Subset equals Image of Preimage implies Surjection

Theorem
Let $g: S \to T$ be a mapping.

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

Similarly, let $f_{g^{-1}}: \mathcal P \left({T}\right) \to \mathcal P \left({S}\right)$ be the mapping induced by the inverse $g^{-1}$.

Let:
 * $\forall B \in \mathcal P \left({T}\right): B = \left({f_g \circ f_{g^{-1}}}\right) \left({B}\right)$

Then $g$ is a surjection.

Proof
Let $g$ be such that:
 * $\forall B \in \mathcal P \left({T}\right): B = \left({f_g \circ f_{g^{-1}}}\right) \left({B}\right)$

In particular, it holds for $T$ itself.

Hence:

So:
 * $T \subseteq \operatorname{Im} \left({g}\right) \subseteq T$

and so by definition of set equality:
 * $\operatorname{Im} \left({g}\right) = T$

So, by definition, $g$ is a surjection.

Also see

 * Image of Premage of Subset under Surjection equals Subset
 * Subset equals Image of Preimage iff Mapping is Surjection