# Subset equals Image of Preimage iff Mapping is Surjection

## Theorem

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

Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.

Similarly, let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$.

Then:

$\forall B \in \powerset T: B = \map {\paren {f^\to \circ f^\gets} } B$

if and only if $f$ is a surjection.

## Proof

### Sufficient Condition

Let $f$ be such that:

$\forall B \in \powerset T: B = \map {\paren {f^\to \circ f^\gets} } B$

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

$\Box$

### Necessary Condition

Let $f$ be a surjection.

$B = \map {f^\to} {\map {f^\gets} B}$

$\blacksquare$