Involution is Permutation

Theorem
Let $S$ be a set, and let $f: S \to S$ be an involution.

Then $f$ is a bijection.

Proof
By definition of involution, for each $x \in S$:


 * $f \left({f (x)}\right) = x$

so that by Equality of Mappings, $f \circ f = \operatorname{id}_S$.

Here $\operatorname{id}_S$ is the identity mapping on $S$.

Thus $f$ is both a left inverse and a right inverse of itself.

The result follows from Bijection iff Left and Right Inverse.