Direct Image Mapping of Surjection is Surjection/Proof 3

Theorem
Let $S$ and $T$ be sets.

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

Let $f'': \mathcal P(S) \to \mathcal P(T)$ be the mapping induced by $f$.

Then $f''$ is also a surjection.

Proof
Let $f^*$ be the mapping induced by the inverse relation of $f$.

Let $X \in \mathcal P(T)$.

Let $Y = f^* (X)$.

By Image of Preimage of Surjection, $f''(Y) = X$.

As such a $Y$ exists for each $X \in \mathcal P(T)$, $f''$ is surjective.