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 1
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.

Proof 2
Suppose $g: S \to T$ is a surjection.

By definition, $f_g$ is defined by sending subsets of $S$ to their image under $g$.

That is, for all $X\subseteq S$, $f_g(X) = \{g(x):x\in X\}\subseteq T$

To prove that $f_g$ is a surjection, we need to show that every subset of $T$ is the image under $f_g$ of some subset of $S$.

Let $Y\subseteq T$.

Since $g$ is a surjection, by definition we have that $\forall t \in T: \exists s \in S: g \left({s}\right) = t$.

Hence (possibly requiring the Axiom of Choice), we  can select for each $y\in Y$ some $s_y\in S$ such that $g(s_y)=y$.

Define $X_Y$ to be $\{s_y : y\in Y\}$.

Then $f_g (X_Y) = \{g(s_y) : s_y \in X_Y \} = \{g(s_y) : y\in Y \} = \{y : y\in Y \} = Y$.

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