Locally compact hausdorff space homeomorphism












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$begingroup$


Suppose that $X, Y$ are locally compact Hausdorff space. A bijective $f:Xto Y$ is also a homeomorphism?



Let $X^+=Xcup {infty_x}$ and $Y^+=Ycup{infty_y}$. By one point compactification, $X^+$ and $Y^+$ are compact.
Define function $f^*:X^+to Y^+$ such that $f^*|_X=f$ and $f^*(infty_x)=f^*(infty_y)$.



Then, $f^*:X^+to Y^+$ is well defined. Since $f$ is bijective function, $f^*$ is bijective function.



For open subset $U$ not containing $infty_y$ in $Y^+$, obviously $f^{-1}(U)$ is open in Y. Then, $f^{-1}(U)$ is open in $Y^+$.
For open subset $U$ containing $infty_y$ in $Y^+$, $f^{-1}(Y^+setminus U)$ is closed in X, because $Y^+ setminus U$ is closed and $f$ is continuous.



$f^*$ is homeomorphic if $f^*$ is continuous.



How can I show that $f^*$ is continuous and $X$ and $Y$ are homeomorphic?






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are wrong that $f^*$ is continuous. For example if $X=[0,1)$ and $Y=[0,1]$ then $X^+=[0,1]$ and $Y^+=[0,1]cup{infty}$. Such $f^*$ cannot be continuous because $X^*$ is connected while $Y^*$ is not. The particular mistake is that you claim that $f^{-1}(Y^+backslash U)$ is closed. It is closed in $X$ but not necessarily in $X^+$. The same is for $f^{-1}(U)$. It is closed in $X$, not necessarily in $X^+$.
    $endgroup$
    – freakish
    Dec 13 '18 at 16:15


















0












$begingroup$


Suppose that $X, Y$ are locally compact Hausdorff space. A bijective $f:Xto Y$ is also a homeomorphism?



Let $X^+=Xcup {infty_x}$ and $Y^+=Ycup{infty_y}$. By one point compactification, $X^+$ and $Y^+$ are compact.
Define function $f^*:X^+to Y^+$ such that $f^*|_X=f$ and $f^*(infty_x)=f^*(infty_y)$.



Then, $f^*:X^+to Y^+$ is well defined. Since $f$ is bijective function, $f^*$ is bijective function.



For open subset $U$ not containing $infty_y$ in $Y^+$, obviously $f^{-1}(U)$ is open in Y. Then, $f^{-1}(U)$ is open in $Y^+$.
For open subset $U$ containing $infty_y$ in $Y^+$, $f^{-1}(Y^+setminus U)$ is closed in X, because $Y^+ setminus U$ is closed and $f$ is continuous.



$f^*$ is homeomorphic if $f^*$ is continuous.



How can I show that $f^*$ is continuous and $X$ and $Y$ are homeomorphic?






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are wrong that $f^*$ is continuous. For example if $X=[0,1)$ and $Y=[0,1]$ then $X^+=[0,1]$ and $Y^+=[0,1]cup{infty}$. Such $f^*$ cannot be continuous because $X^*$ is connected while $Y^*$ is not. The particular mistake is that you claim that $f^{-1}(Y^+backslash U)$ is closed. It is closed in $X$ but not necessarily in $X^+$. The same is for $f^{-1}(U)$. It is closed in $X$, not necessarily in $X^+$.
    $endgroup$
    – freakish
    Dec 13 '18 at 16:15
















0












0








0





$begingroup$


Suppose that $X, Y$ are locally compact Hausdorff space. A bijective $f:Xto Y$ is also a homeomorphism?



Let $X^+=Xcup {infty_x}$ and $Y^+=Ycup{infty_y}$. By one point compactification, $X^+$ and $Y^+$ are compact.
Define function $f^*:X^+to Y^+$ such that $f^*|_X=f$ and $f^*(infty_x)=f^*(infty_y)$.



Then, $f^*:X^+to Y^+$ is well defined. Since $f$ is bijective function, $f^*$ is bijective function.



For open subset $U$ not containing $infty_y$ in $Y^+$, obviously $f^{-1}(U)$ is open in Y. Then, $f^{-1}(U)$ is open in $Y^+$.
For open subset $U$ containing $infty_y$ in $Y^+$, $f^{-1}(Y^+setminus U)$ is closed in X, because $Y^+ setminus U$ is closed and $f$ is continuous.



$f^*$ is homeomorphic if $f^*$ is continuous.



How can I show that $f^*$ is continuous and $X$ and $Y$ are homeomorphic?






general-topology







share|cite|improve this question











$endgroup$




Suppose that $X, Y$ are locally compact Hausdorff space. A bijective $f:Xto Y$ is also a homeomorphism?



Let $X^+=Xcup {infty_x}$ and $Y^+=Ycup{infty_y}$. By one point compactification, $X^+$ and $Y^+$ are compact.
Define function $f^*:X^+to Y^+$ such that $f^*|_X=f$ and $f^*(infty_x)=f^*(infty_y)$.



