Showing Hausdorff and Compact spaces are homeomorphic

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November 28th 2010, 04:28 PM
okor

Showing Hausdorff and Compact spaces are homeomorphic

Ok, I am totally stuck on this problem, I just am not really sure where to go with it.

Let X and Y be topological spaces and let $f:X \rightarrow Y$ be a one-to-one, onto function with a continuous inverse. Prove that if X is Hausdorff and Y is compact, then f is a homeomorphism.

The only thing I could really get is that having a continuous inverse takes the place of the function needing to be open, so all that is left is showing that the function is continuous. So I have to show for any set V in Y, $f^{-1}(V)$ is in X. I am not sure how compactness and Hausdorff can be used to prove this.
November 28th 2010, 06:59 PM
Drexel28

Quote:

Originally Posted by okor

Ok, I am totally stuck on this problem, I just am not really sure where to go with it.

Let X and Y be topological spaces and let $f:X \rightarrow Y$ be a one-to-one, onto function with a continuous inverse. Prove that if X is Hausdorff and Y is compact, then f is a homeomorphism.

The only thing I could really get is that having a continuous inverse takes the place of the function needing to be open, so all that is left is showing that the function is continuous. So I have to show for any set V in Y, $f^{-1}(V)$ is in X. I am not sure how compactness and Hausdorff can be used to prove this.

Recall that $f^{-1}$ is continuous if and only if the preimage of closed sets is closed, said differently if and only if $\left(f^{-1}\right)^{-1}\left(C\right)=f\left(C\right)$ is closed for each closed $C\subseteq X$ . Recall that a closed subspace of a compact space is compact and thus $C$ is compact. That said, compactness is invariant under continuous maps so that $f\left(C\right)\subseteq Y$ is compact, but since $Y$ is Hausdorff it follows that $f\left(C\right)$ is closed (since compact subspaces of Hausdorff spaces are closed). The conclusion follows by previous commment.
November 30th 2010, 04:04 PM
okor

Quote:

Originally Posted by Drexel28

Recall that $f^{-1}$ is continuous if and only if the preimage of closed sets is closed, said differently if and only if $\left(f^{-1}\right)^{-1}\left(C\right)=f\left(C\right)$ is closed for each closed $C\subseteq X$ . Recall that a closed subspace of a compact space is compact and thus $C$ is compact. That said, compactness is invariant under continuous maps so that $f\left(C\right)\subseteq Y$ is compact, but since $Y$ is Hausdorff it follows that $f\left(C\right)$ is closed (since compact subspaces of Hausdorff spaces are closed). The conclusion follows by previous commment.

Thank you a ton for the help. I had a little trouble following at first, but I read through the theorems on compactness in my textbook and it makes perfect sense now.

Here is what I came up with:

Let A be a closed subset of Y. Since Y is compact, A is compact. Then $f^{-1}(A)$ is compact because $f^{-1}$ is continuous. Since X is Hausdorff, $f^{-1}(A)$ is closed, hence f is continuous and therefore a homeomorphism.
November 30th 2010, 05:03 PM
Drexel28

Quote:

Originally Posted by okor

Thank you a ton for the help. I had a little trouble following at first, but I read through the theorems on compactness in my textbook and it makes perfect sense now.

Here is what I came up with:

Let A be a closed subset of Y. Since Y is compact, A is compact. Then $f^{-1}(A)$ is compact because $f^{-1}$ is continuous. Since X is Hausdorff, $f^{-1}(A)$ is closed, hence f is continuous and therefore a homeomorphism.

Hmm...I feel as though you've mixed up the functions. You know that since $f$ is continuous and $A$ closed that $f\left(A\right)=\left(f^{-1}\left(A\right)\right)$ is closed since compact subspaces of Hausdorff spaces are closed. Thus, the preimage of closed sets under $f^{-1}$ are closed...so $f^{-1}$ is continous and thus $f$ is a homeomorphism.
November 30th 2010, 05:14 PM
okor

Quote:

Originally Posted by Drexel28

Hmm...I feel as though you've mixed up the functions. You know that since $f$ is continuous and $A$ closed that $f\left(A\right)=\left(f^{-1}\left(A\right)\right)$ is closed since compact subspaces of Hausdorff spaces are closed. Thus, the preimage of closed sets under $f^{-1}$ are closed...so $f^{-1}$ is continous and thus $f$ is a homeomorphism.

Err, I am a little confused now.

I know $f^{-1}$ is continuous, I need to show f is continuous. I apologize if I have gotten things mixed up.
November 30th 2010, 05:21 PM
Drexel28

Quote:

Originally Posted by okor

Err, I am a little confused now.

I know $f^{-1}$ is continuous, I need to show f is continuous. I apologize if I have gotten things mixed up.

Oh, my apologies. I did misread. The same concept works taking $g=f^{-1}$ :)

Your proof is correct then!
November 30th 2010, 05:23 PM
okor

Quote:

Originally Posted by Drexel28

Oh, my apologies. I did misread. The same concept works taking $g=f^{-1}$ :)

Your proof is correct then!

Awesome, thank you! I appreciate the help working through this problem.