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I want to show that the fundamental group of the Hawaiian earring is uncountable. We construct the Hawaiian earring $X$ as the union of the circles $C_n$ of radius $\frac{1}{n}$ centered at $\left(\frac{1}{n}, 0\right)$.

Let $x_0 =\left(0,0\right)$, we define the continuous map $r_n:X \rightarrow C_n$, by $$x\mapsto \begin{cases}x\text{, if }x \in C_n \\ x_0 \text{, otherwise}\end{cases}$$

For $n\geq 1 $ let $\gamma_n:[0, 1]\to C_n$ be a loop at $x_0$ and define $\gamma:[0,1]\rightarrow X$ by $t\mapsto \gamma_n\left(\frac{t-(1-\frac{1}{n})}{(1-\frac{1}{n+1})-(1-\frac{1}{n})}\right)$ if $1-\frac{1}{n}\leq t < 1-\frac{1}{n+1}$ and $\gamma(1)=x_0$. Then $\gamma$ is well defined because the intervals $[1-\frac{1}{n},1-\frac{1}{n+1})$ form a partition of $[0,1)$. Why is $\gamma$ continuous? Why is the class of $r_{n*}(\gamma)$ in $\pi_1(C_n, x_0)$ equal to the class of $\gamma_n$ in $\pi_1(C_n, x_0)$?

I am aware of Fundamental group of Hawaiian earring is uncountable, but in the answer apparently the above is "clear".

Jakobian
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user1164922
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  • Do you mean $r_n : X\to C_n$? – Arctic Char May 16 '24 at 13:22
  • Did you mean to write $[1-\frac{1}{n}, 1-\frac{1}{n+1})$? Then the answer for why $\gamma$ is continuous is simple, its continuous on $[1-\frac{1}{n}, 1-\frac{1}{n+1}]$ for all $n$, which is a locally finite family of closed sets of $[0, 1)$, so $\gamma$ is continuous on $[0, 1)$ by gluing lemma. And you should be able to check that $\gamma$ is continuous at $x = 1$ separately – Jakobian May 16 '24 at 13:23
  • Did you try to show that $\gamma$ is continuous? Pick any open subset of $C$ and split into two cases: $x_0\in U$ and $x_0 \notin U$. In both cases the argument that $\gamma^{-1} (U)$ is open does not seem difficult. – Arctic Char May 16 '24 at 13:26

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I answered the question Fundamental group of Hawaiian earring is uncountable. My argument for the continuity of $\gamma$ was that each neigborhood of $x_0$ contains all but finitely many $C_n$.

In fact, this show that $\gamma$ is continuous at $1$:

Let $U$ be an open neighborhood of $x_0$. Then $C_n \subset U$ for $n \ge n_0$. Since $\gamma(t) \in C_n$ for $t \in I_n :=[1-\frac{1}{n}, 1 -\frac{1}{n+1}]$ and $\gamma(1) = x_0$, we see that $\gamma(t) \in U$ for $t \in [1 - \frac{1}{n_0},1] = \bigcup_{n \ge n_0}I_n \cup \{1\}$. This shows that $V = (1 - \frac{1}{n_0},1]$ is an open neighborhood of $1$ with $\gamma(V) \subset U$.

Continuity in all $t_0 < 1$ is fairly obvious:

Clearly $\gamma \mid_{[0,1 - \frac 1 n]}$ is continuous because $[0,1 - \frac 1 n]$ is covered by the finitely many closed subsets $I_1,\ldots, I_{n-1}$ on which $\gamma$ is continuous. Thus $\gamma \mid_{[0,1 - \frac 1 n)}$ is continuous. Hence $\gamma \mid_{[0,1)}$ is continuous because the $J_n:= [0,1 - \frac 1 n)$ are open in $[0,1)$, $\gamma$ is continuous on each $J_n$ and $\bigcup_{n=1}^\infty J_n = [0,1)$. Since $[0,1)$ is open in $[0,1]$, we see that $\gamma$ is continuous in all $t_0 < 1$.

Why is $(r_n)_*([\gamma]) = [\gamma_n]$? By definition $(r_n)_*([\gamma]) = [r_n \circ \gamma]$ and $$(r_n \circ \gamma)(t)= \begin{cases} x_0 & t \le 1- \frac{1}{n+1} \\ \gamma_n\left(\frac{t-(1-\frac{1}{n})}{(1-\frac{1}{n+1})-(1-\frac{1}{n})}\right) & 1-\frac{1}{n}\leq t < 1-\frac{1}{n+1} \\ x_0 & t \ge 1- \frac 1 n \end{cases}$$ The map $t \mapsto \frac{t-(1-\frac{1}{n})}{(1-\frac{1}{n+1})-(1-\frac{1}{n})}$ maps $I_n$ linearly onto $[0,1]$. You can explicitly write down a homotopy from $r_n \circ \gamma$ to $\gamma_n$; this just follows the usual proofs showing that $\pi_1(X,x_0)$ is a group (associativity and existence of neutral element $e$). You can also apply Proving Munkres Theorem 51.3 which shows that $[r_n \circ \gamma] = e \circ [\gamma_n] \circ e = [\gamma_n]$.

Paul Frost
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