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?
Composition of coverings of path connected spaces
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algebraic-topology
covering-spaces
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2Well it certainly can't fail if $Z$ has a universal cover. – 2011-07-29
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0The amusing aspect of this is that a composite of covering morphisms of groupoids is a covering morphism. – 2012-04-23
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0The 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! – 2012-06-27
1 Answers
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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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0Good 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. – 2011-07-29
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0@J.T. - opps sorry I had some typos in the edit - should be correct now – 2011-07-29
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1Don't you mean the fiber is finite? – 2011-07-29
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0@Jacob - yes that is exactly what I meant! Of course covering space fibers are always discrete... – 2011-07-29