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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 $M\backslash D$?

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  • $\begingroup$ Every ball around the origin contains points with $x^m>0$ and points as well with $x^m<0$. $\endgroup$
    – Berci
    Commented Mar 2, 2020 at 22:08
  • $\begingroup$ @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 $M \backslash D$. $\endgroup$ Commented Mar 2, 2020 at 22:15
  • $\begingroup$ What is a 'boundary chart' and how does $U$ relate to $V$? $\endgroup$
    – Berci
    Commented Mar 2, 2020 at 23:34
  • $\begingroup$ @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? $\endgroup$ Commented Mar 2, 2020 at 23:40
  • $\begingroup$ 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 $0$. $\endgroup$
    – Berci
    Commented Mar 3, 2020 at 0:33

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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 $M\setminus D$.

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  • $\begingroup$ I think it should be $x^k \ge 0$ and $x^n \ge 0$ since at $p$ we should have $x^n=0$? $\endgroup$ Commented Mar 3, 2020 at 2:41
  • $\begingroup$ @nomadicmathematician yes, indeed. Thanks! $\endgroup$ Commented Mar 3, 2020 at 2:59

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