Direct Image Mapping of Surjection is Surjection/Proof 1

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

Then:
 * $\forall y \in T: \exists x \in S: f \paren x = y$

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

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

Thus as $r: S / \mathcal R_f \to T$ is a bijection, and hence by definition also an injection:
 * $\forall X_1, X_2 \in \powerset S: r \paren {X_1} = r \paren {X_2} \implies X_1 = X_2$

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

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

Let the partition defined by $\mathcal R_f$ 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:
 * $f^\to \left({X_r}\right) = Y_r$

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

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