Surjection Induced by Powerset is Induced by Surjection

Theorem
Let $$\mathcal{R} \subseteq S \times T$$ be a relation.

Let $$f_{\mathcal{R}}: \mathcal{P} \left({S}\right) \to \mathcal{P} \left({T}\right)$$ be the mapping induced on $\mathcal{P} \left({S}\right)$ by $\mathcal{R}$.

Let $$f_{\mathcal{R}}$$ be a surjection.

Let $$X = \mathrm {Im}^{-1} \left({\mathcal{R}}\right)$$, that is, the preimage of $$\mathcal{R}$$.

Then $$\mathcal{R}|_X: X \to T$$, that is, the restriction of $\mathcal{R}$ to $X$, is a surjection.

Proof
Let $$X$$ be the preimage of $$\mathcal{R}$$.


 * Suppose $$\mathcal{R}: X \to T$$ is a mapping, but not a surjection.

Then $$\exists y \in T: \lnot \exists x \in S: \mathcal{R} \left({x}\right) = y$$.

Because no element of $$S$$ maps to $$y$$, no subset of $$S$$ contains any element of $$S$$ that maps to the subset $$\left\{{y}\right\} \subseteq T$$.

Thus, $$\exists \left\{{y}\right\} \in \mathcal{P} \left({T}\right): \lnot \exists X \in \mathcal{P} \left({S}\right): f_{\mathcal{R}} \left({X}\right) = \left\{{y}\right\}$$

So we see that $$f_{\mathcal{R}}: \mathcal{P} \left({S}\right) \to \mathcal{P} \left({T}\right)$$ is not a surjection.


 * Now, suppose $$\mathcal{R}: X \to T$$ is not even a mapping. This could happen, according to the definition of a mapping, in one of two ways:


 * 1) $$\exists x \in X: \lnot \exists y \in T: \left({x, y}\right) \in \mathcal{R}$$.
 * 2) $$\exists x \in S: \left({x, y_1}\right) \in f \land \left({x, y_2}\right) \in f : y_1 \ne y_2$$

Because $$X$$ is already the preimage of $$\mathcal{R}$$, the first of these can't happen here.

For the second, it can be seen that neither $$\left\{{y_1}\right\}$$ nor $$\left\{{y_2}\right\}$$ can be in $$\mathrm{Rng} \left({f_{\mathcal{R}} \left({\mathcal{P} \left({S}\right)}\right)}\right)$$.

Therefore $$f_{\mathcal{R}}: \mathcal{P} \left({S}\right) \to \mathcal{P} \left({T}\right)$$ can not be a surjection.

Thus, by the Rule of Transposition, the result follows.