# Subset equals Preimage of Image iff Mapping is Injection

## Theorem

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

Let $f^{-1}$ denote the inverse of $f$.

Then:

$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$ if and only if $f$ is an injection

where:

$f \sqbrk A$ denotes the image of $A$ under $f$
$f^{-1} \circ f$ denotes the composition of $f^{-1}$ and $f$.

This can be expressed in the language and notation of direct image mappings and inverse image mappings as:

$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$ if and only if $f$ is an injection

## Proof

### Sufficient Condition

Let $g$ be such that:

$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$

Then by Subset equals Preimage of Image implies Injection, $f$ is an injection.

$\Box$

### Necessary Condition

Let $f$ be an injection.

$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$

$\blacksquare$