Definition: Let $\gamma_0,\gamma_1:[0,1]\to G$ two rectifiable curves and $G\subseteq\mathbb{C}$ an open connected set. We say $\gamma_0$ and $\gamma_1$ are homotopic in $G$ if there exists $\Gamma:[0,1]\times[0,1]\to G$ continuous such that:
\begin{cases}
\Gamma(s,0)=\gamma_0(s), \Gamma(s,1)=\gamma_1(s) & 0\le s\le 1 \\
\Gamma(0,t)=\Gamma(1,t) & 0\le t\le 1 \\
\end{cases}
The question is:
Show that if we remove the condition "$\Gamma(0,t)=\Gamma(1,t)$" in the above definition, then the curves $\gamma_0(s)=e^{2\pi i s}$ and $\gamma_1(s)=1$ if $0\le s\le 1$ would be homotopic in $\mathbb{C}\setminus\{0\}$.
I defined by $\Gamma(s,t) = t + (1-t)e^{2\pi is}$.
This function satisfies the definition, but I saw in a topology article that the unit circumference is not homotopic with point 1. So where is my error?
Best Answer
$\Gamma$ doesn't stay inside $\Bbb C\setminus\{0\}$. So no homotopy, and no contradiction.
For the question itself, how about $\Gamma(s,t)=e^{2\pi i s(1-t)}$.