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Do there exist covering maps $p:X\rightarrow Y$ and $q:Y\rightarrow Z$ such that $X$ is path connected and the composition $q\circ p$ fails to be a covering map?

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Well it certainly can't fail if $Z$ has a universal cover. – jspecter Jul 29 '11 at 1:43
The amusing aspect of this is that a composite of covering morphisms of groupoids is a covering morphism. – Ronnie Brown Apr 23 '12 at 11:01
The paper : J. Brazas, "semicoverings: a generalisation of covering space theory", Homology, Homotopy and Applications, 14 (2012) 33-63, shows that semicoverings satisfy the "$2$ out of $3$ property". So $q \circ p$ will be a semicovering! – Ronnie Brown Jun 27 '12 at 14:38
up vote 5 down vote accepted

I think a cover of a cover of the Hawaiian earring gives an example where the composition fails to be a covering space, and the space $X$ is path conencted

See Exericse 6 on page 79 of Hatcher's book

Edit: The composition will be a covering map if the fiber $q^{-1}(z)$ is finite (proof) or, equivalently, the space $Z$ is semi-locally simply connected (In particular the Hawaiian earring is not semi-locally simply connected).

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Good call, I should have looked at Hatcher first. Yes, this example is right but your edit doesn't seem to be right. The fibers of each map are definitely discrete (even of the composite which is not a covering map) and no covering of the Hawaiian earring can be semi-locally simply connected. – J.K.T. Jul 29 '11 at 2:57
@J.T. - opps sorry I had some typos in the edit - should be correct now – Juan S Jul 29 '11 at 4:19
Don't you mean the fiber is finite? – JSchlather Jul 29 '11 at 4:33
@Jacob - yes that is exactly what I meant! Of course covering space fibers are always discrete... – Juan S Jul 29 '11 at 4:57

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