# Inverse image of path connected subspace under a covering map

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.

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

## 1 Answer

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.

• 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. – Mike Dec 29 '19 at 17:36
• Yes, $\tilde X$ has to be path connected to assure that there exists a path from $a_2$ to $a_3$. – Paul Frost Dec 30 '19 at 0:28
• Okay, thank you for your comments and suggestions. – Mike Dec 30 '19 at 3:31