I have the following question:

Assume $p:\widetilde{X} \to X$ is a covering map with $\widetilde{X},X$ both path-connected. Assume $A$ is a path connected subset of $X$ so that $i_*:\pi_1(A,a) \to \pi_1(X,a)$ is onto for some $a \in A$ where $i$ is the inclusion map. Prove that $p^{-1}(A)$ is path connected.

I realize that this question has been asked in this post if $p:\widetilde{X}\rightarrow X$ is a covering space and $\widetilde{X}$ is path connected ,show that $p^{-1}(A)$ is path connected. and users were saying that the claim was false. However, in the question I have we have that $i_*$ is onto which seems to make it work. Here is what I have come up with:

let $a_1,a_2 \in p^{-1}(A)$. Consider the elements $p(a_1),p(a_2) \in A$. Since $A$ is path connected there is a path $f$ in $A$ from $p(a_1)$ to $p(a_2)$. By the path lifting property, we can lift $f$ to a path $\tilde{f}$ starting at $a_1$ and ending at some point in the fiber $p^{-1}(a_2)$. Call this point $a_3$ (so $\tilde{f}$ is a path in $p^{-1}(A)$ from $a_1$ to $a_3$ where $a_3 \in p^{-1}(A)$).

Now, since $\widetilde{X}$ is path connected, the lifting correspondence is surjective. Thus, there is some loop $g$ based at $p(a_3)$ in $X$ such that the lift $\tilde{g}$ is a path in $\widetilde{X}$ from $a_3$ to $a_2$. Now, since $i_*$ is onto there is a loop $h$ based at $p(a_3)$ such that $i\circ h$ is homotopic to $g$. By the homotopy lifting property, $\widetilde{i\circ h}$ is a path in $\widetilde{X}$ that begins at $a_3$ and ends at $a_2$. This means that $\tilde{h}$ is a path in $p^{-1}(A)$ from $a_3$ to $a_2$ (I am not sure if this follows so directly from what I have said). Then, $\tilde{f}\cdot\tilde{h}$ is a path in $p^{-1}(A)$ from $a_1$ to $a_2$.

Does the above argument seem to make sense? Any comments or suggestions would be helpful.

  • $\begingroup$ It looks correct to me. $\endgroup$ – lEm Dec 28 '19 at 6:58

Your proof is correct, it only has a minor gap. Here are some suggestions.

  1. In my opinion the phrase "the lifting correspondence is surjective" seems to be somewhat unclear and I would omit it. In fact you have a path $\tilde g$ in $\tilde X$ from $a_2$ to $a_3$ and thus $g = p \circ \tilde g$ is a loop based at $a' = p(a_2) = p(a_3)$.

  2. You know that $\pi_1(A,a) \to \pi_1(X,a)$ is onto. However, you do not know that $a' = a$, thus you have to add an argument that also $\pi_1(A,a') \to \pi_1(X,a')$ is onto. This is fairly trivial, but you can avoid this by starting your proof with a fixed $a_2 \in p^{-1}(a)$. This will show that any point $a_1 \in p^{-1}(A)$ is connected by a path to this $a_2$, so you are done.

  • $\begingroup$ For your first point, doesn't the covering space ($\widetilde{X}$ in this case) need to be path connected for the correspondence to be surjective? If not, then the hypothesis that $\widetilde{X}$ is path connected can be omitted. For you second suggestion, thank you for pointing this out, I had not noticed it. $\endgroup$ – Mike Dec 29 '19 at 17:36
  • 1
    $\begingroup$ Yes, $\tilde X$ has to be path connected to assure that there exists a path from $a_2$ to $a_3$. $\endgroup$ – Paul Frost Dec 30 '19 at 0:28
  • $\begingroup$ Okay, thank you for your comments and suggestions. $\endgroup$ – Mike Dec 30 '19 at 3:31

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