# Category exponent isomorphism

The book Categories for Working Mathematician - Mac Lane have a exercise described thus: For categories $A$, $B$, and $C$ establish natural isomorphisms $\displaystyle C^{A \times B} \cong (C^B)^A$. I`m not hitting the natural isomorphism.

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Is the question "What is the natural isomorphism?" –  rschwieb Sep 5 '12 at 18:33
Can you do the similar problem where $A$, $B$ and $C$ are simply sets? –  Mariano Suárez-Alvarez Sep 5 '12 at 18:38
I would advise to change the title of the question for something similar to: "The exponential law". –  a.r. Sep 5 '12 at 20:33
@AgustíRoig: I just went ahead and changed it –  Ben Millwood Sep 5 '12 at 21:10

I would start like this: let's try to define the isomorphism on objects first.

So, an object of $C^{A\times B}$ is a functor $f: A\times B \longrightarrow C$. And we have to associate it with a functor $\Phi (f) : A \longrightarrow C^B$.

Which means that, first of all, for each object $a \in A$, I have to find an object $\Phi (f) (a) \in C^B$; that is, a functor $\Phi(f) (a) : B \longrightarrow C$.

That is, for each object $b \in B$, I have to define an object $\Phi (f) (a) (b) \in C$. Right?

So, if I just have the functor $f: A \times B \longrightarrow C$, who could be that $\Phi (f) (a) (b) \in C$?

EDIT. Dear Marcelo, how is it going? Assuming you've found who is $\Phi (f) (a) (b) \in C$, you've just got your isomorphism

$$\Phi : C^{A\times B} \longrightarrow (C^B)^A$$

defined on objects. Before talking about defining it on morphisms (btw, who are the morphisms of $C^{A\times B}$?), let's talk about its inverse (just on objects too, for the moment)

$$\Psi: (C^B)^A \longrightarrow C^{A \times B} \ .$$

So, we would start with an object $g\in (C^B)^A$ -which means a functor $g: A \longrightarrow C^B$- and we would look for an object $\Psi (g) \in C^{A \times B}$ to associate with. That is, $\Psi (g)$ must be a functor

$$\Psi (g) : A\times B \longrightarrow C \ .$$

That is, given an object $(a,b) \in A\times B$, we have to define $\Psi (g) (a,b) \in C$. And the only thing at our disposal is the functor $g: A \longrightarrow C^B$.

Well, after thinking for a while, sure enough everybody finds the only way to play with these data:

$$\Psi (g) (a,b) = g(a)(b) \ .$$

Which means: you apply the functor $g(a): B \longrightarrow C$ to the object $b\in B$.

Now it is easy to check that $\Phi \circ \Psi$ and $\Psi \circ \Phi$ are identities on objects (assuming you've already found the definition of $\Phi$).

If you are willing to do it and you have troubles about defining $\Phi$ and $\Psi$ on morphisms, just let me know it.

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Of course, you need to check that this in fact gives an isomorphism on the categories themselves, not just on their objects, but that should be a matter of what many of my lecturers would call machinery. –  Arthur Sep 5 '12 at 19:58
Of course. And I'm too old to check this machinery. :-) –  a.r. Sep 5 '12 at 20:23
Didn't change your question -nor your explanations: what are you talking about? As for your request: if I'm not wrong the natural isomorphism is a big identity. –  a.r. Sep 6 '12 at 23:30