:
Let $X,Y$ be two simplicial sets. Assume that $X_0$ is empty. Then $X_n$ is empty for all $n$ since there is a morphism $[0] \to [n]$, hence a map $X_n \to X_0$. So there is a morphism $X \to Y$ for trivial reasons. If $X_0$ is non-empty, pick an element. By the Yoneda Lemma it corresponds to a morphism $\Delta^0 \to X$. Since $\Delta^0 = 1$ is terminal, there is a morphism $Y \to \Delta^0$, and these compose to a morphism $Y \to X$.


structure_id: sSet
property_id: generator
type: category
proof:
Let $\Delta^n \coloneqq \Hom([n],-)$ be the standard $n$-simplex for $n \geq 0$. The set $\{\Delta^n : n \geq 0\}$ is a <a href="/category-property/generating_set">generating set</a> by the Yoneda Lemma. For all $n,m$ there is a morphism $[n] \to [m]$ in $\Delta$ and hence a morphism $\Delta^m \to \Delta^n$ in $\sSet$. Then by <a href="/content/generator_construction">this lemma</a> the coproduct $\coprod_{n \geq 0} \Delta^n$ is a generator in $\sSet$.


structure_id: sSet
property_id: locally strongly finitely presentable
type: category
proof:
This follows from the fact that every category of presheaves over a small category is locally strongly finitely presentable.


structure_id: sSet
property_id: finitary algebraic
type: category
proof:
A one-sorted finitary algebraic category has an object $F$ (the free algebra on one generator) such that $F$ is finitely presentable and every object $X$ admits an epimorphism $\coprod_{s \in S} F \to X$ for some index set $S$. Assume that such a simplicial set $F$ exists. By using the sequence of $n$-skeletons of $F$, we see that there is some $n$ such that every $n$-simplex in $F$ is degenerate. Now take $X = \Delta^n$, which has a non-degenerate $n$-simplex. Then there cannot be an epimorphism $\coprod_{s \in S} F \to X$.


structure_id: walking_commutative_square
property_id: locally strongly finitely presentable
type: category
proof:
This is because the <a href="/category/walking_morphism">walking morphism</a> has this property. Alternatively, we may represent this category as the category of algebras for the finitary algebraic theory with two sorts $S_1,S_2$, the equation $x=y$ for $x,y \in S_1$, and the equation $x=y$ for $x,y \in S_2$.


structure_id: walking_composable_pair
property_id: locally strongly finitely presentable
type: category
proof:
Take the two-sorted (finitary) algebraic theory with exactly one unary operation between them and the equation $x=y$ for each sort. There are exactly three algebras for this theory up to isomorphism: the identities on the empty set and the singleton, the morphism from the empty set to the singleton. Hence we get the equivalence to $\{0 \to 1 \to 2\}$.


structure_id: walking_coreflexive_pair
property_id: terminal object
type: category
proof:
The object $[0]$ is terminal since it is already terminal in <a href="/category/Delta">$\Delta$</a>.


structure_id: walking_coreflexive_pair
property_id: generator
type: category
proof:
The object $[0]$ is generator since this is already true in <a href="/category/Delta">$\Delta$</a>. A direct proof is also possible.


structure_id: walking_coreflexive_pair
property_id: cogenerator
type: category
proof:
The object $[1]$ is cogenerator since this is already true in <a href="/category/Delta">$\Delta$</a>. A direct proof is also possible.


structure_id: walking_coreflexive_pair
property_id: epi-regular
type: category
proof:
The only non-identity epimorphism is $p$, which is the coequalizer of $\id, ip : [1] \rightrightarrows [1]$ (since $pi = \id$).


structure_id: walking_coreflexive_pair
property_id: mono-regular
type: category
proof:
The only non-identity monomorphisms are $i$ and $j$. The morphism $i$ is the equalizer of $\id, ip : [1] \rightrightarrows [1]$ (since $pi = \id$), and for $j$ it is the same argument.


