Question about a Hawaiian earring space There is one question in a bigger exercise I can't quite get my head around. For all $n \in \mathbb{N}$, let
$$\begin{align} S_n^+ &= \{(x,y,0) \in \mathbb{R}^3 : (x - 1/n)^2 + y^2 = 1/n^2\},\\
S_n^- &= \{(x,y,0) \in \mathbb{R}^3: (x + 1/n)^2 + y^2 = 1/n^2\}\\ 
D_n^+ &= \{(1 - t)p + t(0,0,1) : p \in S_n^+ , 0 \leq t \leq 1\},\\
D_n^-&= \{(1 - t)q + t(0,0,-1) : q \in S_n^- , 0 \leq t \leq 1\}.
\end{align}$$
Now we define the spaces
$$
X_\infty = \bigcup_{n \in \mathbb{N}} (D_n^+ \cup D_n^-)\\
X_k = \bigcup_{1 \leq n \leq k} (D_n^+ \cup D_n^-).
$$
Let now $R_k$ be the retraction of $X_\infty$ to $X_k$ sending $p \mapsto p$ if $p \in X_k$ and $(x,y,z) \mapsto (0,0,z)$ otherwise. We note by $\alpha_j^{\pm}: [0,1] \to S_j^{\pm}$ the loop going once around the circle $S_j^{\pm}$. We suppose now by contradiction, that the infinite concatenation $\gamma = \dotsm * \alpha_2^- * \alpha_2^+ * \alpha_1^- * \alpha_1^+$ is nullhomotopic to the constant path at the origin via the homotopy $F : [0,1] \times [0,1] \to X_\infty$.
Eventually, we want to show that $X_\infty$ is not simply connected. As a first step we have to show the following: Let $U$ be a connected component of $(R_k \circ F)^{-1}(X_k\setminus \{(0,0,0)\})$ intersecting $[0,1] \times \{0\}$ (where we have $F_{\mid [0,1] \times \{0\}} = \gamma$). Show that $R_k \circ F(U)$ contains $(0,0,1)$ or $(0,0,-1)$.
I tried to use the fact that $(R_k \circ F)(U)$ must lie in one of the connected components of $X_k\setminus (0,0,0)$ but I have a hard time formalizing why either $(0,0,1)$ or $(0,0,-1)$ is contained in it. Any help is appreciated.
 A: The space $X_{\infty}$ is sometimes called the Griffiths twin cone or the double cone of the Hawaiian earring $HE=\bigcup_{n}S_{n}^+\cup \bigcup_{n}S_{n}^-$. This seems to be the main example folks give of a one-point union of contractible spaces, which is not contractible. The simplest idea to deal with your question, I think, is to use the van Kampen theorem twice as is done in this blog post. This tells you that $\pi_1(X_{\infty})$ is the quotient of $\pi_1(\bigcup_{n}S_{n}^+\cup \bigcup_{n}S_{n}^-)$ by the normal subgroup $N$ generated by $\pi_1(\bigcup_{n}S_{n}^+)\cup \pi_1(\bigcup_{n}S_{n}^-)$. So the remaining work is to show your infinite product does not lie in $N$. This should boil down to a question only about the fundamental group of the Hawaiian earring and free groups, although this group is tricky enough as it is.
The post here (and others from this blog) on the Hawaiian earring $HE$ might be helpful. There are retractions $r_n:HE\to\bigcup_{k\leq n}S_{k}^+\cup S_{k}^{-}$. Notice that $\pi_1(\bigcup_{k\leq n}S_{k}^+\cup S_{k}^{-})=F_{2n}$ is the free group on $2n$-letters. The main insight that is not so easy to prove is that an element $a\in \pi_1(HE)$ is trivial if and only if $(r_n)_{\#}(a)$ reduces to the trivial word in $F_{2n}$ for all $n\in\mathbb{N}$. So suppose, to obtain a contradiction, that $g$ (which you know is non-trivial in $\pi_1(HE)$ using retractions) is your alternating infinite word and $g\in N$. Then $g$ can be written as a finite product $\prod_{j=1}^{m}g_jh_jg_{j}^{-1}$ where $g_j\in\pi_1(HE)$ and $h_j\neq 1$ lies in $\pi_1(\bigcup_{n}S_{n}^+)$ or $\pi_1(\bigcup_{n}S_{n}^-)$. Thus $(r_n)_{\#}(g)=(r_n)_{\#}(\prod_{j=1}^{m}g_jh_jg_{j}^{-1})$ for all $n\in\mathbb{N}$. Using shape-injectivity, find $n$ large enough so that $(r_n)_{\#}(h_j)\neq 1$ for all $1\leq j\leq m$,. Then by analyzing the word structure of the projections $(r_n)_{\#}(g)$ and $(r_n)_{\#}(\prod_{j=1}^{m}g_jh_jg_{j}^{-1})$ in the free group $F_{2n}$, you should find your contradiction.
