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Let $M$ be a smooth manifold (without boundary) and $S \subseteq M$ be a smooth embedded submanifold with non-empty boundary $\partial S$. Does there exist a smooth, embedded extension $\tilde S$ of $S$ in the sense that $\tilde S$ is an embedded, smooth submanifold of $M$ without boundary of same dimension as $S$ and $S \subset \tilde S$?

I am quite certain that there is one, but I have not been able to prove it. I also have not found anything on google and don't have any advanced books on differential topology lying around. The proposition would be of use to prove smoothness of mappings defined on $S$, so maybe there's been done something like that before. Either way, those were the two approaches I tried:

1) Use the existence of boundary slice charts and canonically extend the slices of $S$ in each such chart. As long as the extension is "small enough" there should be no issue regarding non-smooth intersections of the different slices and mutual "bad touching" preventing embeddedness. My idea was to use the tubular neighborhood theorem to find a sufficiently "nice" neighborhood of $\partial S$ preventing those bad effects, but I gave up here.

2) Use the flowout theorem with a (smooth) vector field tangent to $S$ and outward pointing on $\partial S$ to "generate" $\tilde S$. My problem here is that the flowout theorem as I know it is too weak, i.e. it does not yield a smooth embedding.

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  • $\begingroup$ I dont think it is true. Seifert 3 manifold with non zero Eular number can give you some counter example. Baically other than a few bad point, its a trivial circle bundle over a surface with boundary, but non-zero Eular number condition will obbstruct you from the extension of such sub-surfaces. $\endgroup$ – Anubhav Mukherjee Dec 11 '17 at 22:21
  • $\begingroup$ Basically consider the circle bundle over a genus 2 surface with a single boundary. Then the resultant manifold has a bounday torus. Now attach a solid torus with it with a (1,1) attaching map. Now try to think....Can you extend the surface of genus two with a single boundary inside it anymore? $\endgroup$ – Anubhav Mukherjee Dec 11 '17 at 22:26
  • $\begingroup$ What is your definition of a smoothly embedded submanifold with boundary? $\endgroup$ – Moishe Kohan Dec 12 '17 at 0:39
  • $\begingroup$ @Moishe Cohen A smooth submanifold with boundary $(S, \varphi)$ of a manifold $M$ (with boundary) is a manifold $S$ with boundary and a smooth injective immersion $\varphi \colon S \to M$. Smoothness of $\varphi$ means that all local representatives are smooth. In particular, if $f$ is a map from an open subset of the half space of $\mathbb{R}^k$ to an open subset of the half space of $\mathbb{R}^n$, then it is smooth, if it has a smooth extension as a map from an open subset of $\mathbb{R}^k$ into $\mathbb{R}^n$. $(S, \varphi)$ is embedded, if $\varphi$ is also a topological embedding. $\endgroup$ – user510186 Dec 12 '17 at 17:15
  • $\begingroup$ If $S \subseteq M$, then $\varphi$ should be understood as the inclusion mapping. As shown in the book by Lee, an embedded submanifold with boundary $S$ admits boundary slice charts in a neighborhood of each point of $\partial S$, in which $S$ is locally given by $x^{k+1}= \dots = x^{n} = 0 , x^k \geq 0$. @AnubhavMukherjee Could you please give a reference where this is explained more carefully? I am not very familiar with Seifert 3-manifolds. Also, I wish to emphasize that $S$ should be embedded. $\endgroup$ – user510186 Dec 12 '17 at 17:27

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