Proving John Lee's ISM proposition 5.47. For a smooth real function, each regular sublevel set is a regular domain. I have a hard time proving the following proposition from John Lee's Introduction to Smooth Manifolds. I have found this question :If $b$ is a regular value of $f$, $f^{-1}(-\infty,b]$ is a regular domain? but I cannot understand the answer.

I think I need to show that $f^{-1}(-\infty, b]$ satiesfies the local $m$-slice condition for submanifolds with boundary, where $m= \dim M$. 
Since $f^{-1}(-\infty,b)$, as an open subset of $M$, is an embedded submanifold, it must satisfy the local $m$-slice condition. The problem is $f^{-1}(b)$, which I predict satisfies the local half-slice condition. My attempt so far has been : since for all $p \in f^{-1}(b)$, $df_p$ is surjective, by Theorem 4.1 of the text, $p$ has a neighborhood $U$ such that $f|U$ is a smooth submersion. Then by the rank theorem, for $p$, there exist smooth charts $(W, \phi)$ for $U$ centered at $p$ and $(V, \psi)$ for $\mathbb{R}$ centered at $f(p)=b$ such that $f(W) \subset V$, in which $f$ has a coordinate representation of the form $f(x^1, \dots, x^m) = x^m$. How can I use this to show that $p$ is contained in the domain of a smooth chart $(A, \varphi =(x^i))$ such that $f^{-1}(b) \cap A$ is a $m$-dimensional half-slice, i.e. $\{(x^1, \dots, x^m) \in \varphi(A): x^m \ge 0\}$?

 A: Maybe this is not relevant for you anymore but i'll post some of my work here.
Denote $R=f^{-1}((\infty,b]) \subseteq M$. We need to show that $R$ is a regular domain, which is by definition $R\subseteq M$ is a smooth manifold with boundary with the inclusion map $i : R \hookrightarrow M$ is a smooth embedding and also $i : R \hookrightarrow M$ is a proper map.
It is easy to show that if $i : R \hookrightarrow M$ is a topological embedding, then the topology of $R$ is the subspace topology. So let's equip $R$ with its subspace topology. Next we will find smooth (boundary) charts for $R=f^{-1}(-\infty,b) \cup f^{-1}(\{b\})$.  Let $B = f^{-1}(\{b\})$.  

$\textbf{Interior Charts for } f^{-1}(-\infty,b)  $  : This is easy since $f^{-1}(-\infty,b) = R\setminus B$ is open in $M$. So (as you may have notice), smooth charts for any $p \in R \setminus B$ comes from those charts in $M$ restricted to $R \setminus B$.
$\textbf{Boundary Charts for }f^{-1}(\{b\}) :$ Suppose $p \in B = f^{-1}(b)$. Since $p$ is a regular point then we can find neighbourhood $U$ of $p$ so that $f|_{U}$ is a local height function of $R$, that is $(U, \varphi)$ is a smooth chart of $M$ such that the representation of $f$ is $$\widehat{f}(x^1,\dots,x^n) = x^n .$$ Note that any $x \in \varphi(B \cap U)$, $\widehat{f}(x) = \widehat{f}(x^1,\dots,x^n)=x^n=b$. that is points of $B \cap U$ have the last coordinate $x^n=b$. Since we want these kind of points become boundary points of $f^{-1}(-\infty,b]$, we translate them using diffeomorphism $\psi : \widehat{U} \to \psi(\widehat{U})$, where $\varphi(U)=\widehat{U}$, defined as 
      $$\color{green}{
 \psi(x^1,\dots,x^n) = (x^1,\dots,x^{n-1},b-x^n).}
$$
  So we have new chart $(U,\phi)$ as $\phi :=\psi \circ \varphi : U \to \phi(U)$. Certainly $\phi(p) = (x_p^1,\dots,x_p^{n-1},0) \in \partial \mathbb{H}^n$ any other points in $f^{-1}(\infty,b) \cap U$ have positive last coordinates. This is our desired boundary chart for $p$.

So $R$ is a smooth $n$-manifold with boundary, and it is not hard to see that $i : R \hookrightarrow M$ is proper smooth embedding.
A: What you want to show is that the boundary of $f^{-1}(-\infty,b]$ is smooth. The boundary is given by $f^{-1}(b)$ 
(Can you argue this? Hint: Can you see why $f^{-1}(-\infty,b)$ is the interior? The boundary is the closure minus the interior. What is the closure?)
Finally, $f^{-1}(b)$ is smooth by the regular level set theorem (level sets of regular values are smooth).
