# proof of inverse function theorem (topological spaces)

We only have to prove that whenever $A\subset X$ is an open set, then also $B={({f}^{-1})}^{-1}(A)=f(A)\subset Y$ is open ($f$ is an open mapping). Equivalently it is enough to prove that ${B}^{\prime}=Y\setminus B$ is closed.

Since $f$ is bijective^{} we have ${B}^{\prime}=Y\setminus B=f(X\setminus A)$

As ${A}^{\prime}=X\setminus A$ is closed and since $X$ is compact^{} ${A}^{\prime}$ is compact too (this and the following are well know properties of compact spaces).
Moreover being $f$ continuous^{} we know that also ${B}^{\prime}=f({A}^{\prime})$ is compact. Finally since $Y$ is Hausdorff^{} then ${B}^{\prime}$ is closed.

Title | proof of inverse function theorem (topological spaces^{}) |
---|---|

Canonical name | ProofOfInverseFunctionTheoremtopologicalSpaces |

Date of creation | 2013-03-22 13:31:55 |

Last modified on | 2013-03-22 13:31:55 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 5 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 54C05 |