# 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 |