Fundamental group of Hawaiian earring is uncountable

Question

I have to show fundamental group of Hawaiian earring($H=\cup^{\infty}_{n=1}K_{n}$, where $K_{n}$ is the circle centered at $\frac{1}{n}$ with radius $\frac{1}{n}$) is uncountable, without using Seifert-van Kampen theorem. So I have come up two ideas of proof:

1.Denote $[n]_{m}$ be the loop that travels counter-clockwise n times in $K_{m}$. Then $\{[n_{1}]_{1}[n_{2}]_{2}...|n_{i}\in\mathbb{Z}, i\in\mathbb{N}\}$ is uncountable, since every element in this set belongs to $\pi_{1}(H,0)$, the fundamental group is thus uncountable.

2.Using the same notation above, the set $\{[1]_{f(1)}[1]_{f(2)}...|f $ is any bijective map from $\mathbb{N} $ to itself$\}$ is uncountable, since $f$ is a reordering of natural numbers and there are uncountable many reordering exist. Thus this set as a subset of fundamental group, the group itself is uncountable.

Are these valid idea of proof?

Answer 1

Your ideas are correct, but you have to make explicit how you regard an element of $\mathbb Z^{\mathbb N}$ as an element of $\pi_1(H)$ and that the resulting function $\phi : \mathbb Z^{\mathbb N} \to \pi_1(H)$ is injective. Let us elaborate 1.

Let us write $l_n^m : [0,1] \to K_n$ for the loop based at $0$ that travels counter-clockwise $m$ times around $K_n$. Explicitly, $l_n^m(t) = \frac{1}{n}(1- e^{2m\pi i t})$. Define

$$\psi((m_n)) : [0,1] \to H, \psi((m_n))(t) = \begin{cases}l_n^{m_n} (n(n+1)t - n) & t \in [\frac{1}{n+1},\frac{1}{n}] \\ 0 & t = 0 \end{cases}$$ This is a well-defined continous map (since each neigborhood of $0$ contains all but finitely many $K_n$). Let $\phi((m_n)) = [\psi((m_n))]$, where $[-]$ denotes homotopy class of paths.

Let us show that $\phi$ is injective. There is a retraction $r_n : H \to K_n$ which maps all $K_r$, $r \ne n$, to $0$. Let $i_n : K_n \to H$ denote inclusion. The map $F_n = (r_n)_* \circ \phi$ has the property that the sequence $(m_n)$ is sent to the homotopy class of the path given by $l_n^{m_n} (n(n+1)t - n)$ for $t \in [\frac{1}{n+1},\frac{1}{n}]$ and maps all other $t$ to $0$. This path is clearly homotopic to $l_n^{m_n}$. Thus, if $\phi((m_n)) = \phi((m'_n))$, then $F_n((m_n)) = F_n((m'_n))$ for all $n$, i.e. $[l_n^{m_n}] = [l_n^{m'_n}]$ for all $n$. But this implies $m_n = m'_n$ for all $n$.

Identifying $\pi_1(K_n)$ with $\mathbb Z$ via the isomorphism $\iota_n : \mathbb Z \to \pi_1(K_n), \iota_n(m) = [l_n^m]$, we can express this alternatively as follows: The homomorphism $$R : \pi_1(H) \to \mathbb Z^{\mathbb N}, R(u) = ((\iota_n)^{-1}(r_n)_*(u))$$ has the property $R \circ \phi = id$.

Also have a look at Fundamental group of mapping cone of quotient map from suspension to reduced suspension .