Is $\mathbb{R}^4-\{0\}=S^3\times \mathbb{R}$?

Question

In D. Bleecker's book "Gauge Theory and Variational Principles", he writes on p. 75 \begin{equation} \mathbb{R}^4-\{0\} \simeq S^3\times \mathbb{R} \end{equation} without further explanation. I would like to know why/if this is the case.

I know that a stereographic projection can send $S^n-p$ to $\mathbb{R}^n$ (for $p$ some point of $S^n$) but I do not know the map that establishes the diffeomorphism above.

And if the above formular is right, how do you picture it? As a (3-)sphere which an infinitely thick shell?

Answer 1

The way to picture it is $\Bbb R^n\setminus \{0\} =\bigcup_{r>0} \Bbb S^{n-1}(0,r)$, but you need to fix up the radii since he is using $\Bbb S^{n-1}\times \Bbb R$ instead of $\Bbb S^{n-1}\times \Bbb R_{>0}$. The map $$\Bbb R^n\setminus\{0\}\ni x\mapsto \left(\frac{x}{\|x\|},\log \|x\|\right)\in \Bbb S^{n-1}\times \Bbb R $$does the magic.

Answer 2

Consider the map$$\begin{array}{ccc}\mathbb{R}^4\setminus\{0\}&\longrightarrow&S^3\times\mathbb{R}\\v&\mapsto&\left(\frac v{\|v\|},\log\|v\|\right).\end{array}$$