John Lee's ISM Problem 5-23 (smooth structure of regular level sets for manifolds with boundary)

Question

The problem statement:

Suppose $M$ is a smooth manifold with boundary, $N$ is a smooth manifold, and $F: M \to N$ is a smooth map. Let $S = F^{-1}(c)$, where $c \in N$ is a regular value for both $F$ and $F\big\vert_{\partial M}$. Prove that $S$ is a smooth submanifold with boundary in $M$, with $\partial S = S \cap \partial M$.

While searching for hints, I found this question whose answer was very helpful. However I am still stuck at the last step of the proof. The answer shows that every $p \in S$ is contained in the domain of a smooth chart $(U, \varphi)$ for $M$ such that $U \cap S$ is diffeomorphic to a smooth manifold with boundary. But is this sufficient to show that $S$ itself is a smooth manifold with boundary? In particular, what exactly is the smooth structure on $S$? I can come up with an atlas for $S$ by taking the union of the smooth structures for each $U \cap S$, but it's not obvious to me that the smooth charts coming from different $U \cap S$, $U' \cap S$ will be smoothly compatible.

In fact, I don't think this sort of argument can work in general, because there are topological manifolds that do not admit any smooth structure, and yet we can always cover them with coordinate balls, which can obviously be given smooth structures. So there must be something else in the construction of the regular level set that allows us to show it has a global smooth structure.

My other question is, I suspect the set $S$ satisfying the assumptions in the problem statement is also a properly embedded submanifold with boundary of $M$. Is that true? In my attempts to prove the result, I don't see a fundamental obstruction to $S$ having the subspace topology, which would make it an embedded submanifold with boundary.

Answer 1

Let $M$ be a smooth manifold (with or without boundary), and let $S$ be a subset of $M$ satisfying the following: for each $p\in S$, there exists a neighborhood $U$ of $p$ in $M$ such that $U\cap S$ is an embedded submanifold of $U$. Then $S$ can be endowed with a topology and smooth structure that makes it an embedded submanifold (with or without boundary) of $M$.

If $\partial M = \varnothing$, then this follows from the local slice condition (Theorem $5.8$ or Theorem $5.51$ in Lee's Introduction to Smooth Manifolds), so we may assume that $M$ has a non-empty boundary. Let $p\in S$ and let $S\cap U$ be embedded in $U$ for some neighborhood $U$ of $p$ in $M$. Embed $M$ in its double $D(M)$ and let $U = \widetilde{U}\cap M$ for some open set $\widetilde{U}$ in $D(M)$. Then $S\cap U = S\cap\widetilde{U}$ is embedded in $\widetilde{U}$. Since $\partial D(M) = \varnothing$, this implies that $S$ is embedded in $D(M)$ and since $S\subseteq M$, we have that $S$ is embedded in $M$.

Edit: We can, in fact, directly show that if $U_{j}\cap S_{j}$ is embedded in $U_{j}$ for $j = 1, 2$, then the charts of $U_{1}\cap S_{1}$ and $U_{2}\cap S_{2}$ are smoothly compatible. Let $(V_{j}, \psi_{j})$ be charts in $U_{j}\cap S_{j}$ for $j = 1, 2$. The map $\psi_{1}^{-1}: \psi_{1}(V_{1}\cap V_{2})\rightarrow M$ is smooth. Since $V_{2}$ is embedded in $M$ and $\psi_{1}^{-1}(V_{1}\cap V_{2})\subseteq V_{2}$, the map $\psi_{1}^{-1}: \psi_{1}(V_{1}\cap V_{2})\rightarrow V_{2}$ is smooth. (Note that if $\partial M\neq\varnothing$, then this follows from Theorem $5.53$(b) in Lee's book, so we are implicitly using the fact that $M$ can be embedded in its double.) Therefore, $\psi_{2}\circ\psi_{1}^{-1} : \psi_{1}(V_{1}\cap V_{2})\rightarrow\psi_{2}(V_{1}\cap V_{2})$ is smooth.