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The twisted arrow category of $\cal C$ is the category of elements of $\hom_{\cal C}$.

When is this category cofiltered?

This is equivalent to ask that the hom functor, taken as a presheaf on ${\cal C}^\text{op}\times \cal C$, commutes with finite limits, so

When does $\hom_{\cal C}$ commute with finite lims?

Does it happen only in trivial cases?

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  • $\begingroup$ E. g. if(f?) $\cal C$ has a zero object its tac has a terminal object, but you need the opposite direction? $\endgroup$ Oct 1, 2016 at 7:57

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I think that it only happens in trivial cases. Let me assume that $\mathcal{C}$ has binary products and coproducts. The functor $\mathrm{Hom}_{\mathcal{C}}$ preserves products if and only if for all $A,A',B,B' \in \mathcal{C}$ the canonical map $$\mathrm{Hom}(A+A',B \times B') \longrightarrow \mathrm{Hom}(A,B) \times \mathrm{Hom}(A',B')$$ is an isomorphism. However, the left hand side decomposes as $$\mathrm{Hom}(A,B) \times \mathrm{Hom}(A,B') \times \mathrm{Hom}(A',B) \times \mathrm{Hom}(A',B')$$ and the canonical map is simply the projection. In particular, the projection $\mathrm{Hom}(A,B)^4 \to \mathrm{Hom}(A,B)^2$ is injective, which shows that $\mathrm{Hom}(A,B)$ has at most one element. Thus, $\mathcal{C}$ is a preorder. The surjectivity of $$\mathrm{Hom}(A+A',A \times A') \longrightarrow \mathrm{Hom}(A,A) \times \mathrm{Hom}(A',A')$$ shows that $A+A' \leq A \times A'$, hence $A' \leq A$ for all $A,A'$. Thus, $\mathcal{C}$ is a trivial (codiscrete) preorder.

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  • $\begingroup$ I think this resolves the problem and shows that this is too strong a condition. Thanks $\endgroup$
    – fosco
    Oct 1, 2016 at 16:02

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