I need some help with a topological proof, hopefully there is someone out there that can help.

Prove that a bijection f: X→Y is a homeomorphism if and only if f and f^-1 map closed sets to closed sets.

Thanks

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- Mar 25th 2008, 02:21 PMreagan3ncTopology Proof
I need some help with a topological proof, hopefully there is someone out there that can help.

Prove that a bijection f: X→Y is a homeomorphism if and only if f and f^-1 map closed sets to closed sets.

Thanks - Mar 25th 2008, 03:30 PMiknowone
The definition of a homeomorphism is a map which is bijective continuous and open (maps open sets to open sets).

Assume $\displaystyle f: X \rightarrow Y$ is a homeomorphism.

$\displaystyle f$ is continuous and open so $\displaystyle f^{-1}$ is continuous and open. But then $\displaystyle f$ is closed (maps closed sets to closed sets) and so is $\displaystyle f^{-1}$ since let $\displaystyle C$ be a closed subset of $\displaystyle Y$. Then $\displaystyle C^c$ is open in $\displaystyle Y$ and so $\displaystyle f^{-1}(C^c)$ is open in $\displaystyle X$. But $\displaystyle f^{-1}(C^c)=(f^{-1}(C))^c$ and therefore $\displaystyle f^{-1}(C) = (f^{-1}(C)^c)^c$ is closed in $\displaystyle X$. Same argument shows $\displaystyle f$ is a closed map.

Now assume that $\displaystyle f$ and $\displaystyle f^{-1}$ are bijective maps which take closed sets to closed sets. We need to show that $\displaystyle f$ is continuous and open. Consider an arbitrary open set $\displaystyle U \subset Y$. $\displaystyle U^c$ is closed and $\displaystyle f^{-1}(U^c) = f^{-1}(U)^c$ is closed in $\displaystyle X$. Therefore $\displaystyle f^{-1}(U)$ is open in $\displaystyle X$. So $\displaystyle f$ is continuous. Similar argument shows $\displaystyle f$ is open. - Mar 26th 2008, 04:37 AMreagan3nc
THanks, but i have one question are C^c and U^c the compliments of C and U? Just want to make sure.

Thanks - Mar 26th 2008, 06:58 AMThePerfectHacker