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I've read the following exercise.

Let $p:\tilde X\to X$ be finite connected covering map. Show that there exists a loop in $X$ none of whose lifts is a loop.

I can't understand why it's supposed to be true. Can't I take $\tilde X=X$ and $p=\mathrm{id}_X?$ Then if I take any loop $\omega$ in $X$, I have $\omega\circ p=\omega$, so $\omega$ is its own lift. Why is it incorrect?

(I'm not sure what a connected covering map is, but I think my example must satisfy this condition, whatever it means, if $X$ is connected.)

Added. Here's what I've got after David Speyer's comments. First of all, we should add to the hypothesis that the degree of $p$ is not $1$. Now, there are the following two theorems.

  • If $p:\tilde X\to X$ is a covering map, $x_0\in X,\ \tilde x_0\in p^{-1}(x_0),$ then the induced homomorphism $p_*:\pi(\tilde X,\tilde x_0)\to\pi(X,x_0)$ of the fundamental groups is a monomorphism.
  • If $p:\tilde X\to X$ is a connected covering map, then its degree is equal to the index $[\pi_1(X,x_0):p_*(\pi_1(\tilde X,\tilde x_0))].$

Suppose every loop in $X$ has a lift that is a loop. This means that for each loop $\omega$ in $X$ there exists a loop $\tilde\omega$ in $\tilde X$ such that $\tilde\omega\circ p=\omega.$ But this means that $p_*([\tilde\omega])=[\omega],$ where $[\gamma]$ denotes the homotopy class of a loop $\gamma.$ But then $p_*$ is a group epimorphism. By the first theorem it is also a monomorphism, and so it is an isomorphism between $\pi_1(\tilde X,\tilde x_0)$ and $\pi_1(X,x_0).$ Therefore, $[\pi_1(X,x_0):p_*(\pi_1(\tilde X,\tilde x_0))]=1,$ and, by the second theorem, the degree of $p$ is $1$, a contradiction with the added hypothesis.

Is this correct? I think it's probably not, because I'm not using the finiteness of $p$. The finiteness of $p$ means, according to the theorem that the index mentioned before is finite. But I can't see at all why that should be necessary.

The only example I know of an infinite covering map is $e^{it}$ covering the unit circle with the real line. But here there are plenty of loops that have no "loopy" lifts because there are no nontrivial loops in the real line, and there are many in the circle. So if the infinity of the covering map is indeed a problem, this example doesn't show it.

Re David Speyer's answer

Don't the theorems work for any $\tilde x_0\in p^{-1}(x_0)?$ That is, for each such $\tilde x_0,$ we have a monomorphism $p_{*,\tilde x_0} : \pi_1(\tilde X,\tilde x_0) \to\pi_1(X,x_0).$ If the hypotheses of the problem hold, there is, as you say, a $y\in p^{-1}(x_0)$ and a loop $\tilde\omega$ in $\tilde X$ starting and ending in $y$ such that $p\circ\tilde\omega=\omega$. Then can't I just use $p_{*,y}$ as above?