structure_id: walking_coreflexive_pair
property_id: coequalizers
type: category
proof:
We already know that the <a href="/category/Delta">$\Delta$</a> has coequalizers, and the proof has shown that the cardinality does not increase, so we are done. But a direct proof is also possible: There are four non-equal parallel pairs: $(i,j)$, $(ip,jp)$, $(\id,ip)$, and $(\id,jp)$. The first two have the same coequalizer (if it exists) since $p$ is an epimorphism, the last two are symmetric, and we already remarked that $p$ is a coequalizer of $(\id,ip)$. So it suffices to check that $p$ is a coequalizer of $i,j$, which is easy.


structure_id: walking_coreflexive_pair
property_id: cosifted
type: category
proof:
Our proof that the <a href="/category/Delta">$\Delta$</a> is cosifted has only used $[0],[1]$ as auxiliary objects and therefore also shows that $\Delta^{\leq 1}$ is cosifted.


structure_id: walking_coreflexive_pair
property_id: generalized variety
type: category
proof:
This actually holds for every truncated simplex category $\Delta^{\leq n}$. See <a href="https://mathoverflow.net/questions/510760" target="_blank">MO/510760</a> for a proof that sifted colimits exist. See <a href="https://mathoverflow.net/questions/510827" target="_blank">MO/510827</a> for a proof that every object is strongly finitely presentable.


structure_id: walking_coreflexive_pair
property_id: coreflexive equalizers
type: category
proof:
The coreflexive pair $i,j : [0] \rightrightarrows [1]$ has no equalizer, in fact is not equalized by any morphism.


structure_id: walking_coreflexive_pair
property_id: pushouts
type: category
proof:
Assume that $[1] \xleftarrow{i} [0] \xrightarrow{i} [1]$ has a pushout in $\Delta^{\leq 1}$, where $i(0)=0$. This amounts to a universal totally ordered set of cardinality $\leq 2$ with elements $a,b,c$ satisfying $a \leq b$, $a \leq c$. Since a finite totally ordered set has trivial automorphism group, the automorphism defined by $a \mapsto a$, $b \mapsto c$, $c \mapsto b$ must be the identity, i.e., we have $b = c$. However, in $[1]$ the equations $0 \leq 0$, $0 \leq 1$ then show that the universal property fails.


structure_id: walking_coreflexive_pair
property_id: multi-complete
type: category
proof:
This follows directly from existing results: If its dual, the walking reflexive pair, was multi-cocomplete, then since it is also accessible, it would be <a href="/category-property/locally_multi-presentable">locally multi-presentable</a>. But then it would have connected limits, in particular pullbacks.


structure_id: walking_fork
property_id: locally cartesian closed
type: category
proof:
We need to check that every slice category is cartesian closed. The slice category over $0$ is the <a href="/category/1">trivial category</a>. The slice category over $1$ is the <a href="/category/walking_morphism">walking morphism</a>. Finally, the slice category over $2$ ist the <a href="/category/walking_commutative_square">walking commutative square</a>. All of these are cartesian closed, see their pages for details.


structure_id: walking_fork
property_id: binary powers
type: category
proof:
Assume that $X \coloneqq 2 \times 2$ exists. Since there is a diagonal morphism $2 \to X$, we must have $X = 2$, and the two projections $p_1,p_2 : X \rightrightarrows 2$ must be equal to the identity. But $f,g$ induce a morphism $(f,g) : 1 \to X$ with $p_1 (f,g) = f$ and $p_2 (f,g) = g$, so that $f=g$, a contradiction.


structure_id: walking_idempotent
property_id: preadditive
type: category
proof:
The monoid $\{1,e\}$ with $e^2=e$ is the underlying multiplicative monoid of the ring $\IZ/2$, where $e=0$. Thus, the (unique) preadditive structure is given by $1 + e = e + 1 = 1$, $e + e = e$ and $1 + 1 = e$.


