Equivalence of Categories and of Their Functor Categories

Question

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.

Answer 1

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

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

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

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

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

$$ \eta^*(\psi ) : \psi \longrightarrow \psi \circ G\circ F $$

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.) :-)

Answer 2

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.