Let $G$ be a bounded, connected, open set in $\mathbb{R}^2$, and let $\gamma_1$ and $\gamma_1$ denote two paths such that $\gamma_0(0)=\gamma_1(0)$, $\gamma_0(1)=\gamma_1(1)$, and the following hold for each $i=1,2$.

  1. For each $t\in[0,1)$, $\gamma_i(t)\in G$.
  2. $\gamma_i(1)\in\partial G$.

Say that a homotopy $\Gamma:[0,1]\times[0,1]\to\mathbb{R}^2$ from $\gamma_0$ to $\gamma_1$ is a $G$-homotopy if, for all $s\in[0,1)$ and all $t\in[0,1]$, $\Gamma(s,t)\in G$.

Let $w$ denote the final point of the $\gamma_i$ which is in $\partial G$. I believe that if $G^c\setminus\{w\}$ is contained in the unbounded face of $\gamma_0([0,1])\cup\gamma_1([0,1])$, then $\gamma_0$ and $\gamma_1$ are $G$-homotopic.

It seems clearly true, but it is not at all clear to me how to construct such a homotopy. Any help?

EDIT: I replaced "simply connected" with "connected" in the definition of $G$.

  • $\begingroup$ @PeterFranek - $S^1$ is neither simply connected nor open. $\endgroup$ – Paul Sinclair Mar 24 '16 at 20:35
  • $\begingroup$ @PeterFranek The union of two paths $\gamma_1$ and $\gamma_2$ in $\mathbb{R}^2$ is a bounded subset of $\mathbb{R}^2$. Therefore there is some circle $C$ in $\mathbb{R}^2$ which contains $\gamma_1\cup\gamma_2$ in its interior. The "unbounded face of $\gamma_1\cup\gamma_2$ is the component of $(\gamma_1\cup\gamma_2)^c$ which contains that circle $C$. $\endgroup$ – Trevor Richards Mar 24 '16 at 20:43
  • $\begingroup$ @PeterFranek Do you mean if $G$ is the unit disk and $\gamma_1$ and $\gamma_2$ are $\gamma_i(t)=(t^i,0)\in\mathbb{R}^2$? If $G$ is convex there is always just the linear combination homotopy. $\endgroup$ – Trevor Richards Mar 24 '16 at 20:53

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