# The homeomorphism $D^n/S^{n-1}\cong S^n$

I want to show that $$D^n/S^{n-1}\cong S^n$$

Let $p$ be the north pole of $S^n$ and denote $(D^n)^o$ the interior of the disc and let $s:\mathbb{R}^n\rightarrow S^n$ be the stereographic projection. Let the map $f:D^n\rightarrow \mathbb{R}^n\rightarrow S^n$ defined as follows

if $x\not \in S^{n-1}$ then $f(x)=s \circ h$ where $$h:(D^n)^o\longrightarrow \mathbb{R}^n; x\longmapsto {{x}\over {1-||x||}}$$ and if $x\in S^{n-1}$, then $f(x)=p$. Then the quotient by $S^{n-1}$ gives a map $\bar f:D^n/S^{n-1}\rightarrow S^n$ which maps the class of $x$ to $f(x)$

and the map $\bar f$ is a homeomorphism.

Is this the right way to do it and is there any other better way to do it. Thanks!

Yes, this is one of the standard and most obvious way to do it. Note that the continuity of $\tilde{f}$ is not too automatic by construction.