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:




  1. Let $B$ be path connected and semi-locally simply connected. Then $B$ not necessarily has a universal covering.

  2. Let $B$ be path connected and semi-locally simply connected and have a universal covering. Then $B$ is not necessarily locally path connected.

  3. 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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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


















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:




  1. Let $B$ be path connected and semi-locally simply connected. Then $B$ not necessarily has a universal covering.

  2. Let $B$ be path connected and semi-locally simply connected and have a universal covering. Then $B$ is not necessarily locally path connected.

  3. 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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















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:




  1. Let $B$ be path connected and semi-locally simply connected. Then $B$ not necessarily has a universal covering.

  2. Let $B$ be path connected and semi-locally simply connected and have a universal covering. Then $B$ is not necessarily locally path connected.

  3. 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







share|cite|improve this question









$endgroup$




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:




  1. Let $B$ be path connected and semi-locally simply connected. Then $B$ not necessarily has a universal covering.

  2. Let $B$ be path connected and semi-locally simply connected and have a universal covering. Then $B$ is not necessarily locally path connected.

  3. 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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 5:38









ColescuColescu

3,26711136




3,26711136












  • $begingroup$
    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






$begingroup$
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












2 Answers
2






active

oldest

votes


















1












$begingroup$

Here is a partial answer.




  1. 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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





















0












$begingroup$

Here is another partial answer.




  1. 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.






share|cite|improve this answer









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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.




    1. 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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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


















    1












    $begingroup$

    Here is a partial answer.




    1. 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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    1












    1








    1





    $begingroup$

    Here is a partial answer.




    1. 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.






    share|cite|improve this answer









    $endgroup$



    Here is a partial answer.




    1. 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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 6 at 14:56









    Paul FrostPaul Frost

    11.6k3935




    11.6k3935












    • $begingroup$
      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










    • $begingroup$
      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













    0












    $begingroup$

    Here is another partial answer.




    1. 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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Here is another partial answer.




      1. 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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Here is another partial answer.




        1. 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.






        share|cite|improve this answer









        $endgroup$



        Here is another partial answer.




        1. 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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 22:55









        Paul FrostPaul Frost

        11.6k3935




        11.6k3935






























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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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            Gyllenstierna

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