Image of Preimage of Subset under Surjection equals Subset

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

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

where:


 * $f^\to$ denotes the mapping induced on the power set $\mathcal P \left({S}\right)$ of $S$ by $f$
 * $f^\gets$ denotes the mapping induced on the power set $\mathcal P \left({T}\right)$ of $T$ by the inverse $f^{-1}$
 * $f^\to \circ f^\gets$ denotes composition of $f^\to$ and $f^\gets$.

Proof
Let $g$ 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:
 * $f^\to \left({f^\gets \left({B}\right)}\right) \subseteq B$

So:
 * $B \subseteq f^\to \left({f^\gets \left({B}\right)}\right) \subseteq B$

and by definition of set equality:
 * $B = f^\to \left({f^\gets \left({B}\right)}\right)$

Also see

 * Subset equals Image of Preimage implies Surjection
 * Subset equals Image of Preimage iff Mapping is Surjection