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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.

Vim
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    What's your definition of $S^n$? (I know you this is the $n$-sphere, but how do you define $n$-sphere?) – Plutoro Apr 01 '16 at 13:57
  • @AlexS $S^n$ lives in $\Bbb R^{n+1}$ and is the boundary of $D^{n+1}$, meaning it consists of points in $\Bbb R^{n+1}$ whose norm is $1$. – Vim Apr 01 '16 at 14:02
  • @PaulFrost yes exactly. Please vote as duplicate. – Vim Oct 23 '19 at 02:26

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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.

Hagen von Eitzen
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  • I don't quite follow how it smashes $S^{n-1}$ into a pole. It seems to me that $h$ doesn't vanish on $S^{n-1}$, so how come the resulting image is a fixed point? – Vim Apr 01 '16 at 14:27
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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.

  • Thanks. Just allow me to first check out the notions of compactification and stereographic projection, which are ahead of what I've learned in the curriculum. – Vim Apr 01 '16 at 14:14