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I found this proposition in a paper stated as a well known result from topology, but I can neither find this result in my textbooks nor proof it by myself:

Let $p:E \rightarrow B$ be a covering space of degree $d$ and $A \subset B$ a simply connected subset. Then $p^{-1}(A)=U_1 \cup\dots\cup U_d$ decomposes into $d$ disjoint path-connected subsets, such that $p|_{U_i}:U_i\rightarrow A$ is a homeomorphism.

Can anybody help?

Jacob
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1 Answers1

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It is not difficult, depending on the tools you want to use.

First, note that the restriction $p:p^{-1}(A)\to A$ is a covering. Now, let $C$ be a connected component of $p^{-1}(A)$. Then the restriction of $p$ to $C$ $$p|_C:C\to A$$ is a covering. Since $A$ is simply connected, that covering must be the universal covering, whence a homeomorphism.

What we are proved so far is that $p^{-1}(A)$ consists of some number $n$ of connected components such that the restriction of $p$ to any such component is a homeomorphism. Since you know that the degree is $d$, you know now that $n=d$.

user126154
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  • This is not true. For example, the Warsaw circle is simply connected and has connected covering spaces of any degree which have no decomposition as required. See https://math.stackexchange.com/q/3063530. You need the additional assumption that $A$ is locally path connected. – Paul Frost May 26 '21 at 10:50
  • Yes you're right: I always make assumption on spaces to bee "good" (connected, locally path connected and locally semi-locally simply connected) when I read "simply connected"; but you're right: this is not maybe the usual setting. For connected, with trivial fundamental group, but not locally path connected, one can have connected coverings with trivial fundamental groups which are not homeomorphisms. Like the closed topological sinusoid (is that the Warsaw circle?). – user126154 May 26 '21 at 18:14
  • I d not know what the closed topological sinusoid is (probably the closed topologist's sine curve aka Warsaw sine curve). The Warsaw circle $W$ is this space. You can embed $W$ into a plane annulus which is certainly "good" and has a $d$-sheeted covering for all $d \in \mathbb N \cup {\infty}$. This restricts to a $d$-sheeted covering of $W$. But although $W$ is simply connected, this covering is not equivalent to $W \times F$. – Paul Frost May 27 '21 at 08:19