Direct Image Mapping of Surjection is Surjection/Proof 2

Proof
Let $f: S \to T$ be a surjection.

By definition, $f^\to$ is defined by sending subsets of $S$ to their image under $f$.

That is:
 * $\forall X \subseteq S: f^\to \left({X}\right) = \left\{{f \left({x}\right): x \in X}\right\}\subseteq T$

To prove that $f^\to$ is a surjection, we need to show that every subset of $T$ is the image under $f^\to$ of some subset of $S$.

Let $Y \subseteq T$.

Since $f$ is a surjection, by definition we have that:
 * $\forall t \in T: \exists s \in S: f \left({s}\right) = t$

Hence (possibly requiring the Axiom of Choice), we can select for each $y \in Y$ some $s_y \in S$ such that $f \left({s_y}\right) = y$.

Define $X_Y$ to be:
 * $X_Y := \left\{{s_y: y \in Y}\right\}$

Then:
 * $f^\to \left({X_Y}\right) = \left\{ {f \left({s_y}\right): s_y \in X_Y}\right\} = \left\{ {f \left({s_y}\right): y \in Y }\right\} = \left\{ {y : y \in Y}\right\} = Y$

Thus $f^\to: \mathcal P \left({S}\right) \to \mathcal P \left({T}\right)$ is a surjection.