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?
Best Answer
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 .