# 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) ProofOfInverseFunctionTheoremtopologicalSpaces 2013-03-22 13:31:55 2013-03-22 13:31:55 paolini (1187) paolini (1187) 5 paolini (1187) Proof msc 54C05