Classification of covering spaces for spaces that are not locally path connected: counterexamples?
0
$begingroup$
The standard theory treats the case where the base space $B$ is path connected, locally path connected, and semi-locally simply connected. While being path connected and semi-locally simply connected is necessary to have a universal covering (which by definitions is just a covering with simply connected total space), the condition local path connectedness is not so natural. So I'd like to see counterexamples for the following:
- Let $B$ be path connected and semi-locally simply connected. Then $B$ not necessarily has a universal covering.
- Let $B$ be path connected and semi-locally simply connected and have a universal covering. Then $B$ is not necessarily locally path connected.
- Let $B$ be path connected and semi-locally simply connected and have a universal covering $p:Eto B$. Do we still have the usual theory that connected coverings of $B$ correspond to subgroups of $pi_1(B)$? In particular, is the group of deck transformations of $E$ isomorphic to the group $pi_1(B)$?
Thanks in advance!
general-topology algebraic-topology examples-counterexamples covering-spaces
asked Jan 6 at 5:38
ColescuColescu
3,20511136
$endgroup$
add a comment |
0
$begingroup$
The standard theory treats the case where the base space $B$ is path connected, locally path connected, and semi-locally simply connected. While being path connected and semi-locally simply connected is necessary to have a universal covering (which by definitions is just a covering with simply connected total space), the condition local path connectedness is not so natural. So I'd like to see counterexamples for the following:
- Let $B$ be path connected and semi-locally simply connected. Then $B$ not necessarily has a universal covering.
- Let $B$ be path connected and semi-locally simply connected and have a universal covering. Then $B$ is not necessarily locally path connected.
- Let $B$ be path connected and semi-locally simply connected and have a universal covering $p:Eto B$. Do we still have the usual theory that connected coverings of $B$ correspond to subgroups of $pi_1(B)$? In particular, is the group of deck transformations of $E$ isomorphic to the group $pi_1(B)$?
Thanks in advance!
general-topology algebraic-topology examples-counterexamples covering-spaces
asked Jan 6 at 5:38
ColescuColescu
3,20511136
$endgroup$
$begingroup$
There is an alternative (and in my opiniuon better) definition of a universal covering. Consider all connected coverings $p: X' to X$ of a connected X (connected covering means that $X'$ is connected). A map from $p_1$ to $p_2$ is a map $f : X'_1 to X'_2$ such that $p_2 circ f = p_1$. Then call $p_u$ a universal covering if for each $p$ there exists a map $f : p_u to p$. Then it is a theorem that a simply connected covering of a connected locally path connected $X$ is a universal covering.
$endgroup$
– Paul Frost
Jan 6 at 9:21
add a comment |
0
0
0
$begingroup$
The standard theory treats the case where the base space $B$ is path connected, locally path connected, and semi-locally simply connected. While being path connected and semi-locally simply connected is necessary to have a universal covering (which by definitions is just a covering with simply connected total space), the condition local path connectedness is not so natural. So I'd like to see counterexamples for the following:
- Let $B$ be path connected and semi-locally simply connected. Then $B$ not necessarily has a universal covering.
- Let $B$ be path connected and semi-locally simply connected and have a universal covering. Then $B$ is not necessarily locally path connected.
- Let $B$ be path connected and semi-locally simply connected and have a universal covering $p:Eto B$. Do we still have the usual theory that connected coverings of $B$ correspond to subgroups of $pi_1(B)$? In particular, is the group of deck transformations of $E$ isomorphic to the group $pi_1(B)$?
Thanks in advance!
general-topology algebraic-topology examples-counterexamples covering-spaces
asked Jan 6 at 5:38
ColescuColescu
3,20511136
$endgroup$
The standard theory treats the case where the base space $B$ is path connected, locally path connected, and semi-locally simply connected. While being path connected and semi-locally simply connected is necessary to have a universal covering (which by definitions is just a covering with simply connected total space), the condition local path connectedness is not so natural. So I'd like to see counterexamples for the following:
- Let $B$ be path connected and semi-locally simply connected. Then $B$ not necessarily has a universal covering.
- Let $B$ be path connected and semi-locally simply connected and have a universal covering. Then $B$ is not necessarily locally path connected.
