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 $f'': \mathcal P \left({S}\right) \to \mathcal P \left({T}\right)$ 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 $X \in \mathcal P \left({T}\right)$.

Let $Y = f^* \left({T}\right)$.

By Subset equals Image of Preimage iff Mapping is Surjection, $f'' \left({Y}\right) = X$.

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