# Non simply-connected covering space of two other non simply-connected covering spaces

Let $$X$$ be a space with fundamental group $$\mathbb{Z}$$ which is path-connected, locally path-connected, and semilocally simply-connected, and let $$x\in X$$. We have two covering spaces $$p_1: (\tilde{X}_1,\tilde{x}_1) \to (X,x)$$ and $$p_2: (\tilde{X}_2,\tilde{x}_2)\to (X,x)$$ where $$\tilde{X}_1,\tilde{X}_2$$ are not simply connected.

I need to show there is a third covering space $$\tilde{Y}$$ which is also not simply connected, as well as a point $$\tilde{y}\in \tilde{Y}$$, with covering maps $$q_1$$ from $$\tilde{X}_1$$ to $$X$$, $$q_2$$ from $$\tilde{X}_2$$ to $$X$$, and $$f$$ from $$\tilde{Y}$$ to $$X$$ (all w/ basepoints) where $$p_1 \circ q_1 = p_2 \circ q_2 = f$$.

First of all I know that since the covering spaces aren't simply connected, I can't invoke the universal cover in place of $$\tilde{Y}$$.

My next thought was to simply take the union of $$\tilde{X}_1 \bigcup\tilde{X}_2$$ which (I think) would cover the three spaces but I'm not sure how I would choose a basepoint.

No doubt this problem requires one of the propositions from Hatcher, based on the assumptions on $$X$$, but I can't see how either apply yet.

Prop $$1$$: Suppose $$X$$ is path-connected, locally path-connected, and semilocally simply-connected. Then for every subgroup $$H\subset \pi_1(X,x_o)$$ there is a covering space $$p:X_H \to X$$ such that $$p_\ast (\pi_1(X_H,\tilde{x}_0))=H$$ for a suitably chosen basepoint $$\tilde{x}_0\in X_H$$.

Prop $$2$$: Let $$X$$ w/ same assumptions as Prop $$1$$. Then there is a bijection between the set of basepoint-preserving isomorphism classes of path-connected covering spaces $$p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$$ and the set of subgroups of $$\pi_1 (X,x_0)$$ obtained by associating the subgroup $$p_\ast (\pi_1(\tilde{X},\tilde{x}_0))$$ to the covering space $$(\tilde{X},\tilde{x}_0)$$.

Any hints on this problem are appreciated- I am sure I will see the path forward.

Edit: On further reflection, now I am noticing that there will be a bijective correspondence between covering spaces of $$X$$ and subgroups of $$\mathbb{Z}$$, which will look like $$n\mathbb{Z}$$, so $$p_\ast (\pi_1 (\tilde{X}_1))$$ and $$p_\ast (\pi_1 (\tilde{X}_2))$$ will look like $$n\mathbb{Z}$$ and $$m\mathbb{Z}$$ for some $$n,m$$ respectively. If I could then find a subgroup $$r\mathbb{Z}$$ of $$\mathbb{Z}$$ with $$n\mathbb{Z}$$ and $$m\mathbb{Z}\leq r\mathbb{Z}$$ then I could identify a new covering space $$\tilde{Y}$$ whose $$\pi_1$$ corresponds to $$r\mathbb{Z}$$ under $$\pi_\ast$$. And (I think) as a result then $$\tilde{Y}$$ would cover everything? Am I on the right track with this?

• Wow! Excellent timing with your edit, haha. You're on exactly the right track! Be careful, though, you want $r\mathbb{Z} \leq n\mathbb{Z}, m\mathbb{Z}$. See my answer ^_^ Commented Mar 9, 2022 at 0:36

I like your idea to use Prop 1 and 2!

Part of the power of algebraic topology comes from translating the topology (which is frequently quite difficult to get your hands on, since everything is continuous, up to homotpy, etc.) into algebra (which is frequently quite easy to get your hands on, since everything is discrete).

One example of this is the duality of covering spaces and subgroups of $$\pi_1$$.

Covering spaces of $$X$$ (up to isomorphism) are in bijection with subgroups of $$\pi_1 X$$ (up to conjugacy)!

This happens since if $$Y$$ covers $$X$$, then $$\pi_1 Y$$ is a subgroup of $$\pi_1 X$$.

Moreover, this correspondence is "direction reversing" in the sense that bigger covering spaces correspond to smaller subgroups (notice this makes sense, since the universal cover corresponds to the trivial subgroup)

Formally, say $$X_1$$ and $$X_2$$, covering spaces of $$X$$, correspond to $$H_1$$ and $$H_2$$, subgroups of $$\pi_1 X$$.

Then $$X_1$$ covers $$X_2$$ if and only if $$H_1 \leq H_2$$

Now, you're given a space $$X$$ where $$\pi_1 X = \mathbb{Z}$$. So we have a very good understanding of the subgroups of $$\pi_1 X$$! That means, using this correspondence, we must have a very good understanding of the covering spaces of $$X$$!

This is remarkable, since covering spaces are extremely complicated objects, but we can understand them by simply looking at subgroups of $$\mathbb{Z}$$, which are classified by natural numbers!

So then, with all this in mind, here's a hint for solving your problem:

We know that $$X_1$$ and $$X_2$$ cover $$X$$. So $$\pi_1 X_1 = m \mathbb{Z}$$ and $$\pi_1 X_2 = n \mathbb{Z}$$ are subgroups of $$\mathbb{Z}$$.

We want to find a space $$Y$$ which covers $$X_1$$ and $$X_2$$. But by our (direction reversing!) correspondence, it suffices to construct a subgroup of $$\mathbb{Z}$$ which is contained in both $$m \mathbb{Z}$$ and $$n \mathbb{Z}$$. Asking that $$Y$$ not be simply connected is asking that this subgroup is nontrivial.

Can you find such a subgroup?

Once you have a subgroup $$H$$, we can construct the space $$Y$$ by looking at $$\widetilde{X} \big / H$$. That is, take the universal cover of $$X$$, and quotient out by the natural action of $$H$$ on it. You can find more details about this construction here, for instance.

I hope this helps ^_^

• Ah, excellent timing! I had just added my edit where I realized the correspondence between covering spaces and the subgroups of $\mathbb{Z}$, as you mentioned. Commented Mar 9, 2022 at 0:37
• Okay, I believe if $r=$lcm$(n,m)$ then $r\mathbb{Z}$ will be the right subgroup of $n\mathbb{Z}$ and $m\mathbb{Z}$. Then by taking the universal cover $\mathbb{R}$ of $X$ and modding out $r\mathbb{Z}$ we can construct $\tilde{Y}$ as $\mathbb{R}/r\mathbb{Z}$. Is this correct? Commented Mar 9, 2022 at 0:43
• Yup! That sounds great! You have to be a bit careful, though, as there are spaces $X$ other than $S^1$ whose fundamental group is still $\mathbb{Z}$. For these spaces the universal cover might not be $\mathbb{R}$, so it's better to call it $\widetilde{X}$. Commented Mar 9, 2022 at 0:51