# Generating examples of smooth manifold with boundary

I'm having difficulties proving lemma 3 of Milnor, Topology from the differentiable point of view. The lemma intends to give us a way to generate examples of mainfold with boundary. The lemma states the following:

Let $$M$$ be a smooth 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 a boundary equal to $$g^{-1}(0)$$.

What I have so far:

I supposed $$g$$ is a smooth function because of the existence of regular value for $$g$$ and the general context of the book.

Because $$g$$ is smooth, $$g^{-1}((0,+\infty))$$ is open in $$M$$. Therefore, $$g^{-1}((0,+\infty))$$ is a smooth manifold of dimension $$m$$, as is $$M$$. By a preceding lemma, we know $$g^{-1}(0)$$ is a smooth manifold of dimension $$m-1$$. In that case, for every point $$x \in g^{-1}(0)$$ there exists an open neighborhood of $$x$$ diffeomorphic to an open set in $$\mathbb{R}^{m-1}$$.

To conclude:

To conclude the proof that $$g^{-1}([0,+\infty))$$ is a smooth manifold, I need to show that we can map with a diffeomorphism every small enought neighborhood in $$g^{-1}([0,+\infty))$$ to $$V \cap H^{m}$$, with the points of the boundary $$g^{-1}(0)$$ mapped to $$\partial H^m$$. Here, $$H^m$$ is the half-space $$\{(x_1,...,x_m) \in \mathbb{R}^m\vert x_m \geq 0 \}$$ and $$V$$ is an open set of $$\mathbb{R}^m$$.

I do not know how to proceed from here. Any help is greatly appreciated.

## 1 Answer

There are several ways of showing this; here's a sketch of one path:

Fix $$p\in g^{-1}(\{0\})$$. Since you have already showed that $$g^{-1}(\{0\})$$ is an embedded codimension $$1$$ submanifold, you can choose local coordinates $$(x^1,\cdots,x^{n-1})$$ with domain $$U$$, where $$p\in U\subseteq g^{-1}(\{0\})$$. Shrinking $$U$$ as necessary, we can arbitrarily extend the $$x^i$$ to smooth functions $$M\to\mathbb{R}$$. One can then use inverse function theorem to show that $$(x^1,\cdots,x^{n-1},g)$$ forms a coordinate chart on $$M$$ when restricted to a sufficiently small neighborhood of $$p$$, and restricting this map to $$g^{-1}([0,\infty))$$will give the desired boundary chart.

• Thank you for your answer, I have a few questions. i) I am not familiar with the $(x^{1},...,x^{n-1})$ notation. Are the exponent indexes of the functions $x$? ii)How do you know you can appply the inverse function theorem on $(x^1,...,x^{n-1},g)$?
– Momo
May 20 at 17:36
• @Momo i) Yes , they're just indices.; each is a local smooth function $U\to\mathbb{R}$. ii) The point is that, taken together, the functions $(x^1,\cdots,x^{n-1},g)$ form a map $M\to\mathbb{R}^n$, and the differential of this map is invertible at $p$. This can be shown using the fact that $x^i$ are coordinate functions for $g^{-1}(\{0\})$ at $p$, and that $p$ is a regular point of $g$. May 20 at 18:05