Why existence of universal covering implies that the base space be locally path connected?

Question

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.

Answer 1

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.

Answer 2

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?)

Answer 3

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

Answer 4

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).