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Math Help - Homeomorphism

  1. #1
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    Homeomorphism

    Something my lecturer said

    if you have a map x:U->V that is a homeomorphism and you have that x(u_0,v_0) = p.

    Then the map x:U\{(u_0,v_0)}->V\{p}

    Then is this a homeomorphism as well? and if so, why?

    Thanks!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Homeomorphism

    Quote Originally Posted by mathswannabe View Post
    Something my lecturer said

    if you have a map x:U->V that is a homeomorphism and you have that x(u_0,v_0) = p.

    Then the map x:U\{(u_0,v_0)}->V\{p}

    Then is this a homeomorphism as well? and if so, why?

    Thanks!
    I think you should get us started. What do you think. Clearly the resulting map is a bijection, so why is it bicontinuous?
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  3. #3
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    Re: Homeomorphism

    thats the bit where I get worried.

    Im thinking if you know there is cts bijection and cts inverse when you didnt place any restriction on your domain, then when you do place those restrictions, then it will still map the same, just excluding that point. The thing that gets me a little weird is i just keep seeing in my head a straight line thats not defined at a point thus it cant be cts. But i guess its okay since if you think of 1/x that is a cts function at all points that it is defined on R...

    i dunno.

    im so bad at this stuff
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Homeomorphism

    Quote Originally Posted by mathswannabe View Post
    thats the bit where I get worried.

    Im thinking if you know there is cts bijection and cts inverse when you didnt place any restriction on your domain, then when you do place those restrictions, then it will still map the same, just excluding that point. The thing that gets me a little weird is i just keep seeing in my head a straight line thats not defined at a point thus it cant be cts. But i guess its okay since if you think of 1/x that is a cts function at all points that it is defined on R...

    i dunno.

    im so bad at this stuff
    Right, so it suffices to show that the preimage of open sets for X-\{x\} to Y say, since you can just apply the result in reverse. But, this is trivial since f^{-1}(U) when f is restriced to X-\{x\} is just f^{-1}(U)\cap(X-\{x\}) which is open in the subspace topology. Get the idea?
    Last edited by Ackbeet; September 8th 2011 at 03:50 PM.
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