structure_id: walking_idempotent
property_id: ℵ₁-filtered
type: category
proof:
In fact, the walking idempotent is $\kappa$-filtered for every regular infinite cardinal $\kappa$: Every diagram $D : \I \to \Idem$ has the cocone $(e : D(i) \to 0)_{i \in \I}$.


structure_id: walking_morphism
property_id: finitary algebraic
type: category
proof:
Take the algebraic theory with no operations but with the equation $x=y$ that is supposed to hold for all elements $x,y$. The algebras for this theory are the empty set and the singleton set, hence we get the equivalence to $I$.


structure_id: walking_pair
property_id: pullbacks
type: category
proof:
The two morphisms $a,b : 0 \rightrightarrows 1$ have no pullback, since it would have to consist of identities $0 \leftarrow 0 \rightarrow 0$, but $a \neq b$.


structure_id: walking_span
property_id: locally cartesian closed
type: category
proof:
The slice category over $0$ is the <a href="/category/1">trivial category</a>, and the slice category over $1$ is the <a href="/category/walking_morphism">walking morphism</a>, which is cartesian closed. The same holds for $2$ by symmetry.


structure_id: walking_splitting
property_id: self-dual
type: category
proof:
There is an isomorphism $\Split^\op \to \Split$ defined by $0 \mapsto 0$, $1 \mapsto 1$, $i \mapsto p$, $p \mapsto i$.


structure_id: walking_splitting
property_id: pointed
type: category
proof:
The object $0$ is initial and terminal. This also means that $\id_0, i, p, ip$ are zero morphisms. The only non-zero morphism is $\id_1$.


structure_id: walking_splitting
property_id: equalizers
type: category
proof:
The only parallel pair of non-equal morphisms is $\id_1, ip : 1 \rightrightarrows 1$, and their equalizer is $i$.


structure_id: walking_splitting
property_id: generator
type: category
proof:
The object $1$ a generator, since the only parallel pair of non-equal morphisms is $\id_1, ip : 1 \rightrightarrows 1$ with domain $1$.


structure_id: walking_splitting
property_id: preadditive
type: category
proof:
We can define $\id_1 + \id_1 \coloneqq ip$ (and it is clear how to add zero morphisms) and then verify that the axioms of a preadditive category hold. Alternatively, it suffices to find a preadditive category which is isomorphic to the walking splitting: Consider the full subcategory of $\Vect_{\IF_2}$ that consists only of the trivial vector space $\{0\}$ and $\IF_2$. Since <a href="/category/Vect">$\Vect_{\IF_2}$</a> is preadditive, it is preadditive as well. It has two objects, two identities, the morphisms $i : \{0\} \to \IF_2$, $p : \IF_2 \to \{0\}$, and the zero morphism $ip : \IF_2 \to \IF_2$. Clearly, $pi$ is the identity.


structure_id: walking_splitting
property_id: sifted colimits
type: category
proof:
We work with the representation of the category as $\Vect^{\leq 1}_{\IF_2}$, the category of vector spaces over $\IF_2$ of dimension $\leq 1$. It suffices to show that it is closed under sifted colimits in <a href="/category/Vect">$\Vect_{\IF_2}$</a>. More generally, we show this for $\Vect^{\leq d}_K \subseteq \Vect_K$, where $d \in \IN$ and $K$ is a field. So let $X : \I \to \Vect_K$ be a sifted diagram with colimit $(u_i : X_i \to X_\infty)_{i \in \I}$. Since $\I$ is sifted, for finitely many objects $i_1,\dotsc,i_n \in \I$ there is an object $k$ that admits morphisms $i_1 \to k, \dotsc, i_n \to k$; this is all we need to know about $\I$. Assume that each $X_i$ is of dimension $\leq d$, we need to show this for $X_\infty$ as well.
Every element in $X_\infty$ is a finite sum of elements of the form $u_i(x_i)$ with $x_i \in X_i$. Choose an object $k$ with morphisms $i \to k$ for every occurring $i$. If $y_i \in X_k$ denotes the image of $x_i$, we get $\sum_i u_i(x_i) = \sum_i u_k(y_i) = u_k(\sum_i y_i)$. Therefore, every element of $X_\infty$ has the form $u_i(x_i)$ for some $i \in \I$ and $x_i \in X_i$. Moreover, for finitely many elements in $X_\infty$ the index $i$ may be chosen uniformly.
Now, if $X_\infty$ has dimension $> d$, it would have linearly independent vectors $v_0,\dotsc,v_d$, all of which have a preimage in $X_i$ for some $i \in \I$. But then these preimages would be linearly independent as well, which contradicts $\dim(X_i) \leq d$.


