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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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1 Answer 1

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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.

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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
    
Thank you so much :) I think you just answered it. –  JW. Sep 25 '13 at 16:46

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