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Reading through Munkres and he says that for, $$ p: E\to B $$ an opet set $U\subset B$ is $\textbf{evenly covered}$ if there exists a collection of disjoint sets $V_\alpha$ s.t. for each $\alpha$, the restriction of $p$ to $V_\alpha$ is a homeomorphism.

But then, his first example of a non-trivial covering map is $$ p: \mathbb{R}\to S^1 $$ s.t. $$ p(x) = (\cos2\pi x,\sin2\pi x). $$ He then says this "maps each interval $[n, n+1]$ onto $S^1$." Is he trying to say that $$ V_\alpha = V_n = [n, n+1] $$ If so, wouldn't he be stating that $[n, n+1]$ is homeomorphic to the circle (which isn't true)?

Paul Frost
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    No, but each point in $S^1$ has a small interval segment such that the inverse image of it is a countable collection of disjoint open intervals in $\Bbb R$. It's a local homeomorphism of an interval on the circle with an interval in $\Bbb R$, countably many times (by periodicity). – Henno Brandsma Nov 13 '21 at 23:28
  • I.e. one interval inside each $(n,n+1)$, typically. Only for the point $(1,0)$ we have intervals around the $n$'s. – Henno Brandsma Nov 13 '21 at 23:43

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No. He says that $p$ maps $[n,n+1]$ onto $S^1$, not that $p_n = p \mid_{[n,n+1]}$ is a homeomorphism. In fact, $p(n) = p(n+1)$ which shows that $p_n$ is not injective.

But it is easy to see that $p$ maps each open interval $(a,b)$ of length $b - a \le 1$ homeomorphically onto its image $p((a,b))$. See my answer to Open sets on the unit circle $S^1$. Also see Is it an open function?.

Paul Frost
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