Bartek
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    Throw in "non trivial", meaning $p$ has degree $>1$, and the statement is true. It's nontrivial though even when you know what connected covering maps are, and I don't think it's reasonable to attempt without that background. – David E Speyer Mar 29 '13 at 00:17
  • I assume that $p$ is supposed to have several sheets. Then if $x_0$ is a basepoint in $X$ and $\widetilde x_1, \widetilde x_2$ are both mapped to $x_0$ then there is a path between them whose image is a loop. I suppose you have to show that at whatever basepoint in the fiber of $x_0$ you take the lift, it is always a loop. – Stefan Hamcke Mar 29 '13 at 00:20
  • @DavidSpeyer I know what a covering map is. I just don't know that the adjective "connected" means here. Is it crucial? I'm quite sure it's not reasonable for me to try to learn homotopy theory, since I'm having trouble understanding the simplest things in it, but I have to if I want to have a Master's degree. – Bartek Mar 29 '13 at 00:21
  • in particular, if $X$ is simply connected then there's no 'nontrivial' connected covering map so the statement is wrong, right? – suissidle Mar 29 '13 at 00:21
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    Connected means that $\tilde{X}$ is connected. Have you learned the relationship between connected covering spaces and the fundamental group yet? @suissidle If $X$ is simply connected, the (corrected) hypotheses of the statement are impossible to satisfy, so the statement is vacuously true. – David E Speyer Mar 29 '13 at 00:23
  • @suissidle If $X$ is simply-connected and $\widetilde X$ is path-connected, then $p$ is a homeomorphism, see http://math.stackexchange.com/questions/311594/proving-that-a-covering-map-with-certain-domain-and-range-is-homeomorphism – Stefan Hamcke Mar 29 '13 at 00:25
  • i was thinking along the lines if $X$ is simply connected then $X$ with the identity map is its own universal cover, so every covering map is trivial. so the op's counterexample holds for simply connected $X$ – suissidle Mar 29 '13 at 00:28
  • Regarding my comment above, I just realized that if one lift is not a loop, this does not imply that other lifts aren't loops either. David Speyer is right in that this is not so trivial and you should know about the relationship between covering spaces and fundamental groups. – Stefan Hamcke Mar 29 '13 at 00:30
  • @DavidSpeyer I've read the theorem again, and I think I understand more or less how to solve the problem now. (I'm treating the theorem as a black box because I see that I will have spend at least a couple of days on understanding the proof.) I think the theorem tells me that for each loop in $X$ to have a "loopy" lift, the induced monomorphism from the fundamental group of $\tilde X$ to $X$ would have to be an isomorphism. But then, the theorem says, the degree of the covering map would have to be $1$. I'm only not sure where I need the finiteness of the map. – Bartek Mar 29 '13 at 01:11
  • counterexample to 'finite' starting with the exponential map $\mathbb{R}\rightarrow S^1$ (infinite covering) identify the 'endpoints' of the real line: this induces an infinite covering $S^1\rightarrow S^1$. now, a loop in the 'domain' $S^1$ is mapped to a loop whose lift is by construction a loop. – suissidle Mar 29 '13 at 15:39
  • @suissidle This doesn't make sense. Where do the endpoints of the line map to? – David E Speyer Mar 29 '13 at 23:34
  • i realize the problem now. what's your suggestion then? – suissidle Mar 30 '13 at 00:55

2 Answers2

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Your proposed proof doesn't work. Let me write out the key part in more detail, so that you can see the issue.

Recall that an element of $\pi_1(X, x_0)$ is an equivalence class of loops starting from and ending at $x_0$. Let $\tilde{X} \to X$ be a connected covering map, and let $\tilde{x}_0$ in $\tilde{X}$ be a preimage of $x_0$. As you say, $\pi_1(\tilde{X}, \tilde{x}_0) \to \pi_1(X, x_0)$ is injective. You then try to prove that it is surjective, but your proof is flawed as I will now explain.

Let $\omega$ be a loop in $X$, starting and ending at $x_0$. By hypothesis, there is a loop $\widetilde{\omega}$ in $\tilde{X}$ mapping to $\omega$. However, $\widetilde{\omega}$ need not start and end at $\tilde{x}_0$. The start and end of $\widetilde{\omega}$ is some preimage of $x_0$, call it $y$, but we are not assuming that $y=\tilde{x}_0$.

The essence of this problem is (1) to work out what consequence the hypothesis has for the fundamental groups, and (2) do some nontrivial group theory.

A hint for part (1): Let $\eta$ be a path in $\tilde{X}$ from $\tilde{x}_0$ to $y$. Then the concatenation $\eta \widetilde{\omega} \eta^{-1}$ is a class in $\pi_1(\tilde{X}, \tilde{x}_0)$. What implication does this have for the map of fundamental groups $\pi_1(\tilde{X}, \tilde{x}_0) \to \pi_1(X, x_0)$?

But I continue to be confused as to how this problem could be reasonably assigned in a class where the only infinite covering map you have seen is the circle to the real line.

