Show that the bijection $\text{Cat}(A \times B, C) \cong \text{Cat}(A, C^B)$ is natural This is exercise 2.5.1 in Categories for the Working Mathematician, and it's been asked here a few times but I've never seen the part that confuses me clarified.
So we wish to create a bijection $\phi : \text{Cat}(A \times B, C) \rightarrow \text{Cat}(A, C^B)$. These are both functor categories, so it has elements of the form $(F:A \times B \rightarrow C) \rightarrow (T:A \rightarrow C^B)$.
We can define a bijection $\phi(F)(a)(b) = F(a, b)$, with $\phi^{-1}(T)(a, b) = T(a)(b)$, but my issue is when showing that it is natural.
Since we are dealing with functor categories, the morphisms are natural transformations $\eta : F \rightarrow F'$ and $\mu : T \rightarrow T'$, what is the explicit mapping here?
 A: For simplicity, let's just fix $B$ and $C$ and focus on naturality in $A.$ If we want to look at this in terms of functors, the functors in question are $\operatorname{Cat}(-\times B,C)$ and $\operatorname{Cat}(-, C^B)$ and we want to show that $\phi$ is an arrow between these two functors in the appropriate category (in this case $\mathrm{Set}^{\mathrm{Cat}^{\mathrm{op}}}$), which means we want to show that $\phi_A:\operatorname{Cat}(A\times B,C) \to \operatorname{Cat}(A, C^B)$ can be regarded as the $A$-th component of a natural transformation between the functors $\operatorname{Cat}(-\times B,C)$ and $\operatorname{Cat}(-, C^B).$
Given an arrow $g:A'\to A,$ the first functor takes an arrow $f:A\times B\to C$ to $f\circ (g\times id_{B})$ and given an arrow $h: A\to C^B,$ the second functor takes $h$ to $h\circ g.$ So showing naturality means showing that for any $f:A\times B \to C,$ $\phi_{A'}(f\circ (g\times id_B)) = \phi_A(f)\circ g.$ We can verify this by looking at how each side acts on some $a\in A'$ (either an object or arrow.. no matter). The left-hand side is $$ \phi_{A'}(f\circ (g\times id_B)) (a) = f\circ (g\times id_B)(a,-) = f(g(a), -)$$ and the right-hand side is $$ (\phi_A(f)\circ g) (a) = \phi_A(f)(g(a)) = f(g(a),-),$$ which are the same, so we're done.
Note that for most of the formality here, it was completely irrelevant that the arrows in question were functors. For everything up till the verification of the equation at the end, we may as well have replaced Cat with Set or any other cartesian closed category. Only in verifying the equation did we actually think about what the category actually was, and even then not really... we could have done that step just in terms of the eval arrow, etc, it just might have looked a little more obscure.
Naturality in $B$ and $C$ can be shown similarly. It can be tediously shown that in general, verifying each component's naturality separately is all that is required to show the whole thing is natural, i.e. that $\phi$ is an arrow in the functor category $\operatorname{Set}^{\operatorname{Cat}^{\mathrm{op}}\times \operatorname{Cat}^{\mathrm{op}}\times \operatorname{Cat}}.$ (And it can also be shown that all it takes to be an isomorphism in said category is for each component of the natural transformation to be an isomorphism.)
