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The Section on Covering Maps in John Lee's book "Introduction to Smooth Manifolds" starts like this:

Suppose $\tilde{X}$ and $X$ are topological spaces. A map $\pi : \tilde{X} \to X$ is called a covering map if $\tilde{X}$ is path-connected and locally path connected, ... (etc).

I hope this question is not too dumb, but how can a space be path connected, but not locally path connected ?

EDIT: I am aware of spaces that are locally path-connected yet not path-connected, but I cannot come up with a space that is path - connected yet not locally path connected.

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See here –  David Mitra Apr 22 '12 at 19:51
    
This is a great comment. –  Kerry Apr 22 '12 at 19:53
    
@DavidMitra: WOW .. Topology always amazes me, there are so many things that I learn from these counterexamples .. many thanks for pointing me to the link!! –  harlekin Apr 22 '12 at 19:54
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I am to unsure to answer: "because the path witnessing path connectedness might have to pass though a specific point (or be otherwise constrained)". There are other examples. From Steen and Seebach's Counterexamples in Topology: The Alexandroff Square (ex 101), The Extended Topologist's Sine Curve (ex 118), The Closed Infinite Broom (ex. 120), and the Integer Broom (ex 121). –  David Mitra Apr 22 '12 at 20:01

3 Answers 3

up vote 17 down vote accepted

One counterexample is a variant on the famous topologist's sine curve.

Consider the graph of $y = \sin(\pi/x)$ for $0<x<1$, together with a closed arc from the point $(1,0)$ to $(0,0)$:

enter image description here

This space is obviously path-connected, but it is not locally path-connected (or even locally connected) at the point $(0,0)$.

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what is the fundamental group of the picture above? –  Ronald Jul 31 '13 at 15:35
    
@Ronald The picture above is simply connected, so its fundamental group is trivial. –  Jim Belk Nov 14 '13 at 14:18
    
Question to Jim: why is your space simply connected? it looks like the circle, which is not simply connected... –  Hila May 27 at 15:56
    
@Hila Jim never claimed it to be simply connected. You are the first to bring up the term "simply connected" here. –  Behaviour May 27 at 17:59
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@Hila It's simply connected because it isn't possible for a path to make it around the "circle". (The sine wave portion is an impassible road block.) –  Jim Belk May 27 at 19:10

You should consider the opposite question, that how a space could be locally path connected, but not path connected. And this should be simple: consider the union of two open disks.

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I think harlekin's point, then, is why both hypotheses are being made. Why not just say $\widetilde{X}$ is locally path-connected? –  KCd Apr 22 '12 at 19:48
    
Thanks for your comment! I have added my post to clarify what confuses me - in my topology course I have seen spaces that are locally path-connected yet not path-connected, but what I have trouble with is coming up with a path-connected space that is not locally path-connected. Yet this is what Lee's opening part of the definition of a covering map suggests exists - I suppose .. –  harlekin Apr 22 '12 at 19:49
    
$KCd: I understand. If I am not being mistaken I think Hatcher's book has some discussion relevant to this. –  Kerry Apr 22 '12 at 19:51
    
Ok I shall have a look at Hatcher's book as well, I am currently reading about the Comb space, as suggested by David, but thanks a lot for your suggestion ! –  harlekin Apr 22 '12 at 19:55
    
On page 63 he commented that if the space is both path-connected and locally path-connected, then components are the same as path components, which simplifies his discussion on the Galois correspondence on the covering space. –  Kerry Apr 22 '12 at 19:59

$\pi$-Base, an online version of Steen and Seebach's Counterexamples in Topology, lists the following spaces as path-connected but not locally path-connected. You can view the search result for more information about these spaces.

Alexandroff Square

Extended Topologist’s Sine Curve

The Closed Infinite Broom

The Integer Broom

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