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Let $p:X\rightarrow Y$ and $q:Y\rightarrow Z$ be covering maps.

What would be an example that $q\circ p:X\rightarrow Z$ is not a covering map?

I saw a counterexample here, but it was too complex. Is there a relatively easy one?

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Rubertos
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    How is anyone supposed to know *what counterexample* you found "too complex" if you don't tell us? – Zev Chonoles Jul 26 '15 at 07:21
  • There are many different notions of covering map, used in different contexts. For the ones I generally use, the composite of covering maps is a covering map. What do you mean by covering map? – Colin McLarty Jul 26 '15 at 07:27
  • @ZevChonoles http://math.stackexchange.com/questions/146976/composition-of-covering-maps/147164#147164 – Rubertos Jul 26 '15 at 07:30
  • Moreover, I don't get what $X$ he means in the link. – Rubertos Jul 26 '15 at 07:30
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    @ColinMcLarty covering map $p:C\rightarrow X$ is a surjective continuous map such that for each $x\in X$, $x$ has an open neighborhood which is evenly covered by $p$. – Rubertos Jul 26 '15 at 07:31
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    Dear @Colin, I'm curious to know what you mean by covering map since I find quite strange that in your definition the composite of two covering maps is a covering map! – Georges Elencwajg Jul 26 '15 at 08:34
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    @GeorgesElencwajg Etale covers, for example. Open covers in analysis or differential geometry. Covers in any Grothendieck topology. – Colin McLarty Jul 26 '15 at 09:34
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    see Hatcher's book section 1.3 ,problem no-6 on page no 79... – Ripan Saha Jul 26 '15 at 10:20

1 Answers1

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The point is that if $p:X\rightarrow Y$ and $q:Y\rightarrow Z$ are covering maps in this sense, but some $z\in Z$ has infinitely many $y_i\in Y$ with $q(y_i)=z$, then each $y_i$ may have some neighborhood $U_i\subseteq Y$ "evenly covered" by $p$ (I take this to mean the inverse image of $U_i$ is a union of parts each mapped isomorphically to $U_i$) but as $i$ varies the corresponding $U_i$ get smaller and smaller so that the intersection of all the images $q(U_i)$ is just $z$. Then $z$ has no neighborhood $V\subseteq Z$ evenly covered by $qp$.

Bob Arthan's answer to https://math.stackexchange.com/posts/147164/revisions includes reference to a proof that if $p : X \rightarrow Y$ and $q : Y \rightarrow Z$ are covering maps and if $Z$ is locally path-connected and semilocally simply-connected, then the composite $qp$ is a covering map. So you will not get simpler counterexamples than the ones there.

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Colin McLarty
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