# Path connectedness and locally path connected

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
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

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 $(0,0)$:

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
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