- Let $B$ be path connected and semi-locally simply connected and have a universal covering $p:Eto B$. Do we still have the usual theory that connected coverings of $B$ correspond to subgroups of $pi_1(B)$? In particular, is the group of deck transformations of $E$ isomorphic to the group $pi_1(B)$?
Thanks in advance!
general-topology algebraic-topology examples-counterexamples covering-spaces
general-topology algebraic-topology examples-counterexamples covering-spaces
asked Jan 6 at 5:38
ColescuColescu
3,20511136
asked Jan 6 at 5:38
ColescuColescu
3,20511136
asked Jan 6 at 5:38
ColescuColescu
3,20511136
asked Jan 6 at 5:38
ColescuColescu
3,20511136
asked Jan 6 at 5:38
ColescuColescu
3,20511136
3,20511136
$begingroup$
There is an alternative (and in my opiniuon better) definition of a universal covering. Consider all connected coverings $p: X' to X$ of a connected X (connected covering means that $X'$ is connected). A map from $p_1$ to $p_2$ is a map $f : X'_1 to X'_2$ such that $p_2 circ f = p_1$. Then call $p_u$ a universal covering if for each $p$ there exists a map $f : p_u to p$. Then it is a theorem that a simply connected covering of a connected locally path connected $X$ is a universal covering.
$endgroup$
– Paul Frost
Jan 6 at 9:21
add a comment |
$begingroup$
There is an alternative (and in my opiniuon better) definition of a universal covering. Consider all connected coverings $p: X' to X$ of a connected X (connected covering means that $X'$ is connected). A map from $p_1$ to $p_2$ is a map $f : X'_1 to X'_2$ such that $p_2 circ f = p_1$. Then call $p_u$ a universal covering if for each $p$ there exists a map $f : p_u to p$. Then it is a theorem that a simply connected covering of a connected locally path connected $X$ is a universal covering.
$endgroup$
– Paul Frost
Jan 6 at 9:21
$begingroup$
There is an alternative (and in my opiniuon better) definition of a universal covering. Consider all connected coverings $p: X' to X$ of a connected X (connected covering means that $X'$ is connected). A map from $p_1$ to $p_2$ is a map $f : X'_1 to X'_2$ such that $p_2 circ f = p_1$. Then call $p_u$ a universal covering if for each $p$ there exists a map $f : p_u to p$. Then it is a theorem that a simply connected covering of a connected locally path connected $X$ is a universal covering.
$endgroup$
– Paul Frost
Jan 6 at 9:21
$begingroup$
There is an alternative (and in my opiniuon better) definition of a universal covering. Consider all connected coverings $p: X' to X$ of a connected X (connected covering means that $X'$ is connected). A map from $p_1$ to $p_2$ is a map $f : X'_1 to X'_2$ such that $p_2 circ f = p_1$. Then call $p_u$ a universal covering if for each $p$ there exists a map $f : p_u to p$. Then it is a theorem that a simply connected covering of a connected locally path connected $X$ is a universal covering.
$endgroup$
– Paul Frost
Jan 6 at 9:21
add a comment |
2 Answers
2
active
oldest
votes
1
$begingroup$
Here is a partial answer.
- Let $X$ be any path connected simply connected space which is not locally connected (which implies that it is not locally path connected). As an example take the Warsaw circle (see https://de.wikipedia.org/wiki/Datei:Warsaw_Circle.png, https://en.wikipedia.org/wiki/Shape_theory_(mathematics), How to show Warsaw circle is non-contractible?). Then $id : X to X$ is a universal covering.
answered Jan 6 at 14:56
Paul FrostPaul Frost
11.5k3934
$endgroup$
$begingroup$
This seems to be a common example. I wonder are there any results on the classification of its coverings?
$endgroup$
– Colescu
Jan 7 at 3:15
$begingroup$
Interesting question, but I do not know the answer although I guess there are no non-trivial coverings of $X$. Perhaps you should ask an additional question.
$endgroup$
– Paul Frost
Jan 7 at 9:40
$begingroup$
By the way, $id$ is a universal covering in the sense of my comment to your question if and only if $X$ does not have nontrivial coverings. To see this, assume that $id$ is universal and let $p$ any covering of $X$. Then the map $f : id to p$ is a section of $p$. Now see math.stackexchange.com/q/256951.
