3

Let $H$ be the Hawaiian earring, i.e. it is the union of circles $S_n$ with centre $(1/n, 0)$ and radius $1/n$. By a universal cover I mean a covering $p\colon X\to H$ (with $X$ connected) such that for any other covering $p'\colon X'\to H$, there is a map $f\colon X\to X'$ such that $p'\circ f = p$. How do I show that a universal cover doesn't exist?

$H$ is connected, path connected but not semilocally simply connected, so there can't be a simply connected universal cover.

What I tried : mimicking the answer here.

We have a covering projection $p_n\colon \mathbb{R}^n\times H\to H$ given by shrinking $H$ and wrapping the $\mathbb{R}$s around the first $n$ circles. We also have a "projection" map $\pi_n\colon H\to S_n$ given by mapping $S_n$ to $S_n$ and other circles to the origin.

Suppose a universal cover $p\colon X\to H$ exists and $\alpha$ is a non trivial loop in $X$. My idea was to take this $\alpha$, push it to $\mathbb{R}^n\times H$ for a sufficiently large $n$, and then go down to $S_n$. I had hoped that pushing down via $\pi_n\circ p$ would give me a non trivial element, while pushing down by first going to $\mathbb{R}^n\times H$ would give me a trivial loop.

Well, that didn't work because if $a_i, i=1, 2$ are the generators of the fundamental groups of $S_1, S_2$, then the loop $a_1a_2a_1^{-1}a_2^{-1}$ is trivial in all $S_n$s, but is itself not trivial in $H$. Basically, the intersection of kernels of $\pi_{n*}$ is not trivial.

So, how does one prove that a universal cover doesn't exist? And is there any way to salvage the above?

P.S. I have found a couple of solutions on this site, but they assume that a universal cover is simply connected. Following Spanier, I am not assuming that.

red whisker
  • 1,254
  • Perhaps this helps. – Paul Frost Dec 02 '20 at 00:58
  • @PaulFrost thanks for the link, but I don't quite understand it. Let $C_n$ cover $S$, as described above. Look at the preimage of the $n$th circle, in $U$, and in $C_n$. These preimages are a loop and an interval respectively. The morphism from $U$ to $C_n$, if it exists, is continuous. Restrict this function to the loop in $U$, and its image must be an interval in $T$.... Why should the preimage be a loop in $U$, and I guess $T$ should be $C_n$, it hasn't been mentioned before. – red whisker Dec 02 '20 at 02:47
  • I didn't try to understand it, I only found it and thought it could be helpful. Comments are no answers ... – Paul Frost Dec 02 '20 at 11:39
  • Here is the hint. Note that this behaviour cannot occur if the base space has a universal covering. (See Spanier $\S2.1$ co. $13$). tbh I haven't worked out the details, but if I get a chance later I'll have a look. – Tyrone Dec 02 '20 at 16:06
  • @Tyrone I am not sure how to proceed actually. Suppose $E_2\xrightarrow{p_2}E_1\xrightarrow{p_1} H$ is such that $p_1, p_2$ are coverings but their composition isn't, and let $U\xrightarrow{p}H$ be a universal cover. All I know is that there's a lift $U\xrightarrow{f_1}E_1$ (which is a covering). If $U$ is simply connected, then I can lift it again to $E_2$, but I don't see how to continue. – red whisker Dec 04 '20 at 02:34

0 Answers0