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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!

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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.

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