Subset equals Image of Preimage iff Mapping is Surjection

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

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

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

Then:
 * $\forall B \in \mathcal P \left({T}\right): B = \left({f^\to \circ f^\gets}\right) \left({B}\right)$

$f$ is a surjection.

Sufficient Condition
Let $f$ be such that:
 * $\forall B \in \mathcal P \left({T}\right): B = \left({f^\to \circ f^\gets}\right) \left({B}\right)$

From Subset equals Image of Preimage implies Surjection, $f$ is a surjection.

Necessary Condition
Let $f$ be a surjection.

Then by Image of Preimage of Subset under Surjection equals Subset:


 * $B = f^\to \left({f^\gets \left({B}\right)}\right)$