# Composite of Inverse of Mapping with Mapping

## Theorem

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

Then:

$f \circ f^{-1} = I_{\Img f}$

where:

$f \circ f^{-1}$ is the composite of $f$ and $f^{-1}$
$f^{-1}$ is the inverse of $f$
$I_{\Img f}$ is the identity mapping on the image set of $f$.

## Proof

By Inverse of Mapping is One-to-Many Relation, $f^{-1}$ is a one-to-many relation:

$f^{-1} \subseteq T \times S$

whose domain is the image set of $f$.

By definition of composition of relations:

$f \circ f^{-1} := \set {\tuple {x, z} \in T \times T: \exists y \in S: \tuple {x, y} \in \mathcal f^{-1} \land \tuple {y, z} \in f}$

Thus $f \circ f^{-1}$ is a relation on $T \times T$.

Let $x \in \Img f$.

Then:

$\exists y \in S: \tuple {x, y} \in f^{-1}$

As $f$ is a mapping, and so by definition a many-to-one relation;

$\map f y = x$

for all $y$ such that $\tuple {x, y} \in f^{-1}$.

That is:

$\forall x \in \Img f: f \circ \map {f^{-1} } x = x$

Hence the result by definition of identity mapping.

$\blacksquare$