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 $n=2,3,\mathrm{\dots }$. 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 $x=y$. $\mathrm{\square }$

Title	mapping of period $n$ is a bijection
Canonical name	MappingOfPeriodNIsABijection
Date of creation	2013-03-22 13:48:57
Last modified on	2013-03-22 13:48:57
Owner	Koro (127)
Last modified by	Koro (127)
Numerical id	7
Author	Koro (127)
Entry type	Proof
Classification	msc 03E20