structure_id: walking_splitting
property_id: generalized variety
type: category
proof:
Again we work with $\Vect^{\leq 1}_{\IF_2}$. We already know that it has sifted colimits and that the embedding to $\Vect_{\IF_2}$ preserves them. The object $0$ is initial and hence strongly finitely presentable. The object $\IF_2$ is strongly finitely presentable in $\Vect^{\leq 1}_{\IF_2}$ since its hom-functor is the composition of the embedding and the forgetful functor $\Vect_{\IF_2} \to \Set$, and the latter preserves sifted colimits by <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01, Lemma 3.3]</a> applied to $\IF_2 \in \Vect_{\IF_2}$.


structure_id: abelianization
property_id: reflector
type: functor
proof:
This functor is left adjoint to the forgetful functor $U_{\Ab,\Grp} : \Ab \hookrightarrow \Grp$, which is fully faithful.


structure_id: abelianization
property_id: faithful
type: functor
proof:
Both the inclusion $A_3 \hookrightarrow S_3$ and the trivial homomorphism $A_3 \to S_3$ are mapped to the trivial homomorphism $A_3 \to C_2$.


structure_id: abelianization
property_id: preserves products
type: functor
proof:
If $G$ is a group, the canonical homomorphism $(G^{\IN})^{\ab} \to (G^{\ab})^{\IN}$ is surjective, but does not need to be an isomorphism: otherwise, the inclusion $[G^{\IN}, G^{\IN}] \subseteq [G,G]^{\IN}$ would be an equality. But this requires the commutator width of $G$ to be finite, which fails for $G = F_2$ for instance. See also <a href="https://arxiv.org/abs/1608.02220">The abelianization of inverse limits of groups</a>, Remark 0.0.7.


structure_id: binary_coproduct_sets
property_id: right-invertible
type: functor
proof:
The functor $\Set \to \Set \times \Set$, $X \mapsto (X,0)$ provides a right inverse since $X + 0 \cong X$.


structure_id: binary_coproduct_sets
property_id: left adjoint
type: functor
proof:
The coproduct functor is left adjoint to the diagonal functor since $$\Hom(A + B,X) \cong \Hom(A,X) \times \Hom(B,X) \cong \Hom((A,B),(X,X)).$$


structure_id: binary_coproduct_sets
property_id: cofinitary
type: functor
proof:
We already know that <a href="/category/Set">$\Set$</a> has cocartesian cofiltered limits; this is a consequence of <a href="/category-implication/extensive_cocartesian_cofiltered_limits">this result</a>.


structure_id: binary_coproduct_sets
property_id: preserves binary products
type: functor
proof:
The canonical map $(X_1 \times X_2) + (Y_1 \times Y_2) \to (X_1 + Y_1) \times (X_2 + Y_2)$ is injective, but misses the "mixed terms" $X_1 \times Y_2$ and $Y_1 \times X_2$ in the codomain, hence is not surjective when $X_1,Y_2$ (or $Y_1,X_2$) are non-empty.


structure_id: binary_coproduct_sets
property_id: reflector
type: functor
proof:
Its right adjoint, the diagonal functor $\Delta : \Set \to \Set \times \Set$, is faithful, but not full.


structure_id: binary_product_sets
property_id: right-invertible
type: functor
proof:
The functor $\Set \to \Set \times \Set$, $X \mapsto (X,1)$ provides a right inverse since $X \times 1 \cong X$.


