Regular domain of a smooth manifold:topological boundary coincides with manifold boundary.

Question

Here is proposition 5.46 in John Lee's Introduction to Smooth Manifolds:

Proposition 5.46 Suppose $M$ is a smooth manifold without boundary and $D \subseteq M$ is a regular domain. The topological interior and boundary of $D$ are equal to its manifold interior and boundary, respectively.

Here a regular domain of a smooth manifold without boundary is a properly embedded codimension-$0$ submanifold with boundary.

And here is the first part of the proof:

Suppose $p \in D$ is a arbitrary. If $p$ is in the manifold boundary of $D$, Theorem 4.15 shows that there exist a smooth boundary chart $(U,\varphi)$ for $D$ centered at $p$ and a smooth chart $(V,\varphi)$ for $M$ centered at $p$ in which $F$ has the coordinate representation $F(x^1,...x^n)=(x^1,...x^n)$, where $n$= dim$M$=dim$D$. Since $D$ has the subspace topology, $U=D \cap W$ for some open subset $W \subseteq M$, so $V_0=V \cap W$ is a neighborhood of $p$ in $M$ such that $V_0 \cap D$ consists of all the points in$V_0$ whose $x^n$ coordinate is nonnegative. Thus every neighborhood of$p$ intersects both $D$ and $M\backslash D$, so $p$ is in the topological boundary of $D$.

I use the bold font in the step I'm stuck in. I couldn't understand why $V_0$ chosen in this way has the property stated in the proof. I feel it's possible for some points in $V_0 \backslash U$ to be mapped into $\mathbb{R^n}$ with nonnegative coordinate, as long as the "shape" of $V_0 \backslash U$ is weird enough....

Could you give me some hint about this step? Thanks in advance.

Answer 1

I agree with Jeff Rubin's answer, but I think it can be simplified. We'll replace consists of all the points with consists of points. It is true that every point in $V_0\cap D$ has nonnegative $x^n$ coordinate. Since $x^n(p)=0$, every open neighborhood of $p$ must contain some point $q\in V_0$ with $x^n(q)<0.$ Then $q\not\in D$. Therefore $p$ is not in the topological interior of $D$. Since $p\in D$, it must be in the topological boundary of $D$.

Answer 2

First, I hope you know that the errata for the book has the proof starting out with "Let $F\colon D\hookrightarrow M$ denote the inclusion map.

I can't prove the statement in bold exactly as is given, but I can prove that any point in $V_0\cap D$ has nonnegative $x^n$ coordinate, and I'll show at the end that that's all we need to prove the last sentence quoted in the question.

First, since $V_0=V\cap W$, when we talk about the coordinates of a point $q\in V_0$ we are talking about $\psi(q)^n$. Part of Theorem 4.15 states that $U=F(U)\subseteq V$. Writing out the coordinate representation of $F$ in detail \begin{equation} (\psi\circ\phi^{-1})(x^1,\dots,x^n) =(\psi\circ F\circ\phi^{-1})(x^1,\dots,x^n) =(x^1,\dots,x^n) \end{equation} for all $(x^1,\dots,x^n)\in\phi(U)$, so $\psi\circ\phi^{-1}$ is the identity on $\phi(U)$. Then since $U\subseteq V$, we have that $\psi|_U=\phi$. Since $\phi(U)$ is open in $\mathbb{H}^n$, this implies $\psi(q)^n=\phi(q)^n\geq 0$ for all $q\in U$. We then note that $$V_0\cap D=V\cap W\cap D=V\cap U=U.$$ This shows that $V_0\cap D$ consists of points of $V_0$ whose $n$-th coordinate is nonnegative.

As to the last sentence, suppose $A$ is a neighborhood of $p$ in $M$. Then $p\in A\cap D\neq\varnothing$. $V_0$ is a neighborhood of $p$ in $M$, hence $V_0\cap A$ is also. $\psi\colon V\to\psi(V)\text{ open}\subseteq\mathbb{R}^n$ is a homeomorphism and $V_0\cap A$ is open in $V$, so $0=\psi(p)\in\psi(V_0\cap A)$ which is open in $\psi(V)$ and therefore also in $\mathbb{R}^n$, so $(0,\dots,0,-\epsilon/2)\in B_\epsilon(0)\subseteq\psi(V_0\cap A)$ for some $\epsilon>0$, whence $$q=\psi^{-1}(0,\dots,0,-\epsilon/2)\in V_0\cap A.$$ If $q$ were in $D$, then $q\in V_0\cap D$ so $\psi(q)^n\geq 0$, a contradiction. So $q\in A\cap(M\setminus D)$.