Smoothness of the boundary is the only obstruction for being a submanifold with boundary?

Question

Let $M$ be a smooth manifold, and let $S$ be an open smooth submanifold of $N$.

Assume the topological boundary of $S$, $\partial S :=\bar S \setminus S$ is a smooth submanifold of codimension 1 in $N$.

Is it true that $\bar S$ is a submanifold with boundary of $N$?

Somehow, I am not even sure whether $\bar S$ must be a topological manifold with boundary...

I am trying to see if the smoothness of the boundary is the only obstruction for being a submanifold with boundary. (If we do not assume this, we can take $S$ to be the interior of a square).

Answer 1

I'm assuming that your first sentence was meant to say "Let $N$ be a smooth manifold, $\dots$ ."

Yes, it is always the case that $\overline S$ is a smooth submanifold with boundary in $N$. However, the boundary of $\overline S$ need not be equal to the boundary of $S$. See this post for an example in which it's not.

To prove that $\overline S$ is a smooth submanifold with boundary, you just need to show that each point of $\overline S$ has a coordinate neighborhood in $N$ whose image is either an open subset of $\mathbb R^n$ or an open subset of the half-space $\mathbb R^{n-1} \times [0,\infty)$. Certainly each point of $S$ has such a neighborhood. For a point of $\partial S$, you can choose a slice chart $(U,\phi)$ for $\partial S$, such that $\phi(U)$ is a coordinate ball centered at the origin, and $\phi(\partial S\cap U)$ is the subset of $\phi(U)$ where $x^n=0$. Now you have to show that one of the following three possibilities holds:

1. $\phi(\overline S\cap U) = \phi(U) \cap \{x^n\ge 0\}$,
2. $\phi(\overline S\cap U) = \phi(U) \cap \{x^n\le 0\}$, or
3. $\phi(\overline S\cap U) = \phi(U) $.

I'll leave it to you to work out the details.