# Mapping is Involution iff Bijective and Symmetric

## Theorem

Let $S$ be a set.

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

Then $f$ is an involution if and only if $f$ is both a bijection and a symmetric relation.

## Proof

By definition an involution on $S$ is a mapping such that:

$\forall x \in S: \map f {\map f x} = x$

### Necessary Condition

Let $f$ be an involution.

By Involution is Permutation, $f$ is a permutation and therefore by definition a bijection.

Then:

 $\displaystyle \tuple {x, y}$ $\in$ $\displaystyle f$ considering $f$ as a relation: $f \subseteq S \times S$ $\displaystyle \leadsto \ \$ $\displaystyle \map f x$ $=$ $\displaystyle y$ Definition of Mapping $\displaystyle \leadsto \ \$ $\displaystyle \map f {\map f x}$ $=$ $\displaystyle \map f y$ Definition of Mapping: $a = b \implies \map f a = \map f b$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \map f y$ Definition of Involution $\displaystyle \leadsto \ \$ $\displaystyle \tuple {y, x}$ $\in$ $\displaystyle f$ considering $f$ as a relation: $f \subseteq S \times S$

Thus $f$, considered as a relation, is symmetric.

Thus it has been shown that if $f$ is an involution, it is both a bijection and a symmetric relation.

$\Box$

### Sufficient Condition

Let $f$ be a mapping which is both a bijection and a symmetric relation.

Then:

 $\displaystyle \map f x$ $=$ $\displaystyle y$ for some unique $y \in S$ as $f$ is a bijection $\displaystyle \leadsto \ \$ $\displaystyle \tuple {x, y}$ $\in$ $\displaystyle f$ considering $f$ as a relation: $f \subseteq S \times S$ $\displaystyle \leadsto \ \$ $\displaystyle \tuple {y, x}$ $\in$ $\displaystyle f$ Definition of Symmetric Relation $\displaystyle \leadsto \ \$ $\displaystyle \map f y$ $=$ $\displaystyle x$ Definition of Mapping $\displaystyle \leadsto \ \$ $\displaystyle \map f {\map f x}$ $=$ $\displaystyle x$ as $y = \map f x$

and so $f$ is shown to be an involution.

$\blacksquare$