Let $$A=\{ z\in \mathbb{C}|1<|z|<2\}.$$Put $G=A\setminus \{a+i0|a\ge 0\}$. Show that $G$ is simply connected.
Our definition of simply connected is a domain $B\subset \mathbb{C}$ where for any two paths $\gamma_0,\gamma_1$ in $B$ with the same starting point $a$ and the same ending point $b$ there exists a homotopy $$H:[0,1]\times [0,1]\to B,$$ such that $$H(0,t)=\gamma_0(t),H(1,t)=\gamma_1(t),\quad t\in [0,1]$$and $$H(s,0)=a,H(s,1)=b,\quad s\in [0,1].$$
I'm not sure how to show this. First of all I have to find two paths for two given points in $G$ with the same end and starting point. Let $z_0=r_0e^{i\theta_0}$ and $z=re^{i\theta}$ be elements in $G$ and consider the paths $$\gamma_1(t)=\begin{cases} (r(1-2t)+r_02t)e^{i\theta},\quad t\in [0,\frac{1}{2}],\\ r_0e^{i(\theta(2-2t)+\theta_0|\cos(\pi t)|},\quad t\in [\frac{1}{2},1]\end{cases}$$and
$$\gamma_2(t)=\begin{cases}(r(1-2t)r_02t)e^{i\theta},\quad t\in [0,\frac{1}{2}],\\r_0e^{i(\theta|\sin(\pi t)|+\theta_0|\cos(\pi t)|},\quad t\in [\frac{1}{2},1].\end{cases}$$Then $\gamma_1,\gamma_2$ are two paths starting at $z$ and ending at $z_0$ and contained in $G$. However I'm not sure how to proceed and if this is even the correct way. Any help is appreciated!
Best Answer
Hint: Consider the map \begin{align*} \Phi\colon (1,2)\times(0,2\pi) &\longrightarrow G, \\ (r,\theta) &\longmapsto r\, e^{i\theta}. \end{align*} Can you show that $\Phi$ is a homeomorphism?