analysis - Show that the map is bijective

Let $X$ be a set. We consider the map \begin{equation*}\Phi : \ \mathcal{P}(X)\rightarrow \{0,1\}^X, \ \ A\mapsto 1_A\end{equation*} that maps a subset $A\subset X$to its characteristc function $1_A$.

I want to show that $\Phi$ is bijective by givung explicitly an inverse map.

Could you give me a hint how we can show that? I don't really have an idea how to find the inverse one.

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If we want to show the bijectivity by proving that the map is injective and surjective, we do the following, or not?

$\Phi$ is surjective because for every element of in the range, i.e. $0$ and $1$ there is a preimage in $\mathcal{P}(X)$ because either one element is contained in the set $A$ or not.

$\Phi$ is injective because every element of $\Phi (X)$ has an image in $\{0,1\}$.

So, $\Phi$ is bijective.

Is everything correct? Could I improve something?