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Image under covering map of Hausdorff space is Hausdorff?
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Let $overlineX rightarrow X$ be a covering space, where $overlineX$ is a Hausdorff space. Does $X$ have to be Hausdorff?
I assume there is a counterexample, but I couldn't find one yet.
general-topology covering-spaces separation-axioms
edited Jul 14 at 22:00
Eric Wofsey
163k12189301
asked Jul 14 at 20:34
L. Gitin
584
add a comment |Â
up vote
2
down vote
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Let $overlineX rightarrow X$ be a covering space, where $overlineX$ is a Hausdorff space. Does $X$ have to be Hausdorff?
I assume there is a counterexample, but I couldn't find one yet.
general-topology covering-spaces separation-axioms
edited Jul 14 at 22:00
Eric Wofsey
163k12189301
asked Jul 14 at 20:34
L. Gitin
584
Try $X$ nonHausdorff and let the cover be a product with a discrete space.
– Randall
Jul 14 at 21:00
1
Why is $bar X$ Hausdorff then?
– Fan Zheng
Jul 14 at 21:02
You’re right that doesn’t work. Never mind.
– Randall
Jul 14 at 21:27
1
Example here.
– user574380
Jul 14 at 21:33
1
It is true if the fibres $p^-1(x), x in X$ are finite. See math.stackexchange.com/q/396989.
– Paul Frost
Jul 14 at 21:39
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $overlineX rightarrow X$ be a covering space, where $overlineX$ is a Hausdorff space. Does $X$ have to be Hausdorff?
I assume there is a counterexample, but I couldn't find one yet.
general-topology covering-spaces separation-axioms
edited Jul 14 at 22:00
Eric Wofsey
163k12189301
asked Jul 14 at 20:34
L. Gitin
584
Let $overlineX rightarrow X$ be a covering space, where $overlineX$ is a Hausdorff space. Does $X$ have to be Hausdorff?
I assume there is a counterexample, but I couldn't find one yet.
general-topology covering-spaces separation-axioms
edited Jul 14 at 22:00
Eric Wofsey
163k12189301
asked Jul 14 at 20:34
L. Gitin
584
edited Jul 14 at 22:00
Eric Wofsey
163k12189301
edited Jul 14 at 22:00
Eric Wofsey
163k12189301
edited Jul 14 at 22:00
Eric Wofsey
163k12189301
163k12189301
asked Jul 14 at 20:34
L. Gitin
584
asked Jul 14 at 20:34
L. Gitin
584
asked Jul 14 at 20:34
L. Gitin
584
584
Try $X$ nonHausdorff and let the cover be a product with a discrete space.
– Randall
Jul 14 at 21:00
1
Why is $bar X$ Hausdorff then?
– Fan Zheng
Jul 14 at 21:02
You’re right that doesn’t work. Never mind.
– Randall
Jul 14 at 21:27
1
Example here.
– user574380
Jul 14 at 21:33
1
It is true if the fibres $p^-1(x), x in X$ are finite. See math.stackexchange.com/q/396989.
– Paul Frost
Jul 14 at 21:39
add a comment |Â
Try $X$ nonHausdorff and let the cover be a product with a discrete space.
– Randall
Jul 14 at 21:00
1
Why is $bar X$ Hausdorff then?
– Fan Zheng
Jul 14 at 21:02
You’re right that doesn’t work. Never mind.
– Randall
Jul 14 at 21:27
1
Example here.
– user574380
Jul 14 at 21:33
1
It is true if the fibres $p^-1(x), x in X$ are finite. See math.stackexchange.com/q/396989.
– Paul Frost
Jul 14 at 21:39
Try $X$ nonHausdorff and let the cover be a product with a discrete space.
– Randall
Jul 14 at 21:00
Try $X$ nonHausdorff and let the cover be a product with a discrete space.
– Randall
Jul 14 at 21:00
1
1
Why is $bar X$ Hausdorff then?
– Fan Zheng
Jul 14 at 21:02
Why is $bar X$ Hausdorff then?
– Fan Zheng
Jul 14 at 21:02
You’re right that doesn’t work. Never mind.
– Randall
Jul 14 at 21:27
You’re right that doesn’t work. Never mind.
– Randall
Jul 14 at 21:27
1
1
Example here.
– user574380
Jul 14 at 21:33
Example here.
– user574380
Jul 14 at 21:33
1
1
It is true if the fibres $p^-1(x), x in X$ are finite. See math.stackexchange.com/q/396989.
– Paul Frost
Jul 14 at 21:39
It is true if the fibres $p^-1(x), x in X$ are finite. See math.stackexchange.com/q/396989.
– Paul Frost
Jul 14 at 21:39
add a comment |Â
1 Answer
1
active
oldest
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up vote
4
down vote
accepted
Here's a counterexample. Let $X=mathbbNcupinfty,infty'$ where a set is open iff it is either contained in $mathbbN$ or cofinite. Let $overlineX=XtimesmathbbZ$ with the following topology (not the product topology). A set $UsubseteqoverlineX$ is open iff it satisfies the following two properties:
- For all $ninmathbbZ$ such that $(infty,n)in U$, $(m,n+m)in U$ for all but finitely many $minmathbbN$.
