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Let $X\subset \mathbb{R}^2$ be open connected (and let's say bounded), let $x\in X$ and $y\in\partial X$ be such that there is a Jordan curve $\gamma:[0,1]\to X\cup\{y\}$ such that $\gamma(0)=x$ and $\gamma(1)=y$.

Does there always exist a Jordan curve $\delta:[0,1]\to X\backslash\gamma(0,1)\cup\{y\}$ such that $\delta(0)=x$ and $\delta(1)=y$ and there is no holes in between $\gamma$ and $\delta$?

By "no holes in between" I mean that $\gamma$ and $\delta$ are homotopic in $X\cup\{y\}$, or alternatively, if $G$ is the Jordan domain defined by the union of $\gamma$ and $\delta$, then $G\subset X$.

I think, another way of stating this question is: is it true that the set $C([0,1], X\cup\{y\})$ is locally path connected?

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  • $\begingroup$ "Jordan curve" means that $\gamma(0)=\gamma(1)$. You want to say just "curve". $\endgroup$ Commented Nov 7, 2019 at 15:40
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    $\begingroup$ @SamZbarsky I think "simple curve" would be more appropriate. $\endgroup$ Commented Nov 8, 2019 at 16:06

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Yes, this is true because every topological copy of $[0,1]$ in the plane is unknotted and can be transformed by a homeomorphism of the plane into the straight arc $[0,1]\times\{0\}$. In the latter case the existence of the curve $\delta$ is more-or-less obvious.

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