structure_id: binary_product_sets
property_id: preserves reflexive coequalizers
type: functor
proof:
See Prop. 2.4 at the <a href="https://ncatlab.org/nlab/show/reflexive+coequalizer" target="_blank">nLab</a>.


structure_id: binary_product_sets
property_id: full
type: functor
proof:
If $X$ is a non-empty set, the unique map of empty sets $X \times \varnothing \to \varnothing \times X$ is not induced by a map $X \to \varnothing$.


structure_id: binary_product_sets
property_id: faithful
type: functor
proof:
If $f,g : A \rightrightarrows B$ are two maps, we always have $f \times \id_{\varnothing} = g \times \id_{\varnothing} : A \times \varnothing \rightrightarrows B \times \varnothing$.


structure_id: binary_product_sets
property_id: preserves coequalizers
type: functor
proof:
Since the diagonal functor $\Delta : \Set \to \Set \times \Set$ preserves coequalizers (in fact, all colimits), it would follow that the composition $\Set \to \Set$, $X \mapsto X^2$ preserves coequalizers. But this is the <a href="/functor/squaring_sets">squaring functor</a>, from we already know that it does not preserve coequalizers.


structure_id: binary_product_sets
property_id: coreflector
type: functor
proof:
Its left adjoint, the diagonal functor $\Delta : \Set \to \Set \times \Set$, is faithful, but not full.


structure_id: continuous_functions
property_id: continuous
type: functor
proof:
Since the forgetful functor $U : \CAlg(\IR) \to \Set$ is conservative and continuous, it suffices to prove that $U \circ C : \Top^{\op} \to \Set$ is continuous. But this functor is representable.


structure_id: continuous_functions
property_id: preserves initial objects
type: functor
proof:
The initial object in $\Top^{\op}$ is the singleton space, which is mapped to $\IR$, the initial commutative $\IR$-algebra.


structure_id: continuous_functions
property_id: essentially injective
type: functor
proof:
If $X,Y$ are non-empty indiscrete spaces, then $C(X) \cong \IR \cong C(Y)$ since every continuous function on $X$ resp. $Y$ is constant.


structure_id: continuous_functions
property_id: dominant
type: functor
proof:
The algebra $C(X)$ is <a href="https://en.wikipedia.org/wiki/Reduced_ring" target="_blank">reduced</a>, and every subalgebra of a reduced algebra is also reduced. But there exist non-reduced algebras such as $\IR[T]/\langle T^2 \rangle$.


structure_id: continuous_functions
property_id: faithful
type: functor
proof:
If $X,Y$ are non-empty indiscrete spaces, there is only one homomorphism of $\IR$-algebras $C(Y) \to C(X)$ (since both are isomorphic to $\IR$), but there can be many maps $X \to Y$.


structure_id: continuous_functions
property_id: full
type: functor
proof:
Every continuous function $\omega_1 \to \IR$ is eventually constant, where $\omega_1$ denotes the first uncountable ordinal equipped with the order topology. We obtain a homomorphism $$\varphi : C(\omega_1) \to \IR \cong C(1)$$ mapping each continuous function $\omega_1 \to \IR$ to its eventual value. Assume that it is induced by a continuous map $1 \to \omega_1$. This corresponds to an ordinal $\alpha < \omega_1$ such that $\varphi(u) = u(\alpha)$ for all $u \in C(\omega_1)$. But the continuous function $$u(\beta) \coloneqq \begin{cases} 0 & \beta \leq \alpha \\ 1 & \beta > \alpha \end{cases}$$ satisfies $\varphi(u) = u(\alpha + 1) \neq u(\alpha)$.


structure_id: continuous_functions
property_id: preserves finite coproducts
type: functor
proof:
The canonical homomorphism $C(X) \otimes_{\IR} C(Y) \to C(X \times Y)$ need not be surjective. For example, for $X = Y = \IR$ one can show that $(x,y) \mapsto \exp(xy)$ is not contained in its image.


