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 $\mathbb{H}^{m}$.

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    $\begingroup$ I think if you dig back into the proof that the fiber above 0 is a manifold - in particular the local normal form kind of theorem that says that up to diffeos f is locally a linear projection above 0 - then you will have your desired diffeomorphism. $\endgroup$ – Lorenzo Najt Feb 25 '18 at 3:24
  • $\begingroup$ @AreaMan Could you explain more please?I know how to construct a diffeomorphism from $g^{-1}(0)$ to an open set in $\mathbb{R}^{m-1}$, which is described in the Milnor's proof of the Lemma 1. However, I don't know how to construct such map for $M'$, since some points of $M'$ may not be regular point. $\endgroup$ – user124697 Feb 25 '18 at 3:29
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    $\begingroup$ You don't need to study every point in $M'$ -- this is because $M' \setminus g^{-1}(0)$ is an open set, so it's easy to show it is a manifold. You only need to produce charts around the boundary of $M'$... $\endgroup$ – Lorenzo Najt Feb 25 '18 at 4:12
  • $\begingroup$ @AreaMan Yes you right. I forgot to use the property of smoothness of $g$. Thank you! $\endgroup$ – user124697 Feb 25 '18 at 4:24
  • $\begingroup$ How can we assume that $g$ is smooth? $\endgroup$ – mathemather Jul 14 '18 at 8:02

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