Lemma 3 in Milnor's "Topology from the differentiable Viewpoint" (p.12) is stated as below; Let $M$ be a manifold without boundary and let $g:M \to \mathbb{R}$ have 0 as regular value.
The set of $x$ in $M$ with $g(x)\geq 0$ is a smooth manifold, with boundary equal to $g^{-1}(0)$.
To discuss efficiently, let $M' = \{x \in M: g(x) \geq 0 \}$, and $\dim M = m$.
In the book, the proof is omitted. I understand that the Lemma 1 of the book implies that $g^{-1}(0)$ is $(m-1)$-manifold. However, I cannot find a map $f: M' \to \mathbb{H}^{m}$, which gives open set of $M'$ has the diffeomorphism onto open set of $\mathbb{H}^{m}$. Could you help me to understand this lemma?
P.S. I also think that $g|_{M'}:M' \to \mathbb{H}$ already gives a $1$-manifold with boundary on the set $M'$, by the definition of Milnor's book. However, if we regard this as a proof of the Lemma 3, then the dimension of $m$ is not meaningful; for example, if we think $D^{m}$, a unit disk, and a map $g:D^{m} \to \mathbb{R}$ by $1- \sum x_{i}^{2}$, then the lemma gives $M'$ a smooth manifold with boundary structure having dimension $m$. To derive such dimension, I think the lemma should be proved by finding a map from $M'$ to $\mathbb{H}^{m}$.