structure_id: continuous_functions
property_id: preserves regular epimorphisms
type: functor
proof:
The embedding $i : \IR \setminus \{0\} \hookrightarrow \IR$ is a regular monomorphism in $\Top$, i.e. a regular epimorphism in $\Top^{\op}$. It is mapped to the restriction map $i^* : C(\IR) \to C(\IR \setminus \{0\})$, which is not surjective, as witnessed by the map $x \mapsto x^{-1}$. Hence, $i^*$ is not a regular epimorphism.


structure_id: continuous_functions
property_id: preserves epimorphisms
type: functor
proof:
It is not sufficient to find an injective continuous map $i : X \to Y$ such that $i^* : C(Y) \to C(X)$ is not surjective, since epimorphisms in $\CAlg(\IR)$ do not need to be surjective. But a simple counterexample can again be constructed using (in)discrete spaces. Namely, if $X$ is any finite set with at least two elements, we have a continuous bijection $D(X) \to I(X)$, $x \mapsto x$, where $D(X)$ (resp. $I(X)$) denotes the discrete (resp. indiscrete) space on $X$. It induces the algebra homomorphism $\IR \cong C(I(X)) \to C(D(X)) \cong \IR^X$, which is not an epimorphism since $\IR^X \otimes_\IR \IR^X \not\cong \IR^X$.


structure_id: continuous_functions
property_id: finitary
type: functor
proof:
For $n \geq 0$ consider the discrete space $X_n \coloneqq \{0,\dotsc,n\}$. Using the evident truncation maps $X_{n+1} \twoheadrightarrow X_n$, their limit is isomorphic to $\IN \cup \{\infty\}$, the one-point compactification of the discrete space $\IN$. The canonical homomorphism $$\textstyle \colim_n C(X_n) \to C(\lim_n X_n)$$ can then be identified with the inclusion of the algebra of eventually constant sequences in $\IR$ into the algebra of convergent sequences in $\IR$. Thus, it is not surjective.


structure_id: countable_copower_sets
property_id: preserves binary products
type: functor
proof:
The canonical map $\IN \times (X \times Y) \to (\IN \times X) \times (\IN \times Y)$ is given by $(n,(x,y)) \mapsto ((n,x),(n,y))$. It is clearly not surjective when $X$ and $Y$ are non-empty.


structure_id: countable_copower_sets
property_id: essentially injective
type: functor
proof:
For any two non-empty finite sets $X,Y$ there is a bijection $\IN \times X \cong \IN \cong \IN \times Y$, so this does not imply $X \cong Y$.


structure_id: diagonal_sets
property_id: left adjoint
type: functor
proof:
The diagonal functor is left adjoint to the product functor since $$\Hom(X, A \times B) \cong \Hom(X,A) \times \Hom(X,B) \cong \Hom((X,X),(A,B)).$$


structure_id: diagonal_sets
property_id: right adjoint
type: functor
proof:
The diagonal functor is right adjoint to the coproduct functor since $$\Hom(A + B,X) \cong \Hom(A,X) \times \Hom(B,X) \cong \Hom((A,B),(X,X)).$$


structure_id: diagonal_sets
property_id: full on isomorphisms
type: functor
proof:
Take any pair of distinct isomorphisms $f,g : X \rightrightarrows Y$. Then $(f,g) : (X,X) \to (Y,Y)$ is not induced by an isomorphism $X \to Y$.


structure_id: diagonal_sets
property_id: dominant
type: functor
proof:
If $Y$ is a non-empty set, there is no set $X$ with a split monomorphism $(\varnothing,Y) \hookrightarrow (X,X)$. In fact, this would give a pair of epimorphisms $X \to \varnothing$ and $X \to Y$. The first one gives $X = \varnothing$, and the second one then gives $Y = \varnothing$.


structure_id: discrete_topology
property_id: dominant
type: functor
proof:
A subspace of a discrete space is again discrete. Hence, if $Y$ is any non-discrete space such as $\IR$ with the Euclidean topology, there is no set $X$ with a regular (or even split) monomorphism $Y \hookrightarrow D(X)$.


