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 $p:E\to B$. 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!

$\endgroup$
  • $\begingroup$ 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
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.
$\endgroup$
  • $\begingroup$ 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 $e^{2\pi it} : \mathbb{R} \to S^1$.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.