Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
6  
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
    
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
add comment

2 Answers

up vote 12 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 $(0,0)$.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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