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?

  • $\begingroup$ 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

1 Answer 1


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.