4
$\begingroup$

Is the functor $\mathsf{Cat} \to \mathsf{Cat}$, $\mathcal{C} \mapsto \mathcal{C}^{\mathrm{op}}$ representable? By this I mean: Is there a small category $\mathcal{P}$ together with isomorphisms of categories $\alpha_{\mathcal{C}} : \mathrm{Hom}(\mathcal{P},\mathcal{C}) \to \mathcal{C}^{\mathrm{op}}$ which are natural in $\mathcal{C} \in \mathsf{Cat}$?

Some of my thoughts.

What is a little bit strange here is that $\mathsf{Cat} \to \mathsf{Cat}$, $\mathcal{C} \mapsto \mathcal{C}^{\mathrm{op}}$ is not a $2$-functor: If $f,g : \mathcal{C} \to \mathcal{D}$ are two functors and $\alpha : f \to g$ is a morphism of functors, we get a morphism of functors $\alpha^{\mathrm{op}} : g^{\mathrm{op}} \to f^{\mathrm{op}}$ in the other direction. Thus, we only get a $2$-functor when we reverse the $2$-morphisms in, say, the target $\mathsf{Cat}$. But this is not the case for $\mathrm{Hom}(\mathcal{P},-)$, so that we cannot formulate any reasonable compatibility of $\alpha$ with $2$-morphisms.

But the question still makes sense if we view $\mathsf{Cat}$ as a $1$-category which is enriched over itsself, I think.

Somehow I cannot believe that $\mathcal{P}$ exists, but I couldn't disprove it so far. Basically, it is an easy task to reconstruct a category $\mathcal{P}$ from the functor $\mathrm{Hom}(-,\mathcal{P})$, for example $\mathrm{Hom}(1,\mathcal{P})\cong\mathrm{Ob}(\mathcal{P})$ and $\mathrm{Hom}(\{0<1\},\mathcal{P})\cong\mathrm{Mor}(\mathcal{P})$. But I don't know how to reconstruct $\mathcal{P}$ from $\mathrm{Hom}(\mathcal{P},-)$.

I have checked that $\mathcal{C} \mapsto \mathcal{C}^{\mathrm{op}}$ preserves coproducts and I also believe that it preserves coequalizers (can someone confirm this?).

$\endgroup$
1
$\begingroup$

Assume the functor is representable in some sense, and let $C$ be a small category which has an initial object but not a final one. Then $Hom(P,C)$ will have an initial object, and $C^{opp}$ won't.

$\endgroup$
  • $\begingroup$ Every cocomplete small category is a preorder and hence complete. But I think we can simply take a category which has binary coproducts but no binary products. Thanks. $\endgroup$ – HeinrichD Oct 5 '16 at 16:35
  • $\begingroup$ Whoops, my bad. I'll edit it. $\endgroup$ – EstonianFootballerJoelLindpere Oct 5 '16 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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