# Mapping is Involution iff Bijective and Symmetric

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## 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.

$\blacksquare$

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $11$