structure_id: doubling_sets
property_id: preserves equalizers
type: functor
proof:
The doubling functor can also be written as $X \mapsto \{1,2\} \times X$, from which the claim follows.


structure_id: doubling_sets
property_id: essentially injective
type: functor
proof:
Let $X,Y$ be two sets with $X + X \cong Y + Y$. If $X$ is finite, then $X + X$ and hence $Y + Y$ is finite, so that also $Y$ is finite. For the cardinalities we deduce $2 \card(X) = 2 \card(Y)$ and hence $\card(X) = \card(Y)$. This yields $X \cong Y$. If $X$ is infinite, then $Y$ is infinite, and $X \cong X + X \cong Y + Y \cong Y$.


structure_id: doubling_sets
property_id: left-invertible
type: functor
proof:
Assume that there is a functor $F : \Set \to \Set$ with natural isomorphisms  $$\alpha_X : X \to F(X+X).$$ By the Yoneda Lemma, there is some $u \in F(1+1)$ such that $\alpha_X$ is given by $$\alpha_X(x) = F((x;x) : 1+1 \to X + X)(u)$$ for $x \in X$. For the involution $\tau_X : X + X \to X + X$ we have $(x;x) = \tau_X \circ (x;x)$. Since $\alpha_X$ is surjective, it follows that $F(\tau_X) : F(X+X) \to F(X+X)$ is the identity. The inclusions $\iota_1,\iota_2 : X \to X + X$ satisfy $\tau_X \circ \iota_1 = \iota_2$, so that $F(\iota_1) = F(\iota_2)$.
Now let $f : X \to X$ be any endomorphism. It induces a map $\tilde{f} : X + X \to X$ defined by $\tilde{f} \circ \iota_1 = \id_X$ and $\tilde{f} \circ \iota_2 = f$. Thus, $F(\iota_1) = F(\iota_2)$ implies $$F(f) = F(\tilde{f} \circ \iota_2) = F(\tilde{f} \circ \iota_1) = F(\id_X) = \id_{F(X)}.$$ In particular, for every endomorphism $f : X \to X$, the induced endomorphism $F(f+f) : F(X+X) \to F(X+X)$ is the identity. By using naturality of $\alpha_X$, it follows that $f$ is the identity, which is absurd.


structure_id: enveloping_group
property_id: reflector
type: functor
proof:
By definition, this functor is left adjoint to the forgetful functor $U_{\Grp,\Mon} : \Grp \hookrightarrow \Mon$, which is fully faithful.


structure_id: enveloping_group
property_id: preserves finite products
type: functor
proof:
Since $F(1)=1$, we only need to consider binary products. The canonical homomorphism $$\alpha : F(M \times N) \to F(M) \times F(N)$$ is defined by $p_i \alpha = F(p_i)$ and satisfies $$\alpha(i_{M \times N}(m,n)) = (i_M(m),i_N(n)).$$ We construct an inverse homomorphism $$\beta : F(M) \times F(N) \to F(M \times N)$$ as follows. The two canonical homomorphisms $M \rightarrow M \times N \leftarrow N$ induce homomorphisms $F(M) \rightarrow F(M \times N) \leftarrow F(N)$. We claim that for every pair of elements $x \in F(M)$, $y \in F(N)$, their images in $F(M \times N)$ commute. Notice that $x$ is a product of elements of the form $i_M(m)$ or $i_M(m)^{-1}$ for $m \in M$, and likewise $y$ is a product of elements of the form $i_N(n)$ or $i_N(n)^{-1}$ for $n \in N$. Thus, by standard arguments from group theory, it suffices to prove that the images of $i_M(m)$ and $i_N(n)$ commute in $F(M \times N)$. This follows since the images of $m$ and $n$ in $M \times N$ commute. Therefore, the two homomorphisms induce a homomorphism $\beta : F(M) \times F(N) \to F(M \times N)$. It is characterized by $$\beta(i_M(m),1) = i_{M \times N}(m,1),\quad \beta(1,i_N(n)) = i_{M \times N}(1,n).$$ It is straightforward to check that it is inverse to $\alpha$ by arguing with the generators in both groups.


