Let $D \subset \mathbb C$ be a domain and $\gamma : [\alpha, \beta] \to D$ a closed path. Let $a \in D$ be a given point. Assume that $\gamma$ is homotopic to that point $a$. Prove that $\gamma$ is homotopic to a point $b$ in $D$ for any $b \in D$.
My first idea was, to show that $\gamma_a(t) = a$ is homotopic to $\gamma_b(t)=b$. Then the result would follow, since homotopy is a equivalence relation:
Let $H(t,s) = (1-s)a+sb$. $H$ is continuous, and $H(t,0) = a$, $H(t,1) = b$, and $H(\alpha,s) = H(\beta,s)$. The problem is, that I don't know if $H$ maps $[\alpha,\beta] \times [0,1]$ into $D$, I just now that $H(t,s) \in \mathbb C$.
Edit: Following the answer we have the join $\eta = \gamma_a + \gamma_b : [\alpha,2\beta - \alpha] \to D$, \begin{align*} \eta(t) = \begin{cases} a, & t \in [\alpha,\beta], \\ b, & t \in [\beta,2\beta-\alpha]. \end{cases} \end{align*} My first problem is here that $\eta$ is not necessarily continuous, unless $a = b$, which it needs to be in order to be a path.
Your homotopy doesn't work because $D$ needn't be convex. However, being a domain, $D$ is connected, so $a$ and $b$ are joined b some path, say $\eta$. Concatenate your path with this one (in some appropriate way) and see what happens.
Added: What you have so far is a path $\gamma : [\alpha, \beta] \to D$ such that $\gamma \simeq c_a$, where $c_a : [\alpha, \beta] \to D$ is the constant path $c_a(t)=a$. Let $H : [\alpha, \beta] \times [0,1] \to D$ be a homotopy from $\gamma$ to $c_a$.
Let $\eta$ be a path from $a$ to $b$ in $D$. This exists because $D$ is path-connected. The choice of domain for $\eta$ is some arbitrary (real) interval, and for reasons that will help later, suppose $\mathrm{dom}(\eta) = [\frac{1}{2}, 1]$, i.e. $\eta(\frac{1}{2})=a$ and $\eta(1) = b$. We don't know what value $\eta(t)$ will take for $\frac{1}{2}<t<1$, but that doesn't matter.
You need to find a new homotopy $K : [\alpha, \beta] \times [0,1] \to D$ such that $K(s,0)=\gamma(s)$ and $K(s,1)=c_b(s)=b$ for all $s \in [\alpha,\beta]$. You can do this by squashing $H$ into $[0,\frac{1}{2}]$ and defining $K(s,t)=\eta(t)$ for $\frac{1}{2} \le t \le 1$. The details are left to you.