Then, $f^*:X^+to Y^+$ is well defined. Since $f$ is bijective function, $f^*$ is bijective function.



For open subset $U$ not containing $infty_y$ in $Y^+$, obviously $f^{-1}(U)$ is open in Y. Then, $f^{-1}(U)$ is open in $Y^+$.
For open subset $U$ containing $infty_y$ in $Y^+$, $f^{-1}(Y^+setminus U)$ is closed in X, because $Y^+ setminus U$ is closed and $f$ is continuous.



$f^*$ is homeomorphic if $f^*$ is continuous.



How can I show that $f^*$ is continuous and $X$ and $Y$ are homeomorphic?






general-topology




general-topology






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share|cite|improve this question













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edited Dec 13 '18 at 16:09







dlfjsemf

















asked Dec 13 '18 at 11:01









dlfjsemfdlfjsemf

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968












  • $begingroup$
    You are wrong that $f^*$ is continuous. For example if $X=[0,1)$ and $Y=[0,1]$ then $X^+=[0,1]$ and $Y^+=[0,1]cup{infty}$. Such $f^*$ cannot be continuous because $X^*$ is connected while $Y^*$ is not. The particular mistake is that you claim that $f^{-1}(Y^+backslash U)$ is closed. It is closed in $X$ but not necessarily in $X^+$. The same is for $f^{-1}(U)$. It is closed in $X$, not necessarily in $X^+$.
    $endgroup$
    – freakish
    Dec 13 '18 at 16:15




















  • $begingroup$
    You are wrong that $f^*$ is continuous. For example if $X=[0,1)$ and $Y=[0,1]$ then $X^+=[0,1]$ and $Y^+=[0,1]cup{infty}$. Such $f^*$ cannot be continuous because $X^*$ is connected while $Y^*$ is not. The particular mistake is that you claim that $f^{-1}(Y^+backslash U)$ is closed. It is closed in $X$ but not necessarily in $X^+$. The same is for $f^{-1}(U)$. It is closed in $X$, not necessarily in $X^+$.
    $endgroup$
    – freakish
    Dec 13 '18 at 16:15


















$begingroup$
You are wrong that $f^*$ is continuous. For example if $X=[0,1)$ and $Y=[0,1]$ then $X^+=[0,1]$ and $Y^+=[0,1]cup{infty}$. Such $f^*$ cannot be continuous because $X^*$ is connected while $Y^*$ is not. The particular mistake is that you claim that $f^{-1}(Y^+backslash U)$ is closed. It is closed in $X$ but not necessarily in $X^+$. The same is for $f^{-1}(U)$. It is closed in $X$, not necessarily in $X^+$.
$endgroup$
– freakish
Dec 13 '18 at 16:15






$begingroup$
You are wrong that $f^*$ is continuous. For example if $X=[0,1)$ and $Y=[0,1]$ then $X^+=[0,1]$ and $Y^+=[0,1]cup{infty}$. Such $f^*$ cannot be continuous because $X^*$ is connected while $Y^*$ is not. The particular mistake is that you claim that $f^{-1}(Y^+backslash U)$ is closed. It is closed in $X$ but not necessarily in $X^+$. The same is for $f^{-1}(U)$. It is closed in $X$, not necessarily in $X^+$.
$endgroup$
– freakish
Dec 13 '18 at 16:15












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It is not true. Let $X = [0,2pi), Y = S^1$. Then $f : X to Y, f(t) = e^{it}$, is a continuous bijection, but not a homeomorphism.



You may suspect that the reason for this phenomenon is that $Y$ is compact and $X$ is not compact. But you can easily modify the above example to $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.






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    $begingroup$

    It is not true. Let $X = [0,2pi), Y = S^1$. Then $f : X to Y, f(t) = e^{it}$, is a continuous bijection, but not a homeomorphism.



    You may suspect that the reason for this phenomenon is that $Y$ is compact and $X$ is not compact. But you can easily modify the above example to $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      It is not true. Let $X = [0,2pi), Y = S^1$. Then $f : X to Y, f(t) = e^{it}$, is a continuous bijection, but not a homeomorphism.



      You may suspect that the reason for this phenomenon is that $Y$ is compact and $X$ is not compact. But you can easily modify the above example to $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        It is not true. Let $X = [0,2pi), Y = S^1$. Then $f : X to Y, f(t) = e^{it}$, is a continuous bijection, but not a homeomorphism.



        You may suspect that the reason for this phenomenon is that $Y$ is compact and $X$ is not compact. But you can easily modify the above example to $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.






        share|cite|improve this answer











        $endgroup$



        It is not true. Let $X = [0,2pi), Y = S^1$. Then $f : X to Y, f(t) = e^{it}$, is a continuous bijection, but not a homeomorphism.



        You may suspect that the reason for this phenomenon is that $Y$ is compact and $X$ is not compact. But you can easily modify the above example to $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 13 '18 at 12:11

























        answered Dec 13 '18 at 11:19









        Paul FrostPaul Frost

        10.4k3933




        10.4k3933






























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