Direct Image Mapping of Surjection is Surjection/Proof 3

Theorem

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

Then the direct image mapping of $f$:

$f^\to: \powerset S \to \powerset T$

is a surjection.

Proof

Let $f^\gets$ be the inverse image mapping of $f$.

Let $Y \in \powerset T$.

Let $X = \map {f^\gets} Y$.

By Subset equals Image of Preimage iff Mapping is Surjection:

$\map {f^\to} X = Y$

As such an $X$ exists for each $Y \in \powerset S$, $f^\to$ is surjective.

$\blacksquare$