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 iff $f$ is both a bijection and a symmetric relation.

Proof
By definition an involution on $S$ is a mapping such that:
 * $\forall x \in S: f \left({f \left({x}\right)}\right) = x$

Necessary Condition
Let $f$ be an involution.

Consider $f \left({S}\right)$, the image of $f$.

In order for $f$ to be a mapping on $f \left({S}\right)$, then $f \left({S}\right) = S$.

It follows from Surjection iff Image equals Codomain that $f$ must be a surjection.

Suppose $f$ were not injective.

Then:
 * $\exists y \in S: f \left({x_1}\right) = f \left({x_2}\right) = y$

such that $x_1 \ne x_2$.

Suppose $f \left({y}\right) = x_1$.

Then $f \left({f \left({x_2}\right)}\right) = x_1 \ne x_2$.

That is, $f$ is not an involution

So for $f$ to be an involution it must at least be an injection.

Thus $f$ is both an injection and a surjection, and therefore a bijection.

Then:

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.

Sufficient Condition
Let $f$ be a mapping which is both a bijective mapping and a symmetric relation.

Then:

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