Let $S$ be a surface and $U$ an open connected subset of $S$. If the frontier of $U$ in $S$ is a two sided circle $C$, then the closure of $U$ in $S$ is a surface whose boundary is $C$. I would like to know if this result is known. Thanks
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1$\begingroup$ What does "two sided circle" mean? $\endgroup$– WojowuCommented Mar 4, 2021 at 17:59
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2$\begingroup$ By a two sided circle I mean an orientation preserving simple closed curve $\endgroup$– Fernando OliveiraCommented Mar 4, 2021 at 18:10
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1$\begingroup$ I'm not sure what that means either. The only meaning of "orientation preserving" in this context I know if is for a map between oriented manifolds of the same dimension, so it's not clear to me what it would mean for a curve on a surface $\endgroup$– WojowuCommented Mar 4, 2021 at 18:28
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3$\begingroup$ A curve on a surface is said to be two-sided if it has a neighborhood homeomorphic to an annulus. Equivalently, there is a neighborhood basis of C consisting of open sets with U \ C two components. $\endgroup$– mmeCommented Mar 4, 2021 at 18:45
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$\begingroup$ Have you tried reading a textbook that describes the basics of surfaces and their submanifolds? There are a lot of references that usually cover many other topics. Guillemin and Pollack is one of my favourites. $\endgroup$– Ryan BudneyCommented Mar 4, 2021 at 19:42
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I don't think this is true. Let $S$ be a torus, and let $C$ be a non-separating simple closed curve in $S$. Let $U=S\setminus C$. Then $U$ is open and connected, and the frontier of $U$ is $C$. But the closure of $U$ is $S$ which has empty boundary.
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2$\begingroup$ Yes, one needs the additional assumption that C is separating (which implies that it is 2-sided). $\endgroup$– mmeCommented Mar 4, 2021 at 18:57
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2$\begingroup$ It will also be necessary to assume that $S$ has empty boundary. If $\partial S\neq\emptyset$ and $\partial S$ is connected, then one can take $U$ as a regular open neighborhood of $\partial S$. Then the frontier $C$ is a connected 2-sided circle, but $\partial\overline{U}=C\cup\partial S$. $\endgroup$ Commented Mar 4, 2021 at 19:05
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$\begingroup$ All observations were correct, thanks. $𝑆$ can not have boundary. Also, if $N$ is a neighbourhood of $C$ homeomorphic to an annulus, then one component of $N\setminus C$ is not contained in $U$. $\endgroup$ Commented Mar 4, 2021 at 21:30
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$\begingroup$ @Mike Miller I was just wandering if there were a version of Schoenflies Theorem that holds for two sided simple closed curves $C$ that are connected components of the frontier of an open connected subset $U$ of $S$. If $N$ is an annular neighbourhood of $C$ then we need that one component of $N\setminus C$ is not contained in $U$. $S$ can be any boundaryless surface. $\endgroup$ Commented Mar 5, 2021 at 19:22
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1$\begingroup$ The correct statement is that if $C,D$ are separating curve in the interior of a compact connected surface $S$ (with or without boundary), then there is a homeomorphism $\varphi: S \to S$ with $\varphi(C) = D$ if and only if $S \setminus C$ and $S \setminus D$ are homeomorphic. This follows essentially from the classification of compact surfaces and what I have called to you before the "unique disc lemma" though some work needs to be done (take closure on each side, cap off boundaries, etc). As Ryan suggests above this is all much much easier if you deal with smooth surfaces, curves, etc. $\endgroup$– mmeCommented Mar 5, 2021 at 19:34