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It is easy to see that

$p:\mathbb{R} \rightarrow S^1: p(t)=(\cos t, \sin t)$ is a covering map of $S^1$

(Indeed take a point $x_0$ in $S^1$, take $U=S^1\setminus \{-x_0\}$ as an open neighborhood and then $p^{-1}(U)=\cup_{n \in Z} J_n$ where $J_n=\{t\in\mathbb{R} : t_0-n-1/2<t<t+0+n+1/2\}$. $p$ then maps $J_n$ homeomorphically onto $U$.)

I can't see why we can't take a similar $p:(a,b) \to S^1$. For example, Massey states that if we take the open interval $(0,10)$ onto the circle, then some points in $S^1$ fail to have an elementary neighborhood.

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Think about what the preimage of a neighborhood of p(a) looks like. –  Qiaochu Yuan Feb 15 '11 at 23:36
    
You appear to be confusing covering maps with local homeomorphisms. They are not the same. en.wikipedia.org/wiki/Local_homeomorphism –  George Lowther Feb 15 '11 at 23:42
    
@George: this is exactly what Massey is showing - that they are not the same –  Juan S Feb 15 '11 at 23:43
    
Does Massey really say "a similar map" or rather "the same map"? –  a.r. Feb 16 '11 at 5:53
    
@Agusti - Neither - that was all me –  Juan S Feb 16 '11 at 10:19

1 Answer 1

up vote 3 down vote accepted

Look at a preimage of a small neighborhood $U$ of $p(10)$. It will have one component which doesn't map homeomorphically onto $U$.

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