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:
 * $\forall X \subseteq S: \map {f^\to} X = \set{\map f x: x \in X} \subseteq T$

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:
 * $\forall t \in T: \exists s \in S: \map f s = t$

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

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

Then:

Thus $f^\to: \powerset S \to \powerset T$ is a surjection.