7
$\begingroup$

I am reading Chapter 13, the chapter about classification of covering spaces, of J.Munkres' Topology. My confusion raised when I read Corollary 82.2. which says:

the space $B$ has a universal covering space if and only if $B$ is path connected, locally path connected, and semilocally simply connected.

The book does not give a proof so I believe it should be straight forward. But I just can not prove the "only if" part of this corollary. I do not know how to see that a space which has a universal cover must be locally path connected. I understand that by Lemma 80.4., a base space with a universal cover has to be semilocally simply connected. And since covering space is simply connected, the base space must be path connected.

Thank you very much for your attention and really appreciate your helps.

$\endgroup$

4 Answers 4

Reset to default
4
$\begingroup$

It depends what the definition of "universal covering" is.

If the definition is that a universal covering of $B$ is a simply connected covering (as it is in Munkres), then this statement as you're writing it isn't quite true. If $B$ is path-connected and simply connected but not locally path connected, then the identity function $B\to B$ is the universal covering of $B$ (but it exists!). For example, let B be the union of the line segments in $\mathbb{R}^2$ from $(0,0)$ to each point of $\{(1,0),...,(1,1/3),(1,1/2),(1,1)\}$.

But if I recall correctly, Munkres typically assumes spaces in consideration are locally path connected. So what is meant is the following: A path-connected and locally path-connected space $B$ has a universal covering if and only if $B$ is semilocally simply connected.

$\endgroup$
4
  • $\begingroup$ Thank you. I believe you are right. There are simply connected spaces that are not locally path connected. You are right that there is a general assumption that spaces taking into consideration are locally path connected. But the statement I copied is exactly the statement in Munkres' book as you can check. I guess he was not very careful at this point. $\endgroup$
    – Sunday Morning
    Commented Oct 25, 2014 at 13:05
  • $\begingroup$ @JeremyBrazas If we mean a universal covering in the categorical sense, that is, a covering $p :\tilde{X}\to X$ with the property that for every covering $q :\tilde{Y}\to X$ with a path connected space $\tilde{Y}$ there exists a covering $h :\tilde{X} \to \tilde{Y}$ such that $q\circ h = p$, then is there any non-simply connected universal covering for a non-locally path connected space? $\endgroup$
    – M.Ramana
    Commented Oct 18, 2018 at 14:17
  • $\begingroup$ @M.Ramana A space, whether it is locally path-connected or not, might have a non-simply connected ``categorical" universal covering space. See this answer. $\endgroup$
    – Jeremy Brazas
    Commented Oct 18, 2018 at 14:24
  • $\begingroup$ @JeremyBrazas Thank you so much for your help. $\endgroup$
    – M.Ramana
    Commented Oct 18, 2018 at 14:27
2
$\begingroup$

This statement is indeed false for the usual definition of a universal covering space. However, in the beginning of Chapter 13 Munkres makes the following convention:

Convention. Throughout this chapter, the statement that $p:E\to B$ is a covering map will include the assumption that $E$ and $B$ are locally connected and path connected, unless specifically started otherwise.

I agree, however, that formulating this corollary this way is misleading (and for no good reason).

$\endgroup$
0
$\begingroup$

Let $\pi\colon A\to B$ be a universal covering of $B$. Let $b\in B$ and $U\ni b$ open. We want to show that $b$ has a path-connected neighbourhood $U'\subseteq U$. Let $a\in A$ with $\pi(a)=b$ and $V$ the connected component of $\pi^{-1}(U)$ that contains $A$. I.e., $V$ contains all points that are reachable from $a$ via a path in $\pi^{-1}(U)$. But then $U':=\pi(V)$ is a path-connected open neighbourhood of $b$. (Careful: why is $U'$ open?)

$\endgroup$
2
  • $\begingroup$ What is your definition of a universal covering space? The one I know, does not make any assumptions about the connected component $V$ in your proof being path connected or even if it is, yielding an open neighborhood of $A$. $\endgroup$
    – Moishe Kohan
    Commented Oct 24, 2014 at 21:28
  • $\begingroup$ Thank you but I am unable to show your $U'$ is open. I think $V$ is not necessarily open so we can not say $U'$ is open. The openness of path components of any open subset is equivalent to locally path connectedness of the space. But there do exist a simply connected space that is not locally connected. $\endgroup$
    – Sunday Morning
    Commented Oct 25, 2014 at 12:52
0
$\begingroup$

It might be too late to answer it, but I'm sure that Munkres Lemma 80.4 is what you are looking for.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .