mapping of period $n$ is a bijection

Theorem Suppose $X$ is a set.Then a mapping $f:X\to X$ of period (http://planetmath.org/PeriodOfMapping) $n$ isa bijection.

Proof. If $n=1$, the claim is trivial;$f$ is the identity mapping.Suppose $n=2,3,\mathrm{\dots }$.Then for any $x\in X$, we have $x=f\left(f^{n-1}(x)\right)$,so $f$ is an surjection. To see that $f$ is a injection,suppose $f(x)=f(y)$ for some $x,y$ in $X$. Since $f^n$is the identity, it follows that $x=y$. $\mathrm{\square }$