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

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.