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.

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?