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?

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    $\begingroup$ Sorry for not posting correctly. I just posted my solution. Is it right? How can i make it completely precise? $\endgroup$ – user97700 Sep 29 '13 at 23:41
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    $\begingroup$ One minor detail seems wrong, but I have "texified" what you wrote so as to let you get some experience in doing edits yourself. In the next to last paragraph you wrote phi applied to the boundary of $B$ equals phi applied to boundary of $B_r(x)$. But you don't want phi on both sides of this equality, do you? $\endgroup$ – hardmath Sep 30 '13 at 6:25
  • $\begingroup$ The exponential map is a local diffeomorphism. That's all you need. $\endgroup$ – Diesirae92 Dec 13 '17 at 15:57
  • $\begingroup$ @Diesirae92 Please elaborate $\endgroup$ – Perturbative Dec 13 '17 at 16:49
  • $\begingroup$ I answered below, I hope it helps, but it is just a comment which didn't fit here $\endgroup$ – Diesirae92 Dec 13 '17 at 17:25

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.


I am answering here since the comment section does not allow so many characters, but I do not consider this as an answer, it's just a list of observations.

First of all your question is not well posed, or at least the thing you ask is false in general, i.e. consider the 2 sided cone without the vertex. This thing is even a smooth manifold, but if you remove a big enough ball with center near the vertex what you get is a bounduary which is homeomorphic to the disjoint union of 2 spheres.

Consider a hyperbolic surface of rotation https://en.wikipedia.org/wiki/Hyperboloid, once again if you choose the ball big enough, your claim is false.

Thus we reduce ourselves to wander if there is a small enough nhbd of points, on which removing a geodesic ball means getting a manifold with that bonduary (the shere $S^{n-1}$). This is given to you in smooth enoughmaifolds by the exponential map which in manifolds which are at least $C^2$ (as far as I Know), gives you a smooth local diffeomorphism between the tangent and the manifold, thus your claim follows easily by this observation. In the general case you have that your manifold is just locally homeomorphic to some open set in $R^n$, but then you can use charts to perform the same argument as above.

  • $\begingroup$ The question is about topological manifolds, not smooth manifolds $\endgroup$ – Perturbative Dec 13 '17 at 17:29
  • $\begingroup$ indeed in the last part I explain to you why it is true for even topological manifold: by definition you have local omeomorphisms between open sets in $\mathbb{R}^n$ and the manifold itself. Moreover smooth manifolds are topological manifold, therefore if something does not hold for smooth ones it does not hold even for topological ones (i.e. you must take small balls). $\endgroup$ – Diesirae92 Dec 13 '17 at 17:43

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