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If two categories are equivalent, then if one has products, then so does the other. The proof of this is easy enough so I'm guessing the same result holds for exponentials but I am having trouble proving it. Here is what I have so far:

If $F:C\Leftrightarrow D:G$ denotes the equivalence between categories $C$ and $D$, and $f:U$x $V\rightarrow W$, we seek an object $W^{V}$, satisfying the definition of exponential. I know that if $U$ x $V$ is a product in $C$, then it is isomorphic to $G(FU$ x $ FV)$. Applying $F$ and using the natural isomorphism $α:FG\rightarrow 1_{D}$, we get $F(U$ x $ V)≅FU$ x $FV$.Thus we have the arrow $Ff:FU$ x $FV\rightarrow FW$, and the exponential $FW^{FV}$ in $D$.

I want to prove $G(FW^{FV})$ is an exponential object $W^{V}$ in $C$:

We have unique morphism $\tilde{Ff}:FU\rightarrow FW^{FV}$ and the morphisms $\tilde{Ff}$ x $1_{FV}:FU$ x $FV\rightarrow FW^{FV}$ x $FV$ and $ε:FW^{FV}$ x $FV→FW$ s.t. $\varepsilon ∘(\tilde{Ff}$ x $1_{FV})=Ff$.

Now apply $G$ to get $G\varepsilon ∘G(\tilde{Ff}$ x $1_{FV})=GFf. $We have that $GFf$ is an arrow from $G(FU$ x $FV)$ to $GFW$, the domain and codomain of which are isomorphc to $U$ x $V$ and $W$, respectively, so we get $f:U$ x $V\rightarrow W$ back via these isomorphisms. But I do not see how to handle $G(\tilde{Ff}$ x $1_{FV})$, which I want to say gives us the required $\tilde{f}$ x $1_{V}$.

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Can you show us your attempts? Any "categorical" property should be preserved under equivalence - with the caveat that the number of representatives of an isomorphism class may change. That shouldn't really be considered a "categorical" property anyway, which is the point of weakening "isomorphisms of categories" to "equivalence of categories". – Dustan Levenstein Apr 11 '14 at 23:40
Just fyi, editing your question is a more effective way to bump the thread and get people reading it, so that you're not left waiting on the random person who decided on a whim to make a comment. – Dustan Levenstein Apr 12 '14 at 7:29
I'm not sure what $\tilde F$ is in your work above, but here's what I came up with: Let $\lambda: F(U \times V) \to F(W)^{F(V)} \times F(V)$ be the unique morphism given by the universal property of the exponential. Then applying $G$ to the diagram, you get a resulting diagram which contains several instances of $GF$, which lend themselves to the use of $GF \simeq id_C$; in particular, you get selected isomorphisms $G(F(U \times V)) \simeq U \times V$ and $G(F(W)) \simeq W$, which are the objects you want in your final diagram. These selected isomorphisms allow you to substitute them in. – Dustan Levenstein Apr 12 '14 at 7:51
As an aside, Freyd, Scedrov prove that a property of categories is preserved by equivalence if and only if it can be defined with diagrams on a blackboard. – Hurkyl Apr 12 '14 at 16:15

The Yoneda lemma is a very powerful tool, use it. The definition of an exponential object in a category $C$ is (where this is a natural isomorphism in $U$): $$\hom_C(U \times V, W) \cong \hom_C(U, W^V)$$

So here (I suggest you carefully check for each step which property I used): $$\begin{align} \hom_C(U \times V, W) & \cong \hom_C(U \times V, GF(W)) \\ & \cong \hom_D(F(U \times V), F(W)) \\ & \cong \hom_D(F(U) \times F(V), F(W)) \\ & \cong \hom_D(F(U), F(W)^{F(V)}) \\ & \cong \hom_C \left( U, G \left( F(W)^{F(V)} \right) \right) \end{align}$$

Each of these isomorphisms are natural, therefore $G \left( F(W)^{F(V)} \right)$ is an exponential object $W^V$.

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Thanks, it is very clear. The second isomorphism follows by faithfulness of F and then the fact that $FGFW\cong FW$ . Is that correct? I still would like to do it from scratch though. – Chilango Apr 12 '14 at 18:58
No, the second isomorphism is simply the adjunction between $G$ and $F$. – Najib Idrissi Apr 13 '14 at 6:15
It comes to the same thing no? I mean, the equivalence gives an adjunction because you have a unit a counit, that is true but equivalence also implies faithfulness of $F$ (and $G$) and the isomomphism $FGFW\cong FW$. This also would be enough to give the second isomporphism. – Chilango Apr 13 '14 at 14:07
Full faithfulness is necessary I think, otherwise you just get an injection. – Najib Idrissi Apr 13 '14 at 14:15
And $F$ and $G$ are fully faithful, being part of an equivalence of categories. In fact, Isn't essential surjectivity of F enough here? – Chilango Apr 14 '14 at 3:58

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