We consider three curves $c_1:[0, 2\pi] \rightarrow \mathbb{C} \setminus \lbrace -1, 1\rbrace$, $c_2, c_3:[0, \pi] \rightarrow \mathbb{C} \setminus \lbrace -1, 1\rbrace$ in the complex plane, where $$ c_1(t) := 2\exp\left( \frac{it}{2}\right), ~c_2(t):= \exp(it)+1, ~c_3(t) := \exp(it)-1. $$
I was wondering whether there exists a homotopy between $c_1$ and the composition $c_2 \oplus c_3 :[0, 2\pi] \rightarrow \mathbb{C} \setminus \lbrace -1, 1\rbrace$ (see picture).
The homotopy is defined as a continuous mapping $h:[0, 2\pi] \times [0, 1] \rightarrow \mathbb{C} \setminus \lbrace -1, 1\rbrace$ such that $h(t, 0) = c_1(t)$, $h(t, 1) = (c_2 \oplus c_3)(t)$, $h(0, s) = 2$ and $h(1, s) = -2$ for all $t \in [0, 2\pi]$ and $s \in [0, 1]$.
The difficulty is that I want $\lbrace -1, 1\rbrace$ to be cut out of the complex plane.
I tried to show that there exists some homotopy between both curves and the curve that directly connects $2$ and $i$ composed with the curve that connects $i$ and $-2$. I have so far failed to construct such mapping.