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Math Help - Showing Hausdorff and Compact spaces are homeomorphic

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    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by okor View Post
    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.
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    Quote Originally Posted by Drexel28 View Post
    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by okor View Post
    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.
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    Quote Originally Posted by Drexel28 View Post
    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by okor View Post
    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!
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    Quote Originally Posted by Drexel28 View Post
    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.
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