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.

$\blacksquare$


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