# What is a counterexample that a composition of covering maps not a covering map?

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?

• How is anyone supposed to know what counterexample you found "too complex" if you don't tell us? – Zev Chonoles Jul 26 '15 at 7:21
• @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 7:31
• 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 8:34
• @GeorgesElencwajg Etale covers, for example. Open covers in analysis or differential geometry. Covers in any Grothendieck topology. – Colin McLarty Jul 26 '15 at 9:34
• see Hatcher's book section 1.3 ,problem no-6 on page no 79... – Ripan Saha Jul 26 '15 at 10:20

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.