abstract algebra - Powerset bijection problem

Please do not provide a full answer for this.

Let $2^{S} = \{f : S \rightarrow \{0, 1\}\}$. For $A \subseteq S$, define $\chi_{A}\in2^{S}$ by

$$\chi_{A}(s) = \begin{cases} 0 & \text{if } s \notin A \\ 1 & \text{if } s \in A \end{cases}.$$
Show that $\mu : P(S)\rightarrow2^{S}$ given by $\mu(A)=\chi_{A}$ is a bijection.

I know that the standard procedure for showing that a function is bijective is to show that it is both injective and surjective, and the "standard procedures" for those as well. It's just that I don't really know where to start with this.