$(A^B)^C $~ $A^{(B\times C)}$

Question

Here's my attempt:

Tried to find a bijection: if i could define $h : (A^B)^C \rightarrow A^{(B\times C)}$ then finished.

Eventually, all my attempts failed.

I'd like to have some directions for the solution.

Answer 1

Recall that $X^Y$ is the set of functions from $Y$ to $X$. So, we want to find a bijection $f: (A^B)^C \to A^{B\times C}$. Now, an element $\alpha$ of $(A^B)^C$ is a function whose domain is $C$ and which outputs functions from $B$ to $A$. We want $f(\alpha)$ to be a function from $B\times C$ to $A$; therefore, we can define $f(\alpha)$ be specifying its output given an element of $B\times C$. So, define $f$ such that if $b\in B, c\in C$, then $$f(\alpha)(b,c) = (\alpha(c))(b)$$

Answer 2

We're looking for a way to assign to each map $f : C \to (A^B)$ a map $B \times C \to A$, and if you're looking for a natural, explicit map, there simply aren't that many possibilities.

Hint If we fix $c \in C$, $f(c)$ is a map $B \to A$, so for fixed $b \in B$, $f(c)(b)$ is an element of $A$.

This process assigns to each choice of $b \in B$ and $c \in C$ an $a \in A$, which is essentially what a map $B \times C \to A$ is, so we have defined a map $\Phi : (A^B)^C \to A^{B \times C}$, by $$\Phi(f)(b, c) := f(c)(b) .$$ It remains to check that this is a bijection.

See the Wikipedia article on currying.