- For all $ninmathbbZ$ such that $(infty',n)in U$, $(m,n-m)in U$ for all but finitely many $minmathbbN$.
I now claim that the projection map $p:overlineXto X$ is a covering map. First, it is easy to check that $p$ is continuous. Any $minmathbbN$ has an evenly covered neighborhood, namely $m$ (since $p^-1(m)=mtimesmathbbZ$ is discrete in $overlineX$). Finally, I claim $infty$ and $infty'$ have evenly covered neighborhoods. Indeed, if $V=Xsetminusinfty'$, then $p^-1(V)$ is homeomorphic to $VtimesmathbbZ$ (with the product topology) via the map sending $(m,n)$ to $(m,n-m)$ and $(infty,n)$ to $(infty,n)$, so $V$ is an evenly covered neighborhood of $infty$. Similarly, $W=Xsetminusinfty$ is evenly covered since $p^-1(W)cong Wtimes mathbbZ$ by mapping $(m,n)$ to $(m,n+m)$.
Finally, $X$ is not Hausdorff ($infty$ and $infty'$ cannot be separated by open sets) but $overlineX$ is Hausdorff (we can get disjoint neighborhoods of $(infty,n)$ and $(infty',n')$ since the collections of points of the form $(m,n+m)$ and $(m,n'-m)$ only intersect at at most one point).
answered Jul 14 at 21:39
Eric Wofsey
163k12189301
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Here's a counterexample. Let $X=mathbbNcupinfty,infty'$ where a set is open iff it is either contained in $mathbbN$ or cofinite. Let $overlineX=XtimesmathbbZ$ with the following topology (not the product topology). A set $UsubseteqoverlineX$ is open iff it satisfies the following two properties:
- For all $ninmathbbZ$ such that $(infty,n)in U$, $(m,n+m)in U$ for all but finitely many $minmathbbN$.
- For all $ninmathbbZ$ such that $(infty',n)in U$, $(m,n-m)in U$ for all but finitely many $minmathbbN$.
I now claim that the projection map $p:overlineXto X$ is a covering map. First, it is easy to check that $p$ is continuous. Any $minmathbbN$ has an evenly covered neighborhood, namely $m$ (since $p^-1(m)=mtimesmathbbZ$ is discrete in $overlineX$). Finally, I claim $infty$ and $infty'$ have evenly covered neighborhoods. Indeed, if $V=Xsetminusinfty'$, then $p^-1(V)$ is homeomorphic to $VtimesmathbbZ$ (with the product topology) via the map sending $(m,n)$ to $(m,n-m)$ and $(infty,n)$ to $(infty,n)$, so $V$ is an evenly covered neighborhood of $infty$. Similarly, $W=Xsetminusinfty$ is evenly covered since $p^-1(W)cong Wtimes mathbbZ$ by mapping $(m,n)$ to $(m,n+m)$.
Finally, $X$ is not Hausdorff ($infty$ and $infty'$ cannot be separated by open sets) but $overlineX$ is Hausdorff (we can get disjoint neighborhoods of $(infty,n)$ and $(infty',n')$ since the collections of points of the form $(m,n+m)$ and $(m,n'-m)$ only intersect at at most one point).
answered Jul 14 at 21:39
Eric Wofsey
163k12189301
add a comment |Â
up vote
4
down vote
accepted
Here's a counterexample. Let $X=mathbbNcupinfty,infty'$ where a set is open iff it is either contained in $mathbbN$ or cofinite. Let $overlineX=XtimesmathbbZ$ with the following topology (not the product topology). A set $UsubseteqoverlineX$ is open iff it satisfies the following two properties:
- For all $ninmathbbZ$ such that $(infty,n)in U$, $(m,n+m)in U$ for all but finitely many $minmathbbN$.
- For all $ninmathbbZ$ such that $(infty',n)in U$, $(m,n-m)in U$ for all but finitely many $minmathbbN$.
I now claim that the projection map $p:overlineXto X$ is a covering map. First, it is easy to check that $p$ is continuous. Any $minmathbbN$ has an evenly covered neighborhood, namely $m$ (since $p^-1(m)=mtimesmathbbZ$ is discrete in $overlineX$). Finally, I claim $infty$ and $infty'$ have evenly covered neighborhoods. Indeed, if $V=Xsetminusinfty'$, then $p^-1(V)$ is homeomorphic to $VtimesmathbbZ$ (with the product topology) via the map sending $(m,n)$ to $(m,n-m)$ and $(infty,n)$ to $(infty,n)$, so $V$ is an evenly covered neighborhood of $infty$. Similarly, $W=Xsetminusinfty$ is evenly covered since $p^-1(W)cong Wtimes mathbbZ$ by mapping $(m,n)$ to $(m,n+m)$.
