Let $\gamma_0,\gamma_1:[0,1]\to\mathbb{R}^2$ be paths such that $\gamma_0(0)=\gamma_1(0)$ and $\gamma_0(1)=\gamma_1(1)$. I wish to show that there is a homotopy $\Gamma:[0,1]\times[0,1]\to\mathbb{R}^2$ from $\gamma_0$ to $\gamma_1$ that satisfies the following:
- For all $(s,t)\in[0,1]\times[0,1]$, $\Gamma(s,t)$ is not in the unbounded face of $\gamma_0([0,1])\cup\gamma_1([0,1])$.
PS: If the answer is no, are there additional restrictions which we could place on $\gamma_0$ and $\gamma_1$ which would allow this? (Ex. rectifiable, differentiable, etc.)
If you allow the contour of the union of those two paths to form a simple closed curve*, this is true, and can be seen as a consequence of the Schoenflies Theorem together with the fact that the disk is contractible.
*I think this is the case, since you are talking about "unbounded face".