structure_id: enveloping_group
property_id: faithful
type: functor
proof:
Take any non-trivial monoid $M$ with an absorbing element, such as $(\IN,\cdot,1)$. Its enveloping group is trivial. Hence, $\id_M,1 : M \rightrightarrows M$ provide a counterexample.


structure_id: enveloping_group
property_id: full
type: functor
proof:
We have $F(\IN) = \IZ$ (w.r.t. addition). The homomorphism $\IZ \to \IZ$, $x \mapsto -x$ is not induced by a homomorphism $\IN \to \IN$.


structure_id: enveloping_group
property_id: preserves regular monomorphisms
type: functor
proof:
Consider the monoid $N = \langle a,b,c : ab = ac \rangle_{\Mon}$ and the free monoid $M = \langle b,c \rangle_{\Mon}$. The canonical homomorphism $M \to N$ is injective. It is also the equalizer of the two natural maps from $N$ into $\langle a_1,a_2,b,c : a_1 b = a_1 c, \, a_2 b = a_2 c \rangle_{\Mon}$. However, $F(M) = \langle b,c \rangle_{\Grp}$ (free group), and $b,c$ become identified in $F(N) \cong \langle a,b,c : b = c \rangle_{\Grp}$ since we may cancel $a$, so that $F(M) \to F(N)$ is not injective.


structure_id: enveloping_group
property_id: preserves products
type: functor
proof:
We will find a family of monoids $(M_n)_{n \in \IN}$ such that the canonical homomorphism $$\textstyle \alpha : F(\prod_{n \in \IN} M_n) \to \prod_{n \in \IN} F(M_n)$$ is not an isomorphism. We first remark that this <i>is</i> true for commutative monoids, so a counterexample must be non-commutative. For $n \in \IN$ let $M_n$ be the free monoid on two generators $a,b$. Then $F(M_n)$ is the free group on these generators. Define $g = (g_n)_{n \in \IN} \in \prod_{n \in \IN} F(M_n)$ by $g_n \coloneqq (a b^{-1})^n$. Assume that $g$ has a preimage under $\alpha$. It is a finite product $X_1 \cdots X_L$, where either $X_i \in \prod_{i \in \IN} M_n$ or $X_i^{-1} \in \prod_{i \in \IN} M_n$, and $L \in \IN$ is a fixed length. For sufficiently large $n \in \IN$ let $x_i \coloneqq \pi_n(X_i) \in M_n \cup M_n^{-1}$, so that $$(a b^{-1})^n = x_1 \cdots x_L.$$ Each $x_i$ is either a "completely positive" word (such as $a^2 b^3 a$) in $a,b$ or a "completely negative" word in $a,b$ (such as $b^{-1} a^{-1} b^{-2}$). The normal form of $x_1 \cdots x_L$ can therefore have at most $L$ alternations between positive and negative letters. But $(a b^{-1})^n = a b^{-1} \cdots a b^{-1}$ is in normal form and has $2n$ such alternations. It follows that $2n \leq L$. Since $n$ was arbitrary, this is a contradiction.


structure_id: forget_abelian
property_id: preserves coequalizers
type: functor
proof:
This follows from the concrete construction of coequalizers, but it can also be proven as follows: Assume that $f,g : A \rightrightarrows B$ have coequalizer $p : B \to C$ in $\Ab$. If $G$ is any group and $h : B \to G$ is a homomorphism with $hf = hg$, consider the image factorization $h = i h'$ of $h$. We have $h' f = h' g$, and the image $\im(h)$ is abelian. Hence, there is a unique homomorphism $k : C \to \im(h)$ with $kp = h'$. Hence, $ik : C \to G$ satisfies $ikp = h$. Uniqueness follows since we already know that the forgetful functor is preserves epimorphisms.


