Prove homeomorphism

**BenH** · March 12th 2011, 07:32 AM

Let $M$ , $N$ be metric spaces and assume $M$ is compact. Show that if $f:M \rightarrow N$ is a bijective and continious function, then $f$ is a homeomorphism.
And does this hold even if we dont assume that M is compact?

**TheEmptySet** · March 12th 2011, 08:02 AM

Originally Posted by BenH

Let $M$ , $N$ be metric spaces and assume $M$ is compact. Show that if $f:M \rightarrow N$ is a bijective and continious function, then $f$ is a homeomorphism.
And does this hold even if we dont assume that M is compact?

Hint:

The only thing you need to prove is that $f^{-1}:N \to M$ is continuous. or equivalently that $f$ is an open map or again if and only if it is a closed map. In my option the last will be the easiest to show.

**BenH** · March 12th 2011, 09:09 AM

Originally Posted by TheEmptySet

The only thing you need to prove is that $f^{-1}:N \to M$ is continuous.

Yes, I knew that. Through the definition that means I have to prove that for every $a\in N$ the following most hold:
For every $\epsilon > 0$ there exists $\delta > 0$ such that
$f^{-1}(B_{\delta}(a))\subset B_{\epsilon}(f^{-1}(a))$ ,
i.e.
$d_M(f^{-1}(x),f^{-1}(a)) < \epsilon$ if $d_N(x,a)<\delta$ .
But how do I prove that?

**TheEmptySet** · March 12th 2011, 09:20 AM

Originally Posted by BenH

Yes, I knew that. Through the definition that means I have to prove that for every $a\in N$ the following most hold:
For every $\epsilon > 0$ there exists $\delta > 0$ such that
$f^{-1}(B_{\delta}(a))\subset B_{\epsilon}(f^{-1}(a))$ ,
i.e.
$d_M(f^{-1}(x),f^{-1}(a)) < \epsilon$ if $d_N(x,a)<\delta$ .
But how do I prove that?

Yes the point I was trying to make is that $f^{-1}$ being continuous is the same as being an open map or a closed map.

Open and closed maps - Wikipedia, the free encyclopedia

Since this is the cases we just need a few facts.

A map is closed if it maps closed sets to closed sets.

Let $A \subset M$ be closed. Since M is a compact space and A is a closed subset of M A is compact.

Now since $f$ is continuous by hypothesis $f(A)$ is a compact set because the continuous image of a compact set is always compact.
Since compact sets are closed f must be a closed map.

**BenH** · March 12th 2011, 11:09 AM

I wasn't too familiar with open and closed maps, thats why i didn't get your point. But now it makes sense. thank you!
How about if you drop the assumption that M is compact. Could you prove that a continuous bijection not necessarily has a continuous inverse? Any examples?

**Opalg** · March 12th 2011, 01:17 PM

Originally Posted by BenH

How about if you drop the assumption that M is compact. Could you prove that a continuous bijection not necessarily has a continuous inverse? Any examples?

Let M be the unit interval [0,1] with the discrete metric, and let N be the same set with the usual metric. The identity map is then a continuous bijection, but the inverse is not continuous.

The discrete metric is given by $d(x,y) = 1$ whenever $y\ne x$ . In this metric, every set is open. So every mapping from a space with the discrete metric is continuous.

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