Locally compact hausdorff space homeomorphism
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
asked Dec 13 '18 at 11:01
dlfjsemfdlfjsemf
968
add a comment |
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
asked Dec 13 '18 at 11:01
dlfjsemfdlfjsemf
968
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 $X^+$.
– freakish
Dec 13 '18 at 16:15
add a comment |
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
asked Dec 13 '18 at 11:01
dlfjsemfdlfjsemf
968
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
asked Dec 13 '18 at 11:01
dlfjsemfdlfjsemf
968
asked Dec 13 '18 at 11:01
dlfjsemfdlfjsemf
968
edited Dec 13 '18 at 16:09
dlfjsemf
asked Dec 13 '18 at 11:01
dlfjsemfdlfjsemf
968
asked Dec 13 '18 at 11:01
dlfjsemfdlfjsemf
968
asked Dec 13 '18 at 11:01
dlfjsemfdlfjsemf
968
968
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 $X^+$.
– freakish
Dec 13 '18 at 16:15
add a comment |
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 $X^+$.
– freakish
Dec 13 '18 at 16:15
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 $X^+$.
– freakish
Dec 13 '18 at 16:15
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 $X^+$.
– freakish
Dec 13 '18 at 16:15
add a comment |
1 Answer
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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 $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.
answered Dec 13 '18 at 11:19
Paul FrostPaul Frost
9,7853732
add a comment |
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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 $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.
answered Dec 13 '18 at 11:19
Paul FrostPaul Frost
9,7853732
add a comment |
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 $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.
answered Dec 13 '18 at 11:19
Paul FrostPaul Frost
9,7853732
add a comment |
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 $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.
answered Dec 13 '18 at 11:19
Paul FrostPaul Frost
9,7853732
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 $X' = (-1,1), Y' = S^1 cup [1,2) times { 0 }$.
answered Dec 13 '18 at 11:19
Paul FrostPaul Frost
9,7853732
edited Dec 13 '18 at 12:11
answered Dec 13 '18 at 11:19
Paul FrostPaul Frost
9,7853732
answered Dec 13 '18 at 11:19
Paul FrostPaul Frost
9,7853732
answered Dec 13 '18 at 11:19
Paul FrostPaul Frost
9,7853732
9,7853732
add a comment |
add a comment |
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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 $X^+$.
– freakish
Dec 13 '18 at 16:15