# Composite of Bijection with Inverse is Identity Mapping

## Theorem

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

Then:

 $\ds f^{-1} \circ f$ $=$ $\ds I_S$ $\ds f \circ f^{-1}$ $=$ $\ds I_T$

where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.

## Proof

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

From Inverse of Bijection is Bijection, $f^{-1}$ is also a bijection.

Let $x \in S$.

$\exists y \in T: y = \map f x \implies x = \map {f^{-1} } y$

Then:

 $\ds \map {f^{-1} \circ f} x$ $=$ $\ds \map {f^{-1} } {\map f x}$ Definition of Composition of Mappings $\ds$ $=$ $\ds \map {f^{-1} } y$ by hypothesis $\ds$ $=$ $\ds x$ by hypothesis $\ds$ $=$ $\ds \map {I_S} x$ Definition of Identity Mapping

From Domain of Composite Relation and Codomain of Composite Relation, the domain and codomain of $f^{-1} \circ f$ are both $S$, and so are those of $I_S$ by definition.

So all the criteria for Equality of Mappings are met and thus $f^{-1} \circ f = I_S$.

Let $y \in T$.

$\exists x \in S: x = \map {f^{-1} } y \implies y = \map f x$

Then:

 $\ds \map {f \circ f^{-1} } y$ $=$ $\ds \map f {\map {f^{-1} } y}$ Definition of Composition of Mappings $\ds$ $=$ $\ds \map f x$ by hypothesis $\ds$ $=$ $\ds y$ by hypothesis $\ds$ $=$ $\ds \map {I_T} y$ Definition of Identity Mapping

From Domain of Composite Relation and Codomain of Composite Relation, the domain and codomain of $f \circ f^{-1}$ are both $T$, and so are those of $I_T$ by definition.

So all the criteria for Equality of Mappings are met and thus $f \circ f^{-1} = I_T$.

$\blacksquare$