I am struggling to understand the proof of the following theorem.
$\textbf{Theorem}$. For every set $A$, $|P(A)| = |2^A| $ where $P(A)$ denotes the power set of $A$ and $2^A $ denotes the set of all functions from $A$ to $S_2 = \{0, 1\}$.
$\textbf{Proof}$. Let $A$ be any set. Define a map $g : {P}(A) \rightarrow 2^A$ as follows. Given $S \in P(A)$, we define $g(S) \in 2^A$ to be the map $g(S) : A \rightarrow \{0, 1\}$ given by
$$\begin{equation} g(S)(a)=\begin{cases} 1, & \text{if $a\in S$}.\\ 0, & \text{otherwise}. \end{cases} \end{equation}$$
Define a map $h : 2^A \rightarrow P(A)$ as follows. Given $f \in 2^A$, we define $h(f) \in P(A)$ to be the subset
$$h(f)=\{ a\in A\mid f(a)=1 \} \subseteq A.$$
The maps $g$ and $h$ are the inverses of each other which completes the proof.
$\textbf{Question}$
Can someone give a sketch proof?
Why is $g$ defined to be $P(A)\rightarrow 2^A$? Isn't it $P(A) \times A \rightarrow \{0,1\}$?
Why are $g,h$ the inverses of each other?
You need to show $|P(A)|=|2^A|.$ So you need to find a bijection between the two sets.
So for every subset $S \subseteq A, g(S)$ is that function from $A \to S_2$ which takes the value $1$ at the points which are in $S$ and it takes the value $0$ at all other points. Basically, it is the characteristic function of $S$ (if you're familar with the term). I shall henceforth denote it by $1_S. $So indeed $g$ is defined from $P(A)$ to $2^A.$ As you input a subset in $g$ and the output is a function.
On the other hand, for every function $f : A \to S_2, h(f)$ is that set of all points of $A$ for which $f$ takes value $1.$
Now, the proof is done once you show that $g$ and $h$ are inverses of each other.
To that end, let $S \subseteq A.$ Then $$h\circ g(S)=h(1_S)=\{a \in A:1_S(a)=1\}=S.$$ On the other hand, for $f \in 2^A$ $$g\circ h(f)=g\left(\{a \in A:f(a)=1\}\right)=1_{\{a \in A:f(a)=1\}}=f.$$