CatDat

Implication Details

Assumptions: left cancellative, sifted

Conclusions: thin

Proof: For any object $X$ in a left-cancellative category, the connected component containing $$X \xrightarrow{\id} X \xleftarrow{\id} X$$ in the category of cospans from $X$ to $X$ consists only of cospans $$X \xrightarrow{f} Y \xleftarrow{g} X$$ where $f=g$; hence when the category is also sifted, all cospans must be of this form, and so any two parallel morphisms are equal.