Easy way to show $D^n/S^{n-1}\cong S^n$

Question

I can work out an heuristic argument for $n=2$: (homeomorphically) turning the disc $D^2$ to something like a funnel (no pipe of course), then gradually contracting the open "mouth" of the funnel to a smaller and smaller hole which eventually vanishes (or "heals", if you imagine the hole as a open wound), we have now got an $S^2$.

This argument unfortunately doesn't carry over into higher dimensions. So I think I have to find another approach. Hopefully this simple-looking result $D^n/S^{n-1}\cong S^n$ doesn't entail a dreadfully hard proof. So is there any easy way out or is there any theorem that the result is just a few steps away from? Thanks in advance.

Answer 1

There are probably several ways to see this. One way is to realize that the $n$-sphere is homeomorphic to the one point compactification $\mathbb{R}^n\cup\{\infty\}$ of $\mathbb{R}^n$ (you can prove this explicitly using the stereographic projection). Now recall that the open unit $n$-disc $D^n$ in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$. Name this homeomorphism $\phi$. Define the map $$f:D^n/S^{n-1}\to \mathbb{R}^{n}\cup\{\infty\}$$ by $\phi$ for all points not on the boundary of $D^n$ and send every point on the boundary $S^{n-1}$ to the point $\infty$. Its easy to show that this map defines a homeomorphism on the quoitient.

Answer 2

I view $S^{n-1}$ and $D^n$ as subspaces of $\Bbb R^n$ in the obvious manner. With the abbreviations $$\begin{align}r&:=\sqrt{x_1^2+\ldots +x_n^2}\\\alpha&:=2\arcsin r\\h&:=\begin{cases}\frac{\sin \alpha}{r}&r>0\\2&r=0\end{cases}\end{align}$$ (which is continuous!), the map $$(x_1,\ldots,x_n)\mapsto (hx_1,\ldots,hx_n,\cos\alpha) $$ maps $D^n\to S^{n}$ while smashing precisely $S^{n-1}$ into the south pole.