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 $M^*$.
- 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?
- 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 $M \setminus U$?