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

3,20511136




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




















  • $begingroup$
    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

    11.5k3934




    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




















    • $begingroup$
      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

        11.5k3934




        11.5k3934






























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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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