Proving $C^{A \times B} \cong (C^B)^A$ without using the "same cardinality" argument

Question

I'd like to prove that there is a bijection between the two sets $C^{A \times B}$ (the set of all functions from $A \times B$ to $C$) and $(C^B)^A$ (the set of all functions from $A$ to the set of all functions from $B$ to $C$) by showing one, without considering that this is immediate because they have the same cardinality.

Could anyone give me a hint of how to construct such a bijection?

Answer 1

Map a function $f:A\times B\to C$ into a function $\phi:A\to C^B$ by setting, for all $a\in A$, that $\phi(a)$ maps $B$ to $C$ this way: $\phi(a)(b)=f(a,b)$ for all $b\in B$. That is all.

Answer 2

The bijection $\Phi : C^{A \times B} \to (C^A)^B$ is given by

$$\Phi(f)(b)(a) = f(a,b).$$

Now prove this works!