This problem was posted before, but not the proof (because the asker knowed the answer), only a counterexample without the hypothesis of finite fibres. I want to know how to prove this proposition:

Let $q:X\to Y$ and $r:Y\to Z$ be covering maps. Let's suppose that for each $z\in Z$ , the set $ r^{-1}(z)$ is finite, then the composition $p = r\circ q$ is also a covering map. Well I have to consider $z\in Z$ and show that there exist a neighborhood that is evenly covered by $p$, I know that there exist a neighborhood $U$ that is evenly covered by $r$, I think that this will be the desired neighborhood. First of all $ r^{-1}(U) = \cup_{i=1}^{n} V_i $ (it's easy to see that the unions is finite using the fact that the fibres are finite and this is a local homeomorphism between the $V_i$) Then $ p^{-1}(U)=q^{-1} (r^{-1}(U) ) = q^{-1}(\cup_{i=1}^{n} V_i)=\cup_{i=1}^{n} q^{-1}(V_i) $ With $ q^{-1}(V_i) \cong V_i$ under $q$ and so $\cong$ U under $p$ I never used the fact that the preimage is finite, so my proof is obviously not correct, please help me with this

  • $\begingroup$ You do not necessarily have $q^{-1}(V_i)\cong V_i$ $\endgroup$ – Stefan Hamcke Feb 19 '13 at 22:07
  • $\begingroup$ So what can I do? $\endgroup$ – Miguel Feb 20 '13 at 6:34
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    $\begingroup$ Possible duplicate of Question on covering spaces $\endgroup$ – Paul Frost May 8 at 23:48

Let $z\in Z$. There is an open $U\ni z$ with $r^{-1}(U)=\bigsqcup_{i=1}^n U_i$, such that $r_i:U_i\xrightarrow{{\approx}} U$. Let $y_i\in U_i$ be the element in the fiber of $z$. For every $i$ there is an open $V_i$ with $y_i\in V_i\subseteq U_i$ such that $q^{-1}(V_i)=\bigsqcup_j V_i^j$ and $q_i^j:V_i^j\xrightarrow{{\approx}} V_i$.

Let $W:=\bigcap_{i=1}^nr(V_i)$. This is an open subset of $U$ containing $z$. Here you are using the finiteness of fibers. We will show that this is the desired neighborhood.

Let $W_i:=r_i^{-1}(W)$. This is an open subset of $V_i$ containing $y_i$, and $r_i:W_i\xrightarrow{{\approx}} W$. We have $q^{-1}(W_i)=\bigsqcup_j W_i^j$ such that $q_i^j:W_i^j\xrightarrow{{\approx}} W_i$. Composing $q_i^j$ with $r_i$ then gives a homeomorphism between $W_i^j$ and $W$. Additionally, $p^{-1}(W)=\bigsqcup_{i=1}^n\bigsqcup_jW_i^j.$

  • $\begingroup$ Could you elaborate a little bit from "We have..." on-wards in your last paragraph, the notation is confusing me a little bit? $\endgroup$ – EgoKilla Apr 14 '14 at 23:35
  • $\begingroup$ @EgoKilla: You should be more specific about the point that is confusing you. The $W_i$ is a subset of $V_i$, so its preimage under $q$ is the disjoint union of $W_i^j$ where $W_i^j\subseteq V_i^j$. And I'm using $q_i^j$ for both the map on $V_i^j$ and its restriction on $W_i^j$. $\endgroup$ – Stefan Hamcke Apr 15 '14 at 21:46
  • $\begingroup$ Yes, thanks I figured it out. Your notation is actually quite succinct once I figured it out. $\endgroup$ – EgoKilla Apr 16 '14 at 10:01
  • $\begingroup$ @StefanHamcke: Can you explain why for all $i$, there exists open $V_i$ such that $y_i \in V_i \subset U_i$? I don't understand why must $V_i$ subset of $U_i$. $\endgroup$ – Idonknow Apr 22 '14 at 10:05
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    $\begingroup$ @Idonknow he uses there the definition of covering map of $q$ on $U_i$ as the covered space. $\endgroup$ – MathematicalPhysicist Sep 24 '15 at 20:06

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