Below is problem 4.28 from Lee's Introduction to Topological manifolds:
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.]
I can establish the "only if" direction by considering a regular coordinate ball surrounding $\infty$. However, I have difficulty tackling the converse direction. After some searching on the site, I found the following answer:
$\color{red}{\text{Since } M\setminus U \text{ is homeomorphic to } \mathbb{R}^n\setminus \mathbb{B}^n, \text{ it follows that } M\setminus\overline{U} \text{ is homeomorphic to } \mathbb{R}^n\setminus \overline{\mathbb{B}^n}.}$ 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 $\mathbb{B}^n$.
It is exactly the red sentence that bothers me for a long time. Assuming the theorem on the invariance of boundary, since homeomorphism preserves interior points, $\text{int } (M\setminus U)\cong \text{int } (\Bbb R^n\setminus \Bbb B^n)= \Bbb R^n\setminus \overline{\Bbb B^n}$ (Here int should be interpreted as the manifold interior). If I can verify $\text{int } (M\setminus U) =M\setminus\overline{U}$, then I am done. But does it hold for arbitrary $U$ satisfying the hypothesis or do I have to choose $U$ cleverly? Of course, alternative proofs are welcome as well. Thanks in advance.
Best Answer
Lemma 1: Suppose $X$ is a manifold with boundary, $M$ is a manifold, and that $i:X\rightarrow M$ is a codimension $0$ embedding. If $Int(X)$ is the manifold interior of $X$, and $int$ refers to the topological interior then $i(Int(X)) = int(i(X))$.
Proof: Suppose $x\in Int(X)$. Then there is an open neighborhood $V\subseteq X$ of $x$ for which $U$ is disjoint from the manifold boundary of $X$. Then $i(V)$ is open in $M$ by invariance of domain. Since $i(x)\in i(V)\subseteq i(X)$, this proves that $i(Int(X))\subseteq int(i(X))$.
Conversely, assume $x\in int(i(X))$. Then there is an open set $V\subseteq i(X)$ containing $x$. Because $M$ is a manifold, the charts form a basis, so there is a neighborhood $x\in V'\subseteq V$ for which $V'$ is homeomorphic to a ball. Since $i$ is an embedding and $V'\subseteq V\subseteq i(X)$ it follows that $i^{-1}(V')$ is a ball around $i^{-1}(x)$. In particular, $i^{-1}(x)\in Int(X)$. That is, $x\in i(Int(X))$. $\square$
Using lemma 1, when you write "Int" we may interpret that as either the manifold interior, or the topological interior. Now, the claim follows from
Lemma 2: For any topological space $M$ and an subset $U$, $\operatorname{int}(M\setminus U) = M\setminus \overline{U}$. (There is no requirement that $M$ be a topological manifold, or that $U$ be open, or anything like that.)
Proof: If $x\in \operatorname{int}(M\setminus U)$, then there is an open set $V$ with the property that $x\in V\subseteq M\setminus U$. In particular, $x\notin\overline{U}$ because $V\cap U=\emptyset$ and $x\in V$. Thus, $x\in M\setminus \overline{U}$. This proves that $\operatorname{int}(M\setminus U)\subseteq M\setminus \overline{U}$.
On the other hand, suppose $x\in M\setminus \overline{U}$. Since $x\notin \overline{U}$, there is an open set $V$ for which $x\in V$ and $V\cap U = \emptyset$. But this means $V\subseteq M\setminus U$, so $x\in \operatorname{int}(M\setminus U)$. This shows that $M\setminus\overline{U}\subseteq \operatorname{int}(M\setminus U)$. $\square$