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?

  • $\begingroup$ What are the coverings of a simply connected space? $\endgroup$ – Alex Youcis Mar 10 '14 at 11:14

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 $n=d$.


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