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

Then $\mathcal R\restriction_X: X \to T$, 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 $\operatorname{Rng} \left({f_\mathcal R \left({\mathcal P \left({S}\right)}\right)}\right)$.

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.