$S^4\setminus S^2$ is homeomorphic to $\Bbb{R}^4 \setminus\Bbb{R}^2$ I need to show that $S^4\setminus S^2$ is homeomorphic to $\Bbb{R}^4 \setminus\Bbb{R}^2$, with $\Bbb{R}^2 = \{(x,y,0,0):x,y\in\Bbb{R}\}\subseteq\Bbb{R}^4$ and $S^2 = \{(x,y,z,0):x^2+y^2+z^2=1\}$.
Now, the model solution states that the homeomorphism holds by stereographic projection from a point in $S^2$.
To be a model solution, it seems bold to me. I would have not felt confident enough to just state that in an exam. Even though we are talking about 4 dimensions, is there any way to have a clearer intuition about this fact?
 A: You know that the stereographic projection is a homeomorphism between a sphere (of any dimension) minus one point and the Euclidean space of same dimension. Now, if you do the stereographic projection defined in $\Bbb S^4$ minus the point $(1,0,0,0)$ in $\Bbb S^2$, this will give a homeomorphism between that and $\Bbb R^4$. If you restrict that to $\Bbb S^4\setminus \Bbb S^2$, the points you will remove from the image of the projection seem to be precisely the $\Bbb R^2$ corresponding to $\Bbb S^2$ under the "same" projection (realized in the hyperplane $\Bbb R^3\times \{0\}$). Thus it restricts to $\Bbb S^4\setminus \Bbb S^2\cong \Bbb R^4\setminus \Bbb R^2$.

For completeness, let $N \in \Bbb S^n$ be a point. For any $p\in \Bbb S^n$, consider the ray starting at $N$ and passing through $p$. Such a line will cross the hyperplane $N^\perp$ in exactly one point ${\rm St}_N(p)$. This defines a homeomorphism between $\Bbb S^n\setminus\{N\}$ and $N^\perp$, whose inverse is constructed as follows: given $x\in N^\perp$, consider the line passing through $x$ and $N$. This will cross $\Bbb S^n$ in two distinct points. One is $N$ itself -- call the other one ${\rm St}_N^{-1}(x)$.
