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I'm stuck with the following problem (number 12.21.1 of Topologia-Marco Manetti, 2008 edition):

Let $p:E \to F$ and $q:F \to X$ two continuous, surjective functions, where $E,F,X$ are connected and locally path connected topological spaces. Prove that, if $qp:E \to X$ is a covering map, then also $p$ and $q$ are covering maps.

What I was thinking about is to take $x \in X,$ and an open set $V \subset X, x\in V$, which respects the covering condition, and the counterimage through $qp$ of $V.$ Then, since $(qp)^{-1}(V)=\bigcup_{i \in I}C_i,$ where $C_i$ are all connected components such that $qp|_{C_i}:C_i \to V$ are homeomorphisms, I obtained that $p|_{C_i}$ is injective.

I made a quite similar reasoning taking the counterimage of $V$ through $q,$ but I couldn't reach very satisfying results. Someone also told me that it doesn't really seem to be true. Thanks to everybody.

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  • $\begingroup$ Doesn't the non-trivial double cover of the one dimensional sphere factor through the "eight space", i.e., the union of two Spheres in a point? $\endgroup$ – Ben May 30 '17 at 5:51
  • $\begingroup$ Sorry, I've understood the spaces but I don't understand what should be p and q :( $\endgroup$ – Riccardo Ceccon May 30 '17 at 6:25
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Here are some counter-examples:

Let $f\colon Y\to X$ be a non-trivial covering map and suppose that $X$ has a non-isolated point $x\in X$. Furthermore, let $y,y'\in Y$ be two distinct points in the fibre over $x$. Then $f$ factors through the space $Z:=Y/\{y,y'\}$ obtained from $Y$ by identifying $y$ and $y'$, via the projection map $\pi\colon Y\to Z$ and a unique, continuous map $g\colon Z\to X$ such that $g\circ \pi = f$. Both are surjective. However, $g$ is not a local homeomorphism near $\pi(y)=\pi(y')$ (unless $x\in X$ is isolated) and $\pi$ is not a covering map over $\pi(y)=\pi(y')$. (It's particularly easy to see in the special case where $ f$ is finite, since a covering map has to have locally constant number of fibres; clearly, $g$ and $\pi$ don't.)

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  • $\begingroup$ Wow! Thanks, now I understand what did you mean with the one dimensional sphere :). $\endgroup$ – Riccardo Ceccon May 30 '17 at 6:29

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