# Path homotopy in the plane

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)?

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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.
Just to make sure, the other component will be the empty space since $A$ is the boundary of $C$. Am I right? – JW. Sep 25 '13 at 16:39
Oh, I thought the image of the curves lies in $C$. In that case, it's even simpler: just take the homeomorphism of $C$ to the unit disk. – Marek Sep 25 '13 at 16:41
And yeah, the other component would be empty as a component of $C$. But we can equivalently work in $\mathbb R^2$ and get the infinite component as well. – Marek Sep 25 '13 at 16:42