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To state it in full.

By a surface we mean a connected, second countable, Hausdorff topological space such that any point on it has an open neighborhood homeomorphic to $\mathbf{R}^2$.

Let $F$ be a surface, $V$ be an open set in $F$, $\overline{V}$ be the closure of $V$ in $F$. For any $s>0$, let $D_s$ ($\overline{D}_s$) be the open (closed) disk centered around $0$ of radius $s$ in $\mathbf{R}^2$.

Let $h: \overline{D}_1 \rightarrow \overline{V}$ be a homeomorphism such that $h(D_1) = V$.

Question. Is it always possible to find $r>1$, an open neighborhood $U$ of $\overline{V}$ in $F$, and a homeomorphism $h’: D_r\rightarrow U$, making the following diagram commutative?

$$\require{AMScd} \begin{CD} \overline{D}_1 @>h>\sim> \overline{V} \\ @VV\cap V @VV\cap V \\ D_r @>(?)h'>\sim> U \end{CD} $$

Background. (Assuming I am not mistaken) I have seen this question in two places.

  • In 8E, page 15 of Ahlfors and Sairo's Riemann Surfaces, it is stated that every Jordan region is a parametric disk, but this fact will not be needed. No reference is given there as far as I am aware.
  • Same question seems to have been raised in this stack exchange post. The accepted answer assumes we are in smooth category and used tubular neighborhoods in the proof. A comment to the answer questioned why it is legitimate to assume smoothness but there was no follow up. No reference is given in that post either.

To me this seems to be an interesting question.

  • Assume the answer to the question is poistive, is it possible to provide a proof in the topological category?
  • Is it possible to point to references (or even just search terms to point directions, so anyone interested can do follow up research from there)?
  • What about dimensions $>2$?
aaa acb
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  • For why it is fair to switch to the smooth category: every homeomorphism class of a 2- (or 3-) dimensional manifold has a smooth member. https://math.stackexchange.com/questions/2388041/unique-differentiable-structure-on-homeomorphic-manifolds-in-low-dimension . There are 4- (and higher) dimensional manifolds that are not homemomorphic to a smooth manifold; e.g., https://math.stackexchange.com/questions/408221/the-easiest-non-smoothable-manifold . – Eric Towers May 21 '25 at 21:32
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    It's a consequence of the fact that every simple topological loop on a surface admits a tubular neighborhood (which is either the annulus or the Moebius band). – Moishe Kohan May 21 '25 at 22:48
  • For me to learn the proof of the fact, do you mind providing an outline or point me to reference(s) (if any)? Thank you @MoisheKohan – aaa acb May 22 '25 at 10:54
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    I agree that a smooth structure can be imposed on $V$ since $V$ is a surface; but $h({\mathbf{S}^1})$ may not be a smooth submanifold of $V$ (where $\mathbf{S}^1$ is the boundary of $\overline{D}_1$) under the smooth structure hence I failed to see the existence of tubular neighborhood (which, as far as I have read, requires the submanifold to be at least $C^1$). If I am missing something obvious please let me know. Thank you @EricTowers – aaa acb May 23 '25 at 02:07
  • How do you feel about "every 2- or 3-manifold (with boundary) is homeomorphic to a PL 2- or 3-manifold (with PL boundary)". Tubular neighborhoods in PL manifolds are not difficult. In higher dimensions, see https://mathoverflow.net/questions/283467/tubular-neighborhoods-of-embedded-manifolds – Eric Towers May 23 '25 at 17:27
  • Think passing to PL category suffers the same issue as passing to smooth category. Namely, although the ambient surface has a PL structure, $h(\mathbf{S}^1)$ may not be a PL sub-manifold. The embedding can be very wild. If I am missing something obvious please let me know. @EricTowers – aaa acb May 23 '25 at 18:57

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