# Composite of Mapping with Inverse of Another is Identity implies Mappings are Equal

## Theorem

Let $A$ and $B$ be classes.

Let $f$ and $g$ be mappings on $A \times B$.

Let $f$ and $g$ be such that:

$f \circ g^{-1} = I_B$

where:

$g^{-1}$ denotes the inverse of $g$
$I_B$ denotes the identity mapping on $B$
$\circ$ denotes composition of mappings.

Then:

$f = g$

## Proof

Let $\tuple {a, b} \in f$.

Then because $\tuple {b, b} \in I_B$ we must have:

$\tuple {b, a} \in g^{-1}$

and so by definition of inverse of mapping:

$\tuple {a, b} \in g$

Hence:

$f \subseteq g$

Let $\tuple {a, b} \in g$.

Then by definition of inverse of mapping:

$\tuple {b, a} \in g^{-1}$

Then because $\tuple {b, b} \in I_B$ we must have:

$\tuple {a, b} \in f$

Hence:

$g \subseteq f$

Hence by definition of set equality:

$f = g$

$\blacksquare$