Hatcher, p.35, $\mathbb{R^{n}} - \{x\}$ is homeomorphic to $S^{n-1} \times \mathbb{R}$

Question

Could some help me understand how $\mathbb{R^{n}} - \{x\}$ is homeomorphic to $S^{n-1} \times \mathbb{R}$.

Also, if the one point set is replaced by any finite set, how does the argument work.

This is in Hatcher's book, example 1.15, p.35, where he states this as a fact. I could not find an explanation either before or in the example.

Thank you.

Answer 1

It is helpful to break the argument down into two steps. First, show that $\mathbb{R}$ is homeomorphic to $\mathbb{R}_{>0}$, the latter denoting the positive real numbers.

Then show that $\mathbb{R}^n \setminus \{x\}$ is homeomorphic to $S^{n-1} \times \mathbb{R}_{> 0}$. For notational convenience, I will assume that $x = (0,0,\dots,0)$, as it is straightforward to modify the argument for any $x$ (or simply show that $\mathbb{R}^n \setminus \{x\}$ is homeomorphic to $\mathbb{R}^n \setminus \{(0,0,\dots,0)\}$).

The main idea is to think of $S^{n-1}$ as the unit sphere in $\mathbb{R}^n$ and the $\mathbb{R}_{>0}$ as a scaling factor. So I claim that the map $$ f : S^{n-1} \times \mathbb{R}_{> 0} \rightarrow \mathbb{R}^n \setminus \{(0,0,\dots,0)\} $$ given by $f(\vec{x}, c) = c \vec{x}$ is a continuous map, with a continuous inverse. The details are left to you.

Answer 2

Even without reading his book, you should probably know how to do this intuitively. Assume without loss of generality, that $x=0$, and then $\mathbb R^n-\{0\}$ admits a foliation into concentric spheres. The radius is of course non-essential in differential geometry, you can always find a way to map homeomorphically the set to $S^{n-1}\times\mathbb R^+$. But again $\mathbb R^+$ is homeomorphic to $\mathbb R$, so you have your result.

For finite set, you should not expect to have the same result. But that part I returned to my teacher ;P, so I can't say more.

Answer 3

First of all, $y \mapsto y - x$ is a homeomorphism from $\mathbb{R}^n\setminus\{x\}$ to $\mathbb{R}^n\setminus\{0\}$.

Note that $\mathbb{R}^n\setminus\{0\}$ is homeomorphic to $S^{n-1}\times(0, \infty)$ via the map $y \mapsto \left(\frac{y}{\|y\|}, \|y\|\right)$.

As $(0, \infty)$ is homeomorphic to $\mathbb{R}$ via the homeomorphism $z \mapsto \ln z$, we see that $\mathbb{R}^n\setminus\{x\}$ is homeomorphic to $S^{n-1}\times\mathbb{R}$ with an explicit homeomorphism given by $$f(y) = \left(\frac{y-x}{\|y - x\|}, \ln\|y - x\|\right).$$