Subset equals Image of Preimage iff Mapping is 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 $g^{-1}$.

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

iff $f$ is a surjection.

Sufficient Condition
Let $f$ 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:
 * $T \subseteq \operatorname{Im} \left({f}\right) \subseteq T$

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

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

Necessary Condition
Let $f$ be a surjection.

Let $B \subseteq T$.

Let $b \in B$.

Then:

From Subset of Codomain is Superset of Image of Preimage, we already have that:
 * $f_g \left({f_{g^{-1}} \left({B}\right)}\right) \subseteq B$

So:
 * $B \subseteq f_g \left({f_{g^{-1}} \left({B}\right)}\right) \subseteq B$

and by definition of set equality:
 * $B = f_g \left({f_{g^{-1}} \left({B}\right)}\right)$