One-point compactification of manifold

Question

A question from Introduction to Topological Manifolds:

4-28. Suppose $M$ is a noncompact manifold of dimension $n \ge 1$. Show that its one-point compactification is an $n$-manifold if and only if there exists a precompact open subset $U \subseteq M$ such that $M \setminus U$ is homeomorphic to $\mathbb{R}^n \setminus \mathbb{B}^n$. [Hint: you may find the inversion map $f: \mathbb{R}^n \setminus \mathbb{B}^n \to \overline{\mathbb{B}^n}$ defined by $f(x)=x/|x|^2$ useful.]

Here $\mathbb{B}^n$ is the open unit ball in $\mathbb{R}^n$. Can anyone provide a bigger hint for this question?

Denote the one-point compactification of $M$ by $M^*$.

1. If $M^*$ is an $n$-manifold, then we can choose some neighborhood $E$ around $\infty$ such that $E$ is homeomorphic to $\mathbb{B}^n$. We know that $M^* \setminus E$ is compact. Choose $U = \operatorname{Int} (M^* \setminus E)$; then $$M \setminus U = \overline{M \setminus (M^* \setminus E)} = \overline{E \setminus \{\infty\}}.$$ Where do I go from here?
2. For the converse, the set $M^* \setminus \overline{U} = \operatorname{Int} (M^* \setminus U)$ is a neighborhood of $\infty$. But how do I connect this with $M \setminus U$?

Answer 1

For 1, yes you have an $E$ around $\infty$ with $E$ homeomorphic to a ball, but you don't know anything about $\overline{E}$. In fact, $\overline{E}$ is potentially all of $M^\ast$!

On the other hand, inside of $E$ is another open set $E'$, also homeomorphic to a ball, but whose closure is homeomorphic to $\overline{\mathbb{B}^n}$. Use $U = M^\ast \setminus \overline{E'}$.

For 2, since $M\setminus U$ is homeomorphic to $\mathbb{R}^n\setminus \mathbb{B}^n$, it follows that $M\setminus\overline{U}$ is homeomorphic to $\mathbb{R}^n\setminus \overline{\mathbb{B}^n}$. Call such a homeomorphism $g$.

By using the inversion map $f$, one sees that $\mathbb{R}^n\setminus\overline{\mathbb{B}^n}$ is homeomorphic to $\mathbb{B}^n \setminus\{\vec{0}\}$.

Composing $g$ and $f$, we have a homeomorphism between $M\setminus\overline{U}$ and $\mathbb{B}^n\setminus \{\vec{0}\}$. Try to prove that we can use these to find a homeomorphism between $M^\ast \setminus \overline{U}$ and $\mathbb{B}^n$.

Answer 2

Although late, I thought I might post another solution using the notation in Lee's Introduction to Topological Manifolds. First, some useful information from the book:

Definition: A coodrinate ball in a topological $n$-manifiold is any open subset which is homeomorphic to a ball in $\mathbb{R}^n$. A coordinate ball $B$ is called a regular coordinate ball if there is an open neighborhood $B'$ of $\overline{B}$ and a homeomorphism $\varphi:B'\to B_{r'}(x)\subset\mathbb{R}^n$ that takes $B$ to $B_r(x)$ for some $r'>r>0$ and $x\in\mathbb{R}^n$.

Proposition 4.60 Every manifold has a countable basis of regular coordinate balls.

Suppose $M^*$ is an $n$-manifold. Let $B$ be a regular coordinate ball centered at $\infty$ (by the above proposition) and without loss, homeomorphic to the unit $n$-ball $\mathbb{B}^n$. Since $B$ is contained in some larger coordinate ball $B'$, then $\overline{B}$ is homeomorphic to $\overline{\mathbb{B}}^n$ via a homeomorphism $B'\to B_{r'}(0)$ for some $r'>1$.

Set $U=M^*\setminus \overline{B}$. In $M^*$, we have $$\overline{U}=\overline{M^*\setminus\overline{B}}=M^*\setminus\left(\overline{B}\right)^{\circ}=M^*\setminus B$$ which is compact (by openness of $B$ in $M^*$). Since $M$ is an open subset of $M^*$ (in the subspace topology) containing $\overline{U}$, then we do indeed have that $\overline{U}$ is compact in $M$, so that $U$ is precompact in $M$.$^{1}$ Using the inversion map, we have that $$\begin{align*} M\setminus U &=M\setminus\left(M^*\setminus\overline{B}\right)\\ &=\left(M\cap\overline{B}\right)\cup\left(M\setminus M^*\right)\\ &=\overline{B}\setminus\{\infty\}\\ &\cong \overline{\mathbb{B}^n}\setminus\{0\}\\ &\cong\mathbb{R}^n\setminus\mathbb{B}^n \end{align*}$$

Now suppose there is a precompact open subset $U$ of $M$ such that $M\setminus U$ is homeomorphic to $\mathbb{R}^n\setminus\mathbb{B}^n$. As $\left(M\setminus U\right)^{\circ}=M\setminus\overline{U}$ (an identity which follows from $Y^{\circ}=X\setminus\overline{X\setminus Y}$ for any subset $Y$ of a topological space $X$) then using the inversion map, we see that $M\setminus\overline{U}\cong\mathbb{R}^n\setminus\overline{\mathbb{B}^n}\cong\mathbb{B}^n\setminus\{0\}$. Set $V=\left(M\setminus \overline{U}\right)\cup\{\infty\}$ so that $V$ is an open neighborhood of $\infty$ in $M^*$ (we have that $M^*\setminus V=\overline{U}$ is compact). Finally, observe: $$V\cong\left(V\setminus\{\infty\}\right)^*=\left(M\setminus\overline{U}\right)^*\cong\left(\mathbb{R}^n\setminus\overline{\mathbb{B}^n}\right)^*\cong\left(\mathbb{B}^n\setminus\{0\}\right)^*\cong\mathbb{B}^n$$ which demonstrates that $M^*$ is locally Euclidean at $\infty$.$^{2}$ It is easily shown that $M^*$ is Hausdorff, second countable, and locally Euclidean away from $\infty$.

$1$. Follows from exercise 4-23.

$2$. Follows from exercise 4-26.