# 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) $n$ is a bijection.

Proof. If $n=1$, the claim is trivial; $f$ is the identity mapping. Suppose $n=2,3,\ldots$. Then for any $x\in X$, we have $x=f\big{(}f^{n-1}(x)\big{)}$, 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 $x=y$. $\Box$

Title mapping of period $n$ is a bijection MappingOfPeriodNIsABijection 2013-03-22 13:48:57 2013-03-22 13:48:57 Koro (127) Koro (127) 7 Koro (127) Proof msc 03E20