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Hi: I am interested in the "Pearl necklace" space: the space created by, given $n$ copies $S_1^2, \dots, S^2_n$ of the unit sphere, gluing each one to the next one by a single dot and the last one to the first, forming a cycle. It is the space studied here.

In that question it is exposed that a covering space is an infinite chain of spheres, which intuitively is quite clear. However, I don't know how to formally prove it. I would be grateful if someone provided a proof of this fact, or a scheme about how a proof would go. Thanks in advance.

Zanzag
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  • The infinity chain of spheres is simply connected and covers the pearl necklace, in the same fashion, $\mathbb R$ covers $S^1$. Therefore it is its universal cover. – Sven-Ole Behrend May 09 '21 at 10:01
  • @Sven-OleBehrend My problem is I'm having troubles formalizing the result? Could you tell me how would you go about it? – Zanzag May 09 '21 at 10:04

2 Answers2

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Remember what it means to be a covering space. We say $X$ is covered by $\tilde{X}$ exactly when there's a continuous surjection $p : \tilde{X} \to X$ with the following ~bonus property~ (often called the covering condition):

Every point $\tilde{x} \in \tilde{X}$ has a neighborhood $U$ so that $p \restriction U$ is a homeomorphism.

That is, we want a continuous surjection which is locally a homeomorphism.

But now let's look at the infinite chain of spheres. This is a quotient space of $\bigsqcup_\mathbb{Z} S^2$ where we identify the south pole of $S^2_k$ with the north pole of $S^2_{k+1}$ for each $k \in \mathbb Z$. This space will be our $\tilde{X}$.

As for our $X$, it's almost the same thing. We have $\bigsqcup_{\mathbb{Z}/n} S^2$ where we do the same identification (which is now made "mod $n$", so we glue the end of the last sphere to the start of the first).

It should be clear from this description what a good choice of $p$ should be. Let's just send the $k$th sphere in $\tilde{X}$ to the $k \pmod{n}$th sphere in $X$. I'll leave it to you to check that

  • this is surjective (almost too easy to mention)
  • this is continuous (easy enough, if you remember how to glue together functions defined on closed subsets)
  • every point $\tilde{x} \in \tilde{X}$ has a neighborhood homeomorphic to a neighborhood of $p \tilde{x}$ in the image (there are two cases: If $\tilde{x}$ is a random point, this shouldn't be hard. If $\tilde{x}$ is one of the points that we're gluing stuff too, we have to check that $p \tilde{x}$ is also a point that got glued in a similar way. But of course, this will be true)

I hope this helps ^_^

HallaSurvivor
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Theorem: Let $E$ be a simply connected, locally path connected topological space and let $L : E \times G \rightarrow E$ be covering left action (i.e. $\forall e \in E$ there exists an open neighbourhood $U_e$ such that $L_g(U_e)\cap U_e = \emptyset$ $\forall g \in G\setminus\{1\}$). Then $p : E \rightarrow E/G$ is a covering space and for every $e \in E$ $$\pi_1(E/G,p(e)) \simeq G$$

In your case consider $E = \bigcup_{m \in \mathbb{Z}} C_m$ with $C_m := \{x \in \mathbb{R}^3\,:\,\vert x - (\frac{2m+1}{2},0,0)| = \frac12\}$, which is simply connected and locally path connected, and consider the action $$L : E \times \mathbb{Z} \rightarrow E$$ $$(x,k) \mapsto x + nk$$ It is possible to prove that it is a covering action, then the quotient map $p : E \rightarrow E/\mathbb{Z}$ is a covering space and since $E/\Bbb Z$ is homeomorphic to $P$ (Pearl Necklace) $E$ is its universal cover and $$\pi_1(P):=\pi_1(P,p) \simeq \mathbb{Z}$$

ocsecnarf ittorettul
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