Suppose $M$ is a manifold of dimension $n \geq 1$ and $B \subseteq M$ is a regular co-ordinate ball. Show that $M \setminus B$ is a $n$-manifold with boundary, whose boundary is homeomorphic to $\mathbb{S}^{n-1}$
I am trying to prove the above, and I've done the following so far. (Note that $\text{Bd}$ denotes the topological boundary and $\partial$ denotes the manifold boundary).
I've shown that $(M \setminus B) \setminus \text{Bd}(B) = M \setminus \overline{B}$ is a $n$-manifold without boundary and I've shown that $\text{Bd}(B)$ is homeomorphic to $\mathbb{S}^{n-1}$. I've claimed that $\text{Bd}(B) = \partial\left(M \setminus B\right)$ and now I'm attempting to show that.
This was my attempt to prove that claim.
Attempt:
Suppose $x \in \text{Bd}(B)$. We need to construct a neighbourhood $W$ of $x $ in $M \setminus B$ that is homeomorphic to $\mathbb{H}^n$. Again since $M$ is an $n$-manifold, there exists a open set $A$ of $x$ in $M$ that is homeomorphic to an open set $\widehat{A}$ in $\mathbb{R}^n$. Let $\zeta : A \to \widehat{A}$ denote the homeomorphism.
Choose an $\epsilon > 0$ such that $\Psi = B_{(\mathbb{R}^n, d)}(\zeta(x), \epsilon) \subseteq \widehat{A}$. Since all open balls in $\mathbb{R}^n$ are homeomorphic to the unit ball $\mathbb{B}^n$ we have $\zeta^{-1}[\Psi] \cong \Psi \cong \mathbb{B}^n$. Now since $\Psi$ is an open set in $\mathbb{R}^n$, we have that $\zeta^{-1}[\Psi]$ is an open set in $M$. Hence $ \zeta^{-1}[\Psi] \cap M \setminus B$ is open in $M \setminus B$ and $\zeta^{-1}[\Psi]$ is a neighbourhood of $x$....
Let $\Omega = \zeta^{-1}[\Psi]$. Now my reasoning was the following, since $\Omega$ is open in $M$ we'd have $\Omega \cap (M \setminus B)$ open in $M \setminus B$. Hence since $\Omega \cong \mathbb{B}^n$ we'd have $\Omega \cap (M \setminus B)$ homeomorphic to some subset (that is open in $\mathbb{H}^n$) of $\mathbb{B}^n$ that intersects $\partial \mathbb{H}^n$, so we'd have an open set in $M \setminus B$ containing $x$ homoermorphic to an open set in $\mathbb{H}^n$ that has nonempty intersection with $\partial \mathbb{H}^n$ so we could conclude that $x$ was a manifold-boundary point on $M \setminus B$ and that $\text{Bd}(B) = \partial\left(M \setminus B\right)$ since $x \in \text{Bd}(B)$ was arbitrary.
But here's the problem I'm facing, I don't actually know if I can conclude that $\Omega \cap (M \setminus B)$ is homeomorphic to some subset (that is open in $\mathbb{H}^n$) of $\mathbb{B}^n$ that intersects $\partial \mathbb{H}^n$.
So that leads me to the question I have.
Let $\Omega_1= (M \setminus \overline{B}) \cap \Omega$ be the points in $\Omega$ that are not contained in the interior or boundary of $B$.
Let $\Omega_2 = \text{Bd}(B) \cap \Omega$ be the points in $\Omega$ that are contained in the boundary $B$
Let $\Omega_3 = B \cap \Omega$ be the points that are contained in the interior of $B$ (note $\text{Int}(B) = B$ since $B$ is open in $M$).
My question is does there exist a homeomorphism, between $\Omega$ and $\mathbb{B}^n$ that does the following:
- Maps $\Omega_1$ to the upper half of $\Psi$
- Maps $\Omega_2$ to the equator of $\Psi$
- Maps $\Omega_3$ to the lower half of $\Psi$
Why am I interested in such a homeomorphism? Because if there does exists such a homeomorphism $f : \Omega \to \mathbb{B}^n$ then I can conclude that restricting $f$ to the domain $\Gamma = \Omega \cap (M \setminus B) = \Omega_1 \cup \Omega_2 \subseteq \Omega$ would be a homeomorphism between $\Gamma$ and $f[\Gamma]$ and $\Gamma$ would be open in $M \setminus B$ and $f[\Gamma]$ would be open in $\mathbb{H}^n$ (by the fact that $f[\Gamma]$ is the intersection of $\mathbb{B}^n$ with $\mathbb{H^n}$). Thus I could conclude that $x$ is a manifold-boundary point of $M \setminus B$.
It seems what you have written so far is not far from the idea of the proof. So, for the sake of elaboration, I shall write down a cleaner version of it:
Suppose $M$ is an $n$-manifold and $B$ is a coordinate ball on $M$ around $p$. By definition of a coordinate ball, this implies there is an open chart $U$ around $p$ such that the chart homeomorphism $\varphi : U \to \Bbb R^n$ sends $\varphi(p) = \mathbf{0}$ and $\varphi(B) = D(\mathbf{0}, r)$, the open disk of radius $r$ centered at the origin.
Restriction of $\varphi$ to $U \setminus B$ then induces a homeomorphism $\varphi_0 : U \setminus B \to \Bbb R^n \setminus D(\mathbf{0}, r)$. I claim that it is sufficient to check that $\Bbb R^n \setminus D(\mathbf{0}, r)$ is an $n$-manifold with boundary to prove that so is $M \setminus B$.
Proof: Assume $\Bbb R^n \setminus D(\mathbf{0}, r)$ is an $n$-manifold with manifold boundary being the topological boundary of $D(\mathbf{0}, r)$ in $\Bbb R^n$. Consider a point $q \in M$. If $q$ belongs to the topological interior of $M \setminus B \subset M$, then there is a chart neighborhood of $q$ that is homeomorphic to $\Bbb R^n$ as $M$ is a manifold. If $q$ belongs to the topological boundary of $M \setminus B \subset M$, in particular $q \in U\setminus B$ (as $U$ contains $B$), hence $\varphi_0(q)$ lies on the topological boundary of $\Bbb R^n \setminus D(\mathbf{0}, r)$. By hypothesis, there is a boundary chart $V$ around $\varphi_0(q)$, i.e., one which comes with a homeomorphism $\psi : V \to \Bbb H^n$ to the closed upper half plane. Now consider the composition $\psi \circ \varphi_0 : \varphi_0^{-1}(V) \to \Bbb H^n$. Since composition of two homeomorphisms is a homeomorphism, we conclude that $\varphi_0^{-1}(V)$ is a boundary chart at $q$. Hence, $M \setminus B$ is a manifold with boundary.
Finally, to see that $\Bbb R^n \setminus D(\mathbf{0}, r)$ is a manifold with boundary, we'd have to produce boundary charts for points $\mathbf{x} \in \text{Bd}(\Bbb R^n \setminus D(\mathbf{0}, r))$ on the topological boundary. This is visually obvious, but you can try to write down a formula for it.