$endgroup$
– Paul Frost
Jan 8 at 0:59
add a comment |
0
$begingroup$
Here is another partial answer.
- As in the answer to 2., let $W$ be the Warsaw circle which is path connected simply connected. It has a universal covering, $id : W to W$. However, it has infintely many distinct connected coverings, and these cannot be classified by subgroups of $pi_1(W) = 0$. These coverings are obtained by pasting together $n$ copies of the closed toplogist's sine curve $S$ into a "circular" pattern and mapping this space in the obvious way to $W$ by wrapping it $n$-times around $W$. Another covering is obtained by pasting together infinitely copies of $S$ into a "linear" pattern and mapping this space in the obvious way to $W$. This is in complete analogy to the coverings $z^n : S^1 to S^1$ and $e^{2pi it} : mathbb{R} to S^1$.
Note that all these coverings are not path connected. It therefore potentially makes a difference whether we work with connected coverings or with path connected coverings. For a locally path connected base space $X$ this is the same.
answered Jan 11 at 22:55
Paul FrostPaul Frost
11.5k3934
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Here is a partial answer.
- Let $X$ be any path connected simply connected space which is not locally connected (which implies that it is not locally path connected). As an example take the Warsaw circle (see https://de.wikipedia.org/wiki/Datei:Warsaw_Circle.png, https://en.wikipedia.org/wiki/Shape_theory_(mathematics), How to show Warsaw circle is non-contractible?). Then $id : X to X$ is a universal covering.
answered Jan 6 at 14:56
Paul FrostPaul Frost
11.5k3934
$endgroup$
$begingroup$
This seems to be a common example. I wonder are there any results on the classification of its coverings?
$endgroup$
– Colescu
Jan 7 at 3:15
$begingroup$
Interesting question, but I do not know the answer although I guess there are no non-trivial coverings of $X$. Perhaps you should ask an additional question.
$endgroup$
– Paul Frost
Jan 7 at 9:40
$begingroup$
By the way, $id$ is a universal covering in the sense of my comment to your question if and only if $X$ does not have nontrivial coverings. To see this, assume that $id$ is universal and let $p$ any covering of $X$. Then the map $f : id to p$ is a section of $p$. Now see math.stackexchange.com/q/256951.
$endgroup$
– Paul Frost
Jan 8 at 0:59
add a comment |
1
$begingroup$
Here is a partial answer.
- Let $X$ be any path connected simply connected space which is not locally connected (which implies that it is not locally path connected). As an example take the Warsaw circle (see https://de.wikipedia.org/wiki/Datei:Warsaw_Circle.png, https://en.wikipedia.org/wiki/Shape_theory_(mathematics), How to show Warsaw circle is non-contractible?). Then $id : X to X$ is a universal covering.
answered Jan 6 at 14:56
Paul FrostPaul Frost
11.5k3934
$endgroup$
$begingroup$
This seems to be a common example. I wonder are there any results on the classification of its coverings?
$endgroup$
– Colescu
Jan 7 at 3:15
$begingroup$
Interesting question, but I do not know the answer although I guess there are no non-trivial coverings of $X$. Perhaps you should ask an additional question.
$endgroup$
– Paul Frost
Jan 7 at 9:40
$begingroup$
By the way, $id$ is a universal covering in the sense of my comment to your question if and only if $X$ does not have nontrivial coverings. To see this, assume that $id$ is universal and let $p$ any covering of $X$. Then the map $f : id to p$ is a section of $p$. Now see math.stackexchange.com/q/256951.
$endgroup$
– Paul Frost
Jan 8 at 0:59
add a comment |
1
1
1
$begingroup$
Here is a partial answer.
- Let $X$ be any path connected simply connected space which is not locally connected (which implies that it is not locally path connected). As an example take the Warsaw circle (see https://de.wikipedia.org/wiki/Datei:Warsaw_Circle.png, https://en.wikipedia.org/wiki/Shape_theory_(mathematics), How to show Warsaw circle is non-contractible?). Then $id : X to X$ is a universal covering.
answered Jan 6 at 14:56
Paul FrostPaul Frost
11.5k3934
$endgroup$
Here is a partial answer.
