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.


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 $M \setminus B$.

  • $\begingroup$ None of these are true. What you can say is that if $B$ is a coordinate ball then you can choose $A$ such that these statements hold. $\endgroup$ Dec 12 '17 at 15:15
  • $\begingroup$ @MoisheCohen I've edited the OP accordingly, how can I show that these statements hold? (This is the main reason I asked this question) $\endgroup$ Dec 12 '17 at 16:23
  • $\begingroup$ You've chosen $B$ to be a coordinate ball which is fine. But what you want is still not true for all choices of $A$. $\endgroup$
    – Lee Mosher
    Dec 12 '17 at 16:47
  • $\begingroup$ @LeeMosher What choices of $A$ would make it true? $\endgroup$ Dec 12 '17 at 17:50
  • $\begingroup$ @LeeMosher I've edited the OP to make my question clearer, please take a look at it again if possible. $\endgroup$ Dec 12 '17 at 19:05

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.

  • $\begingroup$ I've been worked out this problems too. The only thing left is that to show that $\mathbb{R}^n \smallsetminus D(0,r)$ is an $n$-manifold with boundary, with $\partial D(0,r)$ as the manifold boundary. It is enough if we did this for unit ball $\mathbb{R}^n \smallsetminus \mathbb{B}^n$. A nice homeomorphism is a composing stereographic projection (from the south pole) with a projection other than the last coordinate. $\endgroup$
    – Brown Bear
    Apr 26 '18 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.