Direct Image Mapping of Surjection is Surjection

Theorem
Let $g: S \to T$ be a surjection.

Then the mapping induced by $g$ on $\mathcal P \left({S}\right)$:
 * $f_g: \mathcal P \left({S}\right) \to \mathcal P \left({T}\right)$

is a surjection.

Proof
Suppose $g: S \to T$ is a surjection.

Then $\forall y \in T: \exists x \in S: g \left({x}\right) = y$.

From the Quotient Theorem for Surjections, there is one and only one bijection $r: S / \mathcal R_g \to T$ such that $r \circ q_{\mathcal R_g} = g$.

Each element of $S / \mathcal R_g$ is a subset of $S$ and therefore an element of $\mathcal P \left({S}\right)$.

Thus:
 * $\forall X_1, X_2 \in \mathcal P \left({S}\right): r \left({X_1}\right) = r \left({X_2}\right) \implies X_1 = X_2$.

Because $g$ is a surjection, every $y \in T$ is mapped to by exactly one element of the partition of $S$ defined by $\mathcal R_g$.

Let $T = \left\{{y_1, y_2, \ldots}\right\}$.

Let the partition defined by $\mathcal R_g$ be $\bigcup \left({X_1, X_2, \ldots}\right)$ where $r \left({X_n}\right) = y_n$.

Let $Y_r \in \mathcal P \left({T}\right)$, such that $Y_r = \left\{{y_{r_1}, y_{r_2}, \ldots}\right\}$.

Then $f_g \left({X_r}\right) = Y_r$, where $X_r = \bigcup \left({X_{r_1}, X_{r_2}, \ldots}\right)$.

As $\left\{{X_1, X_2, \ldots}\right\}$ is a partition of $S$, $\forall Y_r \in \mathcal P \left({T}\right): X_r$ is unique.

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