Cfrgtkky

Suppose $A, B, C$ are categories. If $A$ and $B$ are equivalent, is it the case that $C^A$ and $C^B$ are equivalent? Also, is it the case that $A^C$ and $B^C$ are equivalent.

I first conjectured that $C^A$ and $C^B$ are equivalent, but then as I tried to prove the conjecture, I started to realize that it might not be the case. So now I'm trying to find a counterexample. So far I have not been able to find one. Maybe I'm stuck with "too complicated" ideas. I strongly believe that there are very simple counterexamples. Please help. (For more detail, I got to the point where I have a cube that I want to be commutative, but I can only show that 4 of its sides are commutative squares. That's when I realize the other two sides may not necessarily commute.)

And of course, there is the second part to the problem that's not just the dual problem. I would like to know about the special case in which $A$ is a preorder and $B$ is a partial order which is the skeleton of $A$. This may have nothing to do with the validity of the statement in question at all, but it is the setting that originated this question.

Maybe I'm missing something, but I would say that $C^A$ and $C^B$ are indeed equivalent.

In fact, if we have an equivalence of categories between $A$ and $B$, that means we have a couple of functors

$$
F: A longrightarrow B qquad text{and} qquad G : B longrightarrow A ,
$$

together with a couple of isomorphisms of functors

$$
varepsilon : FG longrightarrow mathrm{id}_B qquad text{and} qquad eta : mathrm{id}_A longrightarrow GF .
$$

So, I'm lost: why can't you build another equivalence of categories just like this?

$$
F^* : C^B longrightarrow C^A qquad text{and} qquad G^* : C^A longrightarrow C^B ,
$$

where

$$
F^*(phi) = phicirc F qquad text{and} qquad G^* (psi ) = psi circ G quad text{?}
$$

EDIT 1. In the same vein: why

$$
F_*: A^C longrightarrow B^C qquad text{and} qquad G : B^C longrightarrow A^C ,
$$

defined as

$$
F_*(phi ) = Fcirc phi qquad text{and} qquad G_*(psi ) = G circ psi
$$

do NOT define another equivalence of categories? As far as I can see, there is no "cube" that needs to commute anywhere. Am I wrong?

EDIT 2. More details for the first case. I should have defined $F^*$ and $G^*$ on morphisms too. Ok, let's do it like this: for any natural transformation of functors $f:phi_1 longrightarrow phi_2$ (a morphism of $C^B$), define

$$
F^*(f)_a = f_{Fa} : phi_1(Fa) longrightarrow phi_2(Fa) .
$$

And analogously for $G^*$.

Then, the natural isomorphism $eta$ induces another natural isomorphism

$$
eta^* : mathrm{id}_{C^A} longrightarrow F^*G^*
$$

as follows: for any functor $psi : A longrightarrow C$, you've got a natural transformation

$$
eta^*(psi ) : psi longrightarrow psi circ Gcirc F
$$

defined as

$$
eta^*(psi )_a = psi (eta_a) : psi (a) longrightarrow psi (GFa)
$$

This $eta^*(psi)$ is indeed a natural transformation of functors (that is, a morphism of $C^A$): for any $f: a longrightarrow b$ you've got the needed commutative diagram just applying $psi$ to the commutative diagram you obtain from the fact that $eta$ is a natural transformation.

You have to check that also $eta^*$ is a natural transformation and isomorphism. At this point some diagrams are needed. Let's see if I'm able to put here a link to the right pdf file.

EDIT 3. I forgot to tell you: here and there you'll probably need the word "small" for these categories of functors being actual categories. For instance, if I remember correctly (I don't have my Mac Lane with me right now), you'll need $C$ to be small for $A^C$ and $B^C$ to be categories. Otherwise, you invoke some bigger universe and it will do. (I guess.) :-)

Recall that $mathsf{Cat}$ (the category of all small categories) is a $mathsf{Cat}$-enriched category. Thus we have internal hom functors. Let me spell this out explicitly:

You should be able to check that $A mapsto [A,C]$ (this is what you write as $C^A$) is a contravariant $mathsf{Cat}$-endofunctor of $mathsf{Cat}$ , i.e. that we have functors $[B,A] to [[A,C],[B,C]]$, $F mapsto F^*$ which commute with the identity and composition. As such, it preserves equivalences by formal nonsense.

Similarly, $A mapsto [C,A]$ provides a covariant $mathsf{Cat}$-endofunctor of $mathsf{Cat}$, i.e. have functors $[A,B] to [[C,A],[C,B]]$, $F mapsto F_*$ which commute with the identity and composition. Again it follows that equivalences are preserved.