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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 $p:E\to B$. 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!

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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
    Commented Jan 6, 2019 at 9:21

2 Answers 2

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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.
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  • $\begingroup$ This seems to be a common example. I wonder are there any results on the classification of its coverings? $\endgroup$
    – Yuxiao Xie
    Commented Jan 7, 2019 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
    Commented Jan 7, 2019 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
    Commented Jan 8, 2019 at 0:59
  • $\begingroup$ Update: My second answer shows that $X$ has non-trivial coverings. $\endgroup$
    – Paul Frost
    Commented Apr 5, 2021 at 10:05
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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 infinitely 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 $e^{2\pi it} : \mathbb{R} \to S^1$.

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.

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