Confusion about definition of manifold with boundary

Question

Consider this definition:

A space $M$ is a manifold with boundary if each point $x\in M$ has a neighborhood $U_x$ that is homeomorphic to $\mathbb R^n$ or to $\mathbb R^n_+=\{(x_1,\cdots,x_n)\;|\; x_n \ge 0\}$; the points which have neighborhoods homeomorphic to $\mathbb R^n_+$ form the boundary $\partial M$ of $M$.

I want to understand how precise this definition is. Indeed, a point that has an open neighborhood homeomorphic to $\mathbb R^n$ can also have another open neighborhood that is homeomorphic to $\mathbb R^n_+$ and in that case is this point on the boundary $\partial M$ or in $M\setminus \partial M$ ? How to choose? Is my definition incomplete so that I have to say: "In a manifold with boundary, a point in $M$ has an open neighborhood homeomorphic to $\mathbb R^n$ but if it doesn't have such neighborhood then it has to have an open neighborhood homeomorphic to $\mathbb R^n_+$"

Thank you for your help!

Answer 1

This is about existence of atleast one neighborhood of a point $p\in M$ that is homeomorphic to $\mathbb{R}^n$. Obviously it might have some other neighborhood which intersect with boundary and will be homeomorphic to $\mathbb{R}_{+}^{n}$ but we don't have to worry about those. If a point $p \in M$ has no neighborhood homeomorphic to $\mathbb{R}^n$, then $p \in \partial M$.

Answer 2

If $M$ is a manifold with boundary, $U$ is an open set in $M$ and $\varphi\colon U \to \Bbb \varphi[U]\subseteq \Bbb R^n_+$ is a chart, you can prove that if $p \in U$ is such that $\varphi(p) = (x^1(p),\ldots,x^{n-1}(p),0)$, then if $\widetilde{\varphi}\colon \widetilde{U} \to \widetilde{\varphi}[\widetilde{U}]\subseteq \Bbb R^n_+$ is another chart, then $\widetilde{\varphi}(p)$ has also the form $(\widetilde{x}^1(p),\ldots,\widetilde{x}^{n-1}(p),0)$.

See Lemma 24.2 in page 205 of Munkres' Analysis on Manifolds.

Answer 3

As mentioned in answers and comments, the definition given for a boundary point is not the correct one. A possible one should read:

A point $p \in M$ is a boundary point if it is such that $\pi_n(\phi(p))=\phi_n(p)=0$ for some chart $\phi:U \to \mathbb{R}^n_{+}$.

Then, I think, you would be willing to prove that if this is true for one chart $\phi: U \to \mathbb{R}^n_+$ as per the definition above, then it is true for any such chart. As one other answer suggests and references, this is not a big deal in the smooth setting.

This is true in the topological setting too, but due to a more involved reason: namely, the invariance of domain theorem. I don't know a reference which manages to avoid this.

Now, moving forward: Indeed, let $\psi: V \to \mathbb{R}^n_+$ be another chart. What we must prove then is that $\psi_n(p)=0$.

Instead of doing that, let's prove that if $\phi_n(p)>0$, then $\psi_n(p)>0$. It is easy to see (exchange places of $\phi$ and $\psi$) that this will prove that $\phi_n(p)>0$ if and only if $\psi_n(p)>0$, thus concluding that $\phi_n(p)=0$ if and only if $\psi_n(p)=0$.

Thus, suppose $\phi_n(p)>0$. We have that $\psi \circ \phi^{-1}$ is a continuous injection from $U \cap\mathbb{R}^n_+$ to $\mathbb{R}^n_+$, where $U$ is an open subset of $\mathbb{R}^n$. Restricting, $\psi \circ \phi^{-1}|_{U \cap \mathbb{R}^n_{>0}}: U \cap \mathbb{R}^n_{>0} \to \mathbb{R}^n$ is then injective and continuous. By invariance of domain, we know that such an image is open in $\mathbb{R}^n$. But this image is inside $\mathbb{R}^n_+$, therefore it can't intersect the set $\{x \mid x_n=0\}$. Since $\psi(\phi^{-1}(\phi(p)))=\psi(p)$ is in such image, it follows that it can't be such that $\psi_n(p)=0$.

As a sidenote which may also be of interest, a similar use of invariance of domain also yields that if $p$ is a boundary point, then there is no chart $\phi: U \to \mathbb{R}^n$ around $p$ (which is a homeomorphism with $\mathbb{R}^n$).