# Proof from John Lee on the equivalence of topological boundary and manifold boundary for regular domains

I am having difficulty following the conclusion of the proof below from John Lee's Introduction to Smooth Manifolds. Here a regular domain in $$M$$ is a properly embedded codimension-$$0$$ submanifolds with boundary.

Here $$F$$ is the inclusion map from $$D \hookrightarrow M$$, which is a smooth embedding.

I understand the proof before the final sentence. But how do we conclude from here that every neighborhood of $$p$$ intersects both $$D$$ and $$M\backslash D$$?

• Every ball around the origin contains points with $x^m>0$ and points as well with $x^m<0$. Commented Mar 2, 2020 at 22:08
• @Berci Where is the fact that $V_0 \cap D$ consists of all the points in $V_0$ whose $x^n$ coordinate is nonnegative used? And I can't see why every ball containing points with $x^m>0$ and $x^m<0$ indicate that every ball intersects with $M \backslash D$. Commented Mar 2, 2020 at 22:15
• What is a 'boundary chart' and how does $U$ relate to $V$? Commented Mar 2, 2020 at 23:34
• @Berci I am confused with the construction here. We need an arbitrary neighborhood of $p$ to intersect both $D$ and $M \backslash D$, but instead we have shown that we can find a neighborhood $V_0$ of p in $M$ whose intersection with $D$ is all the points in $V_0$ whose $n$th coordinate is $0$. This suggests some relationship with the boundary chart, perhaps that $V_0$ restricted to $D$ becomes a boundary chart for $D$? But how does this answer the final sentence? Commented Mar 2, 2020 at 23:40
• Note that the coordinate maps together (that is, $\psi$) form a diffeomorphism $M\to\Bbb R^n$, so every open neighborhood of $p$ corresponds to an open neighborhood of $0$, and that contains a ball around $0$. Commented Mar 3, 2020 at 0:33

I found this proof confusing, too. Here is my understanding: we have that $$D$$ is an embedded submanifold of $$M.$$ Take any $$p\in \partial D$$. Then, there is a chart $$\textit{in M},\$$ say, $$(V,(x^1,\cdots,x^n))$$ about $$p$$ such that $$(D\cap V,(x^1,\cdots,x^k))$$ is a boundary (slice) chart for $$D$$ about $$p$$. Thus, $$q\in D\cap V\Rightarrow x^k(q)\ge 0$$ But $$\text{dim}\ D=\text{dim}\ M\Rightarrow k=n$$, and so $$x^n\ge 0.$$ Now, $$M$$ is a manifold without boundary, which means that there must be a point $$q\in V$$ such that $$x^n(q)<0$$, (because $$(V,(x^1,\cdots,x^n))$$ is a chart about $$p$$ in $$M$$), which in turn implies that $$q\notin D$$ (because $$D\cap V$$ has all $$x^n\ge 0$$). Therefore, $$V$$ contains points in $$D$$ and in $$M\setminus D$$.
• I think it should be $x^k \ge 0$ and $x^n \ge 0$ since at $p$ we should have $x^n=0$? Commented Mar 3, 2020 at 2:41