Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The origin of my question arose from a problem: Let $q: X \to Y$ and $r: Y \to Z$ be covering maps, let $p= r \circ q$. Show that if $r^{-1}(z)$ is finite for each $z \in Z$, then $p$ is a covering map.

I am just wondering is there an easy counter example to show the necessary of "inverse image of every point is finite" condition? I have already checked Hatcher's book (the wedged circle example) but I wonder is there a less complex and more obvious example?

I tried to construct a counter example by composition of a covering map from $R_+$ to $S^1$ with $R$ in the middle, but I failed due to I can't find a covering map from $R^+$ to $R$ which does the job.

share|improve this question
4  
compositino has a sweet Italian ring to it. Che ne pensano i nostri amici italiani? –  Georges Elencwajg May 19 '12 at 8:51

2 Answers 2

The condition in your problem is sufficient but not necessary. Taking $q$ to be the identity map gives rather trivial counter-examples. For a less trivial counter-example, take $X = \mathbb{R} \times \mathbb{R}$, $Y = S^1 \times \mathbb{R}$, $Z = S^1 \times S^1$, $q = u \times 1_{\mathbb{R}}$ and $r = 1_{S^1} \times u$, where $u : \mathbb{R} \rightarrow S^1$ is the universal covering map. Then $r \circ q : X \rightarrow Z$ is a covering map and $r^{-1}(\{z\})$ is infinite for every $z \in Z$.

In general, to get examples where $r \circ q$ is a covering map and $r^{-1}(\{z\})$ is not finite, the correspondence between covering maps over $Z$ (for "well-behaved" spaces $Z$) and subgroups of the fundamental group $\pi_1(Z)$ means that you get examples where the conclusion of the theorem holds but the hypothesis does not in cases where $\pi_1(Z)$ has non-trivial subgroups of infinite index (hence my choice of $Z = S^1 \times S^1$ above, so that $\pi_1(Z) = \mathbb{Z} \times \mathbb{Z}$). This correspondence also implies that the only covering maps from $\mathbb{R}_+$ to $\mathbb{R}$ are homeomorphisms, so they won't help here.

In the above "well-behaved" refers to some conditions on $Z$ that make everything work nicely. Connected and locally contractible would do or a bit more generally, connected, locally path-connected and semilocally simply-connected (which is what Hatcher uses). Semilocally simply-connectedness is explained in Hatcher. In, a note by Ethan Jerzak, you will find a proof that if $q : X \rightarrow Y$ and $r : \rightarrow Z$ are covering maps and if $Z$ is locally path-connected and semilocally simply-connected, then the composite $r \circ q$ is a covering map. So to get an example where the composite of two covering maps is not a covering map, you have to work with spaces that don't satisfy these conditions. The wedge of shrinking circles is about as simple an example as you can get of a space that is locally connected but not semilocally simply-connected. Jerzak's note gives a fairly detailed account of how you use that space to get the desired example of covering maps $q$ and $r$ such that $r \circ q$ is not a covering map.

There is an alternative to the shrinking circles given in Spanier's Algebraic Topology, Chap. 2, Example 2.8 that you may find simpler. For this, let $Z = S^1 \times S^1 \times S^1\times \ldots$ be a countably infinite product of circles and let $X_n = \mathbb{R}^n \times S^1 \times S^1 \times \ldots $, $n \in \mathbb{N}$. Then there is a covering map $q_n :X_n \rightarrow Z$ that acts as the universal covering $u: \mathbb{R} \rightarrow S^1$ on the first $n$ factors of the product and as the identity $1 : S^1 \rightarrow S^1$ on the remaining factors. Now let $Y$ be the product space $\mathbb{N} \times Z$ where $\mathbb{N}$ is given the discrete topology. Then the covering maps $x \mapsto (n, q_n(x)) : X_n \rightarrow \{n\} \times Z$ fit to together to give a covering map $q : X \rightarrow Y$ and the projection $(n, y) \mapsto y$ gives a covering map $r : Y \rightarrow Z$. However no open subset of $Z$ can be evenly covered by $r \circ q$, so $r \circ q$ is not a covering map.

share|improve this answer
    
Thanks! Can you also provide an easy example to show the composition of two covering maps is not a cover map? That is actually what I wonder about. –  Harry May 19 '12 at 21:40
1  
I think the example you have in Hatcher of two covering maps whose composition is not a covering map is as simple as you are going to get. –  Rob Arthan May 19 '12 at 23:11

You could also look at the paper

J. Brazas, "Semicoverings: a generalisation of covering space theory", Homology, Homotopy and Applications, Vol. 14 (2012), No. 1, pp.33-63.

which gives a generalisation of covering maps, so that a covering map is a semicovering, but semicoverings, unlike coverings, satisfy the "2 in 3" rule: if $h=f\circ g$ and two of $f,g,h$ are semicoverings, then so is the third.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.