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Thread: Prove homeomorphism

  1. #1
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    Prove homeomorphism

    Let $\displaystyle M$, $\displaystyle N$ be metric spaces and assume $\displaystyle M$ is compact. Show that if $\displaystyle f:M \rightarrow N $ is a bijective and continious function, then $\displaystyle f$ is a homeomorphism.
    And does this hold even if we dont assume that M is compact?
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    Quote Originally Posted by BenH View Post
    Let $\displaystyle M$, $\displaystyle N$ be metric spaces and assume $\displaystyle M$ is compact. Show that if $\displaystyle f:M \rightarrow N $ is a bijective and continious function, then $\displaystyle f$ 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 $\displaystyle f^{-1}:N \to M$ is continuous. or equivalently that $\displaystyle f$ 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.
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    Quote Originally Posted by TheEmptySet View Post
    The only thing you need to prove is that $\displaystyle f^{-1}:N \to M$ is continuous.
    Yes, I knew that. Through the definition that means I have to prove that for every $\displaystyle a\in N$ the following most hold:
    For every $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ such that
    $\displaystyle f^{-1}(B_{\delta}(a))\subset B_{\epsilon}(f^{-1}(a))$,
    i.e.
    $\displaystyle d_M(f^{-1}(x),f^{-1}(a)) < \epsilon$ if $\displaystyle d_N(x,a)<\delta$.
    But how do I prove that?
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    Quote Originally Posted by BenH View Post
    Yes, I knew that. Through the definition that means I have to prove that for every $\displaystyle a\in N$ the following most hold:
    For every $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ such that
    $\displaystyle f^{-1}(B_{\delta}(a))\subset B_{\epsilon}(f^{-1}(a))$,
    i.e.
    $\displaystyle d_M(f^{-1}(x),f^{-1}(a)) < \epsilon$ if $\displaystyle d_N(x,a)<\delta$.
    But how do I prove that?
    Yes the point I was trying to make is that $\displaystyle f^{-1}$ 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 $\displaystyle A \subset M$ be closed. Since M is a compact space and A is a closed subset of M A is compact.

    Now since $\displaystyle f$ is continuous by hypothesis $\displaystyle f(A)$ 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.
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  5. #5
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    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?
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    Quote Originally Posted by BenH View Post
    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 $\displaystyle d(x,y) = 1$ whenever $\displaystyle y\ne x$. In this metric, every set is open. So every mapping from a space with the discrete metric is continuous.
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