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 $\Theta$ which maps $(A^B)^C$ to $A^{B \times C}$. That is

$$\Theta (g) = h$$

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

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

UNSURE

That is, $\Theta$ 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 $\Theta$, as we have defined it, is indeed a bijection. 

Let's show $\Theta$ 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 $\Theta( g(c)) = g(c)(b) = h(b, c)$.  UNSURE

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