David E Speyer
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  • Thank you for the answer. The difference in level between me and the question in not the class's fault but mine. I wasn't paying attention enough. I think trying to solve a problem is a good way of getting to understand a theory. At least it has usually worked for me. Let me try again in the next comment. – Bartek Mar 29 '13 at 14:10
  • Response to @Bartek's edit: What you have shown is that $\bigcup_{y \in p^{-1}(x_0)} p_{\ast}\left( \pi_1(\tilde{X}, y) \right) = \pi_1(X, x_0)$. This does not show $p_{\ast}(\pi_1(\tilde{X}, y))=\pi_1(X, x_0)$ for any single $y$. However, it is on the right track. For $y$ and $z$ two different preimages of $x_0$, what can you say about the relationship between $p_{\ast}(\pi_1(\tilde{X}, y))$ and $p_{\ast}(\pi_1(\tilde{X}, z))$? – David E Speyer Mar 29 '13 at 14:52
  • OK, so I have four isomorphic groups: $\pi_1(\tilde X,y),$ $\pi_1(\tilde X,z),$ $p_(\tilde X,y)$ and $p_(\tilde X,z)$, the latter two being subgroups of $\pi_1(X,x_0).$ We can draw a diagram with four arrows standing for the isomorphisms: $p_*$ pointing downwards both on the left and on the right, $\xi=\eta(\cdot)\eta^{-1}$ on the top, and $\nu=p\circ\xi\circ p^{-1}$ on the bottom. Is that correct so far? – Bartek Mar 29 '13 at 17:28
  • Or rather, I have a diagram that's a rectangle: $1$-arrow high and $(n-1)$-arrow wide, where $n=|p^{-1}(x_0)|<\infty.$ I also know that each element of $\pi_1(X,x_0)$ belongs to one of the groups which are the nodes of the bottom side of the rectangle. – Bartek Mar 29 '13 at 17:36
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    I think I got it! The path $\eta$ is also sent to a loop in $\pi_1(X,x_0).$ That's the missing link I couldn't figure out. So now all of the images of the particular fundamental groups of $\tilde X$ are conjugate to each other in $\pi_1(X,x_0).$ And now is probably the time to do the non-trivial group theory you mentioned in your answer. This is probably what Stefan H. is saying in his answer, but I haven't read it carefully yet because I wanted to figure it out for myself. – Bartek Mar 29 '13 at 21:55
  • You got it! Congratulations! And yes, now it is time for nontrivial group theory. – David E Speyer Mar 29 '13 at 23:33
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If the covering space is path-connected, then for $\widetilde x_0$ and $\widetilde x_1$ in the fiber of $x_0$ there is a path $\widetilde h$. Now the $p_*(\pi_1(\widetilde X,\widetilde x_1))$ and $p_*(\pi_1(\widetilde X,\widetilde x_0))$ are related in a special way. A loop $\widetilde\gamma$ at $\widetilde x_1$ can be written as $\overline{\widetilde h}\cdot\widetilde\beta\cdot\widetilde h$ where $\widetilde\beta=(\widetilde h\cdot\widetilde\gamma\cdot\overline{\widetilde h})$ is a loop at $\widetilde x_0$. Denote with $\beta,\gamma,h$ the homotopy classes of the images under $p$. Then the image of $[\gamma]$ under $p_*$ is $h^{-1}\beta h$. Analogously, the image of $[\beta]$ is $h\gamma h^{-1}$. So we have $p_*(\pi_1(\widetilde X,\widetilde x_1))=h^{-1}p_*(\pi_1(\widetilde X,\widetilde x_0))h$. You want to show that the lift at every point in the fiber of $x_0$ is not a loop. So you have to prove that there is a loop in $\pi_1(X,x_0)$ which is not in any conjugacy class of $p_*(\pi_1(\widetilde X,\widetilde x_0))$. Since the image has finite index, we can apply Arturo's result that the union of the conjugates is not the whole group. That should do it.

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Stefan Hamcke
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  • Why does the group have to be finite? The result in the linked answer seems to hold for any groups of finite index, doesn't it? – Bartek Mar 29 '13 at 14:52
  • @Bartek You are right. I have to confess that I didn't read it carefully enough :-) I'll edit the Edit. – Stefan Hamcke Mar 29 '13 at 15:53
  • This, and other interpretations of the algebraic result, is discussed in the nice elementary paper On a theorem of Jordan by J-P. Serre (Bulletin AMS, 2003). – ACL Apr 13 '14 at 10:26