What does it mean to establish a natural isomorphism between categories?

Question

Consider a problem from Categories for the Working Mathematician:

For categories $A$, $B$, and $C$ establish natural isomorphisms

$$ (A \times B)^C \cong A^C \times B^C, \\ C^{A \times B} \cong (C^B)^A $$

I understand the notation $F_1 \cong F_2$ denotes "$F_1$ is naturally isomorphic to $F_2$" (for functors $F_1$ and $F_2$).

Question: What does this question then mean? Here the expressions on each side of $\cong$ denote categories, not functors. For example, $(A \times B)^C$ is a functor. So I don't know how to parse the meaning of this question. What is a natural isomorphism between categories?

Answer 1

You can think of these as functors with three inputs taken from the category of categories (with some components contravariant): e.g.

$(- \times -)^- :\underline{Cat} \times \underline{Cat} \times \underline{Cat}^{op} \rightarrow \underline{Cat}$

You can then apply what you know about natural isomorphisms between these functors. (Of course, the components of these natural isomorphisms are themselves functors, but it is important not to confuse them with the functors that you are finding isomorphisms between)

Answer 2

Reading "$P \cong Q$" as merely saying "$P$ and $Q$ are isomorphic" is somewhat misleading. Sure, it's what the notation literally means, but in reality one often means (and other times, often should mean) to refer to a particular isomorphism between them.

For the isomorphism to be natural, one can describe it as being analogous to being well-defined — that the variables $A,B,C$ don't just refer to objects, but can also refer to arrows as well, and we insist that the individual isomorphisms selected for each object are also consistent in a particular with all the maps between them when we plug in arrows.

Of course, we don't need to appeal to formal syntax; an equivalent interpretation is that if we define functors by the formulas

$$ F(A,B,C) = (A \times B)^C \qquad G(A,B,C) = A^C \times B^C $$

we are interested in a natural isomorphism $\eta : F \to G$.