Tao Analysis I - 3.6.6

Exercise 3.6.6

Let $A$, $B$, $C$ be sets. Show that the sets $(A^B{)}^C$ and $A^{B\times C}$ have equal cardinality by constructing an explicit bijection between the two sets. Conclude that $(a^b{)}^c=a^{bc}$ for any natural numbers $a$, $b$, $c$. Use a similar argument to also conclude $a^b\times a^c=a^{b+c}$ .

This stack exchange post helped with this exercise (link), as well as this solution (link).

Let's start with $A^B$. This is a set of all the functions from $B$ to $A$. Let's denote an element of this set $f$. 

$$f\in A^B\text{ where }f:B\to A$$

So $f(b)=a$ where $a\in A$ and $b\in B$.

Now $(A^B{)}^C$ is the set of all functions from $C$ to $A^B$. Let's denote an element of this set $g$. 

$$g\in (A^B{)}^C\text{ where }g:C\to A^B$$

So $g(c)=f_c$ where $c\in C$ and $f_c\in A^B$.

Let's now consider $A^{B\times C}$. This is a set of all the functions from $B\times C$ to $A$. Let's denote an element of this set $h$. 

$$h\in A^{B\times C}\text{ where }h:B\times C\to A$$

So $h((b,c))=a$ where $a\in A$, and $(b,c)$ is an ordered pair with $b\in B,c\in C$.

We need a bijection, let's call it $\mathrm{\Theta }$ which maps $(A^B{)}^C$ to $A^{B\times C}$. That is

$$\mathrm{\Theta }(g)=h$$

Noting that $g$ takes one parameter $c$, and $h$ takes two $(b,c)$, helps suggest a possible mapping.

$$\mathrm{\Theta }(g)=g(c)(b):=h(b,c)$$

UNSURE

That is, $\mathrm{\Theta }$ maps $g(c)\in (A^B{)}^C$ to $g(c)(b)\in A^{B\times C}$ which we define to be $h(b,c)\in A$.

We now need to show that $\mathrm{\Theta }$, as we have defined it, is indeed a bijection. 

Let's show $\mathrm{\Theta }$ is surjective. For every $h(b,c):=g(c)(b)\in A^{B\times C}$ there exists a $g(c)\in (A^B{)}^C$ such that $\mathrm{\Theta }(g(c))=g(c)(b)=h(b,c)$.  UNSURE

Let's now show $\mathrm{\Theta }$ is injective. TODO