- Let $X$ be any path connected simply connected space which is not locally connected (which implies that it is not locally path connected). As an example take the Warsaw circle (see https://de.wikipedia.org/wiki/Datei:Warsaw_Circle.png, https://en.wikipedia.org/wiki/Shape_theory_(mathematics), How to show Warsaw circle is non-contractible?). Then $id : X to X$ is a universal covering.
answered Jan 6 at 14:56
Paul FrostPaul Frost
11.5k3934
answered Jan 6 at 14:56
Paul FrostPaul Frost
11.5k3934
answered Jan 6 at 14:56
Paul FrostPaul Frost
11.5k3934
answered Jan 6 at 14:56
Paul FrostPaul Frost
11.5k3934
11.5k3934
$begingroup$
This seems to be a common example. I wonder are there any results on the classification of its coverings?
$endgroup$
– Colescu
Jan 7 at 3:15
$begingroup$
Interesting question, but I do not know the answer although I guess there are no non-trivial coverings of $X$. Perhaps you should ask an additional question.
$endgroup$
– Paul Frost
Jan 7 at 9:40
$begingroup$
By the way, $id$ is a universal covering in the sense of my comment to your question if and only if $X$ does not have nontrivial coverings. To see this, assume that $id$ is universal and let $p$ any covering of $X$. Then the map $f : id to p$ is a section of $p$. Now see math.stackexchange.com/q/256951.
$endgroup$
– Paul Frost
Jan 8 at 0:59
add a comment |
$begingroup$
This seems to be a common example. I wonder are there any results on the classification of its coverings?
$endgroup$
– Colescu
Jan 7 at 3:15
$begingroup$
Interesting question, but I do not know the answer although I guess there are no non-trivial coverings of $X$. Perhaps you should ask an additional question.
$endgroup$
– Paul Frost
Jan 7 at 9:40
$begingroup$
By the way, $id$ is a universal covering in the sense of my comment to your question if and only if $X$ does not have nontrivial coverings. To see this, assume that $id$ is universal and let $p$ any covering of $X$. Then the map $f : id to p$ is a section of $p$. Now see math.stackexchange.com/q/256951.
$endgroup$
– Paul Frost
Jan 8 at 0:59
$begingroup$
This seems to be a common example. I wonder are there any results on the classification of its coverings?
$endgroup$
– Colescu
Jan 7 at 3:15
$begingroup$
This seems to be a common example. I wonder are there any results on the classification of its coverings?
$endgroup$
– Colescu
Jan 7 at 3:15
$begingroup$
Interesting question, but I do not know the answer although I guess there are no non-trivial coverings of $X$. Perhaps you should ask an additional question.
$endgroup$
– Paul Frost
Jan 7 at 9:40
$begingroup$
Interesting question, but I do not know the answer although I guess there are no non-trivial coverings of $X$. Perhaps you should ask an additional question.
$endgroup$
– Paul Frost
Jan 7 at 9:40
$begingroup$
By the way, $id$ is a universal covering in the sense of my comment to your question if and only if $X$ does not have nontrivial coverings. To see this, assume that $id$ is universal and let $p$ any covering of $X$. Then the map $f : id to p$ is a section of $p$. Now see math.stackexchange.com/q/256951.
$endgroup$
– Paul Frost
Jan 8 at 0:59
$begingroup$
By the way, $id$ is a universal covering in the sense of my comment to your question if and only if $X$ does not have nontrivial coverings. To see this, assume that $id$ is universal and let $p$ any covering of $X$. Then the map $f : id to p$ is a section of $p$. Now see math.stackexchange.com/q/256951.
$endgroup$
– Paul Frost
Jan 8 at 0:59
add a comment |
0
$begingroup$
Here is another partial answer.
- As in the answer to 2., let $W$ be the Warsaw circle which is path connected simply connected. It has a universal covering, $id : W to W$. However, it has infintely many distinct connected coverings, and these cannot be classified by subgroups of $pi_1(W) = 0$. These coverings are obtained by pasting together $n$ copies of the closed toplogist's sine curve $S$ into a "circular" pattern and mapping this space in the obvious way to $W$ by wrapping it $n$-times around $W$. Another covering is obtained by pasting together infinitely copies of $S$ into a "linear" pattern and mapping this space in the obvious way to $W$. This is in complete analogy to the coverings $z^n : S^1 to S^1$ and $e^{2pi it} : mathbb{R} to S^1$.
