Is $\Bbb R^2\times\Bbb S^2$ homeomorphic to $\Bbb R^4$ with a line removed?
Solution 1:
Using $\Bbb S^2\subset \Bbb R^3$, we have the homeomorphism $$\begin{align}\Bbb R\times \Bbb S^2&\to \Bbb R^3\setminus\text{pt}\\ (r,v)&\mapsto e^rv\end{align}$$ and from that and this post readily the desired result via the obvious $$\Bbb R\times(\Bbb R^3\setminus\text{pt})\to \Bbb R^4 \setminus\text{line}$$