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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. $p$ 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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    @Agusti - Neither - that was all me2011-02-16

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