Suppose $M$ is a (topological) manifold of dimension $ n \geq 1$ and $B$, is a regular coordinate ball in $M$.
Show that $M\backslash B$ is an $n$-manifold with boundary and whose boundary is homeomorphic to $S^{n-1}$
Ok, so here is what I have. $M\backslash \overline{B}$ is an open space in $M$ and hence is locally euclidean. So what's left is to prove that $\partial B$, the boundary of $B$, is the boundary of $M\backslash B$.
Here is where we use the the fact that $B$ is a regular coordinate ball. We know there exists a a function $\phi: B' \to B_{r'}(x)$ such that $\phi(B)=B_{r}(x)$ and $\phi(\overline{B})=\overline{B_{r}(x)}$. With this information, we know that there is a neighborhood of the boundary of $B$, namely $B' \cap M\backslash B$, which is homeomorphic to a the closure of a a ball in $\mathbb{R}^n$. So we conclude that $M\backslash B$ is a manifold with boundary.
Moreover, the boundary of the manifold is the boundary of B, which is homeomorphic to $S^{n-1}$ because $\phi(\partial B)=\phi(\partial B_r(x))$.
Is this proof airtight? How can I make it more precise? For example exactly what would be the neighborhood of the boundary of B which works to prove that it is the boundary of the manifold?
Best Answer
No, your proof is insufficient (I will explain why in the end of the answer). You have to revise the notion of a coordinate ball in an $n$-dimensional manifold $M$ as follows:
A compact subset $B\subset M$ is called a coordinate ball if there exists an open subset $U\subset M$ containing $B$, and a homeomorphism $\phi: U\to U'\subset R^n$ (where $U'$ is necessarily open in $R^n$) such that the image $\phi(B)$ is the unit ball $B'=\{x: |x|\le 1\}$ in $R^n$.
With this definition the proof that $M-int(B)$ is a manifold with boundary (equal to the topological frontier $fr(B)$ of $B$ in $M$) becomes effortless: The claim is equivalent to the assertion that $U-int(B)$ is a manifold with boundary (where the boundary equals to $fr(B)$), which is equivalent to the claim that $U'- int(B')$ is a manifold with boundary (whose boundary equals the sphere $S'=\{x: |x|=1\}$). The latter can be proven, for instance, by applying stereographic projections (with centers at the north and the south poles of the sphere $S'$ sending $\{x: |x|>1\}$ to the upper half-space $\{x: x_n> 0\}$.
Now, an example showing that your notion of a coordinate ball is insufficient. There exist examples of wild spheres $S\subset R^3$ which separate $R^3$ in two components: The closure $B$ of one component is homeomorphic to the closed unit ball. (Think of this as your "coordinate ball" in $R^3$.) On the other hand, the closure of the other component (I will call this closure $C$) is not a manifold with boundary. The latter is because $S$ is "knotted". A precise way to state this knottedness property is in terms of the fundamental group. There exist points $x\in S$ such that for an arbitrarily small neighborhood $G$ of $x$ in $C$ the map $$ \pi_1(G - S) \to \pi_1(int(C)) $$ is nontrivial. (There are loops in $int(C)$ arbitrarily close to $x$ which cannot be contracted to a point in $int(C)$.) A manifold with boundary cannot have such pathological behavior.