Let $C$ be a closed and simply connected subspace of the Euclidean plane $\mathbb{R}^2$.
Suppose we have two simple paths in $C$, continuous functions $\alpha, \beta : [0,1] \to \partial C$, and $\alpha(0) = \beta(0), \alpha(1) = \beta(1)$.
In other words, $\alpha$ and $\beta$ are simple paths have same endpoints in the boundary of $C$ and go through on the boundary of $C$.
Then, I wonder if there exists a continuous map $\Gamma: [0,1] \times [0,1] \to C$ such that $\Gamma(x,0) = \alpha(x)$, $\Gamma(x,1) = \beta(x)$, $\Gamma(0,y) = \alpha(0) = \beta(0)$ and $\Gamma(1,y) = \alpha(1) = \beta(1)$ as well as $\Gamma(x,t) \neq \Gamma(y,t)$ for any $t \in [0,1], x \neq y$
Basically, self-intersection is not allowed during the deformation. Could you refer me to any references related to this property (self-intersection)?
The set $A = \alpha([0,1]) \cup \beta([0,1])$ is homeomorphic to a circle. If the total curve $\alpha - \beta$ is a Jordan curve (as I think you assume) then Jordan curve theorem tells us that $A$ separates $C$ into two components. The compact component will be homeomorphic to the unit disk with $A$ homeomorphic to the boundary of the unit disk. Reducing to this case, the problem is easily solved by taking the linear homotopy between the curves.