Finally, $X$ is not Hausdorff ($infty$ and $infty'$ cannot be separated by open sets) but $overlineX$ is Hausdorff (we can get disjoint neighborhoods of $(infty,n)$ and $(infty',n')$ since the collections of points of the form $(m,n+m)$ and $(m,n'-m)$ only intersect at at most one point).
answered Jul 14 at 21:39
Eric Wofsey
163k12189301
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Here's a counterexample. Let $X=mathbbNcupinfty,infty'$ where a set is open iff it is either contained in $mathbbN$ or cofinite. Let $overlineX=XtimesmathbbZ$ with the following topology (not the product topology). A set $UsubseteqoverlineX$ is open iff it satisfies the following two properties:
- For all $ninmathbbZ$ such that $(infty,n)in U$, $(m,n+m)in U$ for all but finitely many $minmathbbN$.
- For all $ninmathbbZ$ such that $(infty',n)in U$, $(m,n-m)in U$ for all but finitely many $minmathbbN$.
I now claim that the projection map $p:overlineXto X$ is a covering map. First, it is easy to check that $p$ is continuous. Any $minmathbbN$ has an evenly covered neighborhood, namely $m$ (since $p^-1(m)=mtimesmathbbZ$ is discrete in $overlineX$). Finally, I claim $infty$ and $infty'$ have evenly covered neighborhoods. Indeed, if $V=Xsetminusinfty'$, then $p^-1(V)$ is homeomorphic to $VtimesmathbbZ$ (with the product topology) via the map sending $(m,n)$ to $(m,n-m)$ and $(infty,n)$ to $(infty,n)$, so $V$ is an evenly covered neighborhood of $infty$. Similarly, $W=Xsetminusinfty$ is evenly covered since $p^-1(W)cong Wtimes mathbbZ$ by mapping $(m,n)$ to $(m,n+m)$.
Finally, $X$ is not Hausdorff ($infty$ and $infty'$ cannot be separated by open sets) but $overlineX$ is Hausdorff (we can get disjoint neighborhoods of $(infty,n)$ and $(infty',n')$ since the collections of points of the form $(m,n+m)$ and $(m,n'-m)$ only intersect at at most one point).
answered Jul 14 at 21:39
Eric Wofsey
163k12189301
Here's a counterexample. Let $X=mathbbNcupinfty,infty'$ where a set is open iff it is either contained in $mathbbN$ or cofinite. Let $overlineX=XtimesmathbbZ$ with the following topology (not the product topology). A set $UsubseteqoverlineX$ is open iff it satisfies the following two properties:
- For all $ninmathbbZ$ such that $(infty,n)in U$, $(m,n+m)in U$ for all but finitely many $minmathbbN$.
- For all $ninmathbbZ$ such that $(infty',n)in U$, $(m,n-m)in U$ for all but finitely many $minmathbbN$.
I now claim that the projection map $p:overlineXto X$ is a covering map. First, it is easy to check that $p$ is continuous. Any $minmathbbN$ has an evenly covered neighborhood, namely $m$ (since $p^-1(m)=mtimesmathbbZ$ is discrete in $overlineX$). Finally, I claim $infty$ and $infty'$ have evenly covered neighborhoods. Indeed, if $V=Xsetminusinfty'$, then $p^-1(V)$ is homeomorphic to $VtimesmathbbZ$ (with the product topology) via the map sending $(m,n)$ to $(m,n-m)$ and $(infty,n)$ to $(infty,n)$, so $V$ is an evenly covered neighborhood of $infty$. Similarly, $W=Xsetminusinfty$ is evenly covered since $p^-1(W)cong Wtimes mathbbZ$ by mapping $(m,n)$ to $(m,n+m)$.
Finally, $X$ is not Hausdorff ($infty$ and $infty'$ cannot be separated by open sets) but $overlineX$ is Hausdorff (we can get disjoint neighborhoods of $(infty,n)$ and $(infty',n')$ since the collections of points of the form $(m,n+m)$ and $(m,n'-m)$ only intersect at at most one point).
answered Jul 14 at 21:39
Eric Wofsey
163k12189301
answered Jul 14 at 21:39
Eric Wofsey
163k12189301
answered Jul 14 at 21:39
Eric Wofsey
163k12189301
answered Jul 14 at 21:39
Eric Wofsey
163k12189301
163k12189301
add a comment |Â
add a comment |Â
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Try $X$ nonHausdorff and let the cover be a product with a discrete space.
– Randall
Jul 14 at 21:00
1
Why is $bar X$ Hausdorff then?
– Fan Zheng
Jul 14 at 21:02
You’re right that doesn’t work. Never mind.
– Randall
Jul 14 at 21:27
1
Example here.
– user574380
Jul 14 at 21:33
1
It is true if the fibres $p^-1(x), x in X$ are finite. See math.stackexchange.com/q/396989.
– Paul Frost
Jul 14 at 21:39