Note that all these coverings are not path connected. It therefore potentially makes a difference whether we work with connected coverings or with path connected coverings. For a locally path connected base space $X$ this is the same.
answered Jan 11 at 22:55
Paul FrostPaul Frost
11.5k3934
$endgroup$
add a comment |
0
$begingroup$
Here is another partial answer.
- As in the answer to 2., let $W$ be the Warsaw circle which is path connected simply connected. It has a universal covering, $id : W to W$. However, it has infintely many distinct connected coverings, and these cannot be classified by subgroups of $pi_1(W) = 0$. These coverings are obtained by pasting together $n$ copies of the closed toplogist's sine curve $S$ into a "circular" pattern and mapping this space in the obvious way to $W$ by wrapping it $n$-times around $W$. Another covering is obtained by pasting together infinitely copies of $S$ into a "linear" pattern and mapping this space in the obvious way to $W$. This is in complete analogy to the coverings $z^n : S^1 to S^1$ and $e^{2pi it} : mathbb{R} to S^1$.
Note that all these coverings are not path connected. It therefore potentially makes a difference whether we work with connected coverings or with path connected coverings. For a locally path connected base space $X$ this is the same.
answered Jan 11 at 22:55
Paul FrostPaul Frost
11.5k3934
$endgroup$
add a comment |
0
0
0
$begingroup$
Here is another partial answer.
- As in the answer to 2., let $W$ be the Warsaw circle which is path connected simply connected. It has a universal covering, $id : W to W$. However, it has infintely many distinct connected coverings, and these cannot be classified by subgroups of $pi_1(W) = 0$. These coverings are obtained by pasting together $n$ copies of the closed toplogist's sine curve $S$ into a "circular" pattern and mapping this space in the obvious way to $W$ by wrapping it $n$-times around $W$. Another covering is obtained by pasting together infinitely copies of $S$ into a "linear" pattern and mapping this space in the obvious way to $W$. This is in complete analogy to the coverings $z^n : S^1 to S^1$ and $e^{2pi it} : mathbb{R} to S^1$.
Note that all these coverings are not path connected. It therefore potentially makes a difference whether we work with connected coverings or with path connected coverings. For a locally path connected base space $X$ this is the same.
answered Jan 11 at 22:55
Paul FrostPaul Frost
11.5k3934
$endgroup$
Here is another partial answer.
- As in the answer to 2., let $W$ be the Warsaw circle which is path connected simply connected. It has a universal covering, $id : W to W$. However, it has infintely many distinct connected coverings, and these cannot be classified by subgroups of $pi_1(W) = 0$. These coverings are obtained by pasting together $n$ copies of the closed toplogist's sine curve $S$ into a "circular" pattern and mapping this space in the obvious way to $W$ by wrapping it $n$-times around $W$. Another covering is obtained by pasting together infinitely copies of $S$ into a "linear" pattern and mapping this space in the obvious way to $W$. This is in complete analogy to the coverings $z^n : S^1 to S^1$ and $e^{2pi it} : mathbb{R} to S^1$.
Note that all these coverings are not path connected. It therefore potentially makes a difference whether we work with connected coverings or with path connected coverings. For a locally path connected base space $X$ this is the same.
answered Jan 11 at 22:55
Paul FrostPaul Frost
11.5k3934
answered Jan 11 at 22:55
Paul FrostPaul Frost
11.5k3934
answered Jan 11 at 22:55
Paul FrostPaul Frost
11.5k3934
answered Jan 11 at 22:55
Paul FrostPaul Frost
11.5k3934
11.5k3934
add a comment |
add a comment |
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$begingroup$
There is an alternative (and in my opiniuon better) definition of a universal covering. Consider all connected coverings $p: X' to X$ of a connected X (connected covering means that $X'$ is connected). A map from $p_1$ to $p_2$ is a map $f : X'_1 to X'_2$ such that $p_2 circ f = p_1$. Then call $p_u$ a universal covering if for each $p$ there exists a map $f : p_u to p$. Then it is a theorem that a simply connected covering of a connected locally path connected $X$ is a universal covering.
$endgroup$
– Paul Frost
Jan 6 at 9:21