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