To show that Warsaw circle is simply connected

Question

I am doing exercise from algebraic topology book which says show that fundamental group of Warsaw circle is trivial, where Warsaw circle is C=$A_1\cup A_2\cup A_3 \cup A_4 $, where $A_1=\{(x,\sin\frac{\pi}{x}):0<x \le 1 \}$, $A_2=\{ (0,y):-1\le y \le 1\}$, $A_3=\{(x,1+\sqrt{x-x^2}):0\le x \le 1\}$, $A_4=\{(1,y):0\le y \le 1\}$. This is clearly path connected so we can choose any point. Lets pick point $p\in C$. Now we need to show that any loop at $p$ is path homotopic to identity loop. What I am thinking from looking at the graph is any loop at $p$ will be of this way. We will start at $p$, we will travel upto some point $q$, then we have to return to $p$ via the same path, then there's possibility that we move to some other point $r$, but again we have to return via same path. Lets call this path $w$. Now to construct homotopy between $w$ to identity, we need to continuously deform our path $w$ to identity. What I thought is first we will travel through our $w$. Then next time again we start at $p$ but we travel upto point little behind $q$, come back to $p$, travel to point slightly behind $r$, then return to $p$. This way slowly slowly we will deform our $w$ to identity. But I can't write this as an answer. I am struggling to come up with proper mathematical argument, properly defined homotopy. I would also like to know if any reference book has proof of it. Thanks.

Answer 1

Since $C$ is path connected, we may choose any basepoint $p$ and show that all loops based at $p$ are null-homotopic. So let $u : I \to C$ be a loop based at $p$. We claim that there exists $\xi \in (0,1]$ such $u(I) \cap A_1 \subset A_1(\xi) = \{(x,\sin\frac{\pi}{x}): \xi \le x \le 1 \}$. This will prove that $u(I) \subset A_1(\xi) \cup A_2 \cup A_3 \cup A_4 = C(\xi)$ which is homeomorphic to a closed interval, thus showing that $u$ is null-homotopic.

Assume that no such $\xi$ exists. Then we get a sequence $(t_n)$ in $I$ such that $u(t_n) \in A_1$ and $u_1(t_n) < \frac{1}{n}$, where $u_1(t)$ is the first coordinate of $u(t)$. The sequence $(t_n)$ has a convergent subsequence, hence we may assume w.l.o.g. that $t_n \to \tau \in I$. Thus $u(t_n) \to u(\tau)$. Clearly $u_1(t_n) \to 0$, thus $u(\tau) \in A_2$. There exists $\epsilon > 0$ such that $u(t) \in C' = C \setminus \{(\frac{1}{2},\frac{3}{2})\}$ for $\lvert t - \tau \rvert < \epsilon$. We have $\lvert t_n - \tau \rvert < \epsilon$ for $n \ge n_0$. This implies that $C'$ is path connected since $u$ maps the interval between $t_n$ and $\tau$ to $C'$ if $n \ge n_0$ (recall $u(\tau) \in A_2$, $u(t_n) \in A_1$). This is a contradiction because $C'$ is known to have two path components. See for example Path components of Topologist's Sine Curve.