How to prove $D^n/S^{n-1}\cong S^n$? In my textbook it is said that the quotient space $D^n/S^{n-1}$ is homeomorphic to $S^n$. I can imagine it for $n=2$, but fail to make a mathematical proof for any dimension.
Can anyone provide a rigorous proof?
 A: Let's follow Henno's idea. We will write $D^n = e^n \uplus S^{n-1}$, that is denote the iterior of $D^n$ by $e^n$. $\def\R{\mathbb R}$Then $e^n$ is homeomorphic to $\R^n$ by $f\colon e^n\ni x \mapsto x/(\def\abs#1{\left|#1\right|}1-\abs x)$. Now $\R^n$ is homeomorphic to $S^n - \{(0,\ldots, 0, 1)\}$ via the inverse of the stereographic projection 
$$ g \colon x \mapsto \left(\frac{2x}{\abs x^2 + 1}, \frac{\abs x^2-1}{\abs x^2 + 1}\right) $$
Combining this two maps gives a homeomorphism $e^n \to S^n - \{(0,\ldots, 0, 1)\}$. Now define $F \colon D^n/S^{n-1}\to S^n$ by sending $S^{n-1}$ to $\{(0,\ldots, 0, 1)\}$. 
Let's prove that $F$ is continuous, as both spaces are Hausdorff and compact, $F$ is a homeomorphism then. So suppose $e^n\ni x_k \to x \in S^{n-1}$, then $\abs{f(x_k)} \to \infty$, giving $g\bigl(f(x_k)\bigr) \to (0,\ldots, 0, 1) = F(S^{n-1})$. Hence $F$ is continuous at $S^{n-1}$, on $e^n\subseteq D^n/S^{n-1}$ $F$ is continuous as composition of $g$ and $f$.
A: Theorem 1: Let $X$ be compact Hausdorff, let $p \in X$. If $Y$ is homeomorphic to $X \setminus \{p\}$ then $X$ is homeomorphic to the one-point compactification of $Y$.
Theorem 2: if $X$ is compact Hausdorff and $A \subset X$ is closed, then $X / A$ (the quotient of $X$ under the equivalence relation that identifies $A$ to a point) is compact Hausdorff as well.
Now apply Theorem 2 to $X = D^n$ and $A = S^{n-1} \subset D^n$. Then note that the result with the identified point removed is just the interior of $D^n$, which is homeomorphic to $\mathbb{R}^n$ (use the map $x \to \frac{x}{\|x\|+1}$ e.g.). So Theorem 1 then says that $X / A$ is homeomorphic to the one-point compactification of $\mathbb{R}^n$, which is $S^n$ (via the usual stereographic projection).
A: You can do the following:


*

*First, lift the disk to the upper hemisphere by the map $(x_1,...,x_n)\mapsto\left(x_1,...,x_n,\sqrt{1-(x_1^2+...+x_n^2)}\right)$

*Then you can wrap the northern hemisphere around the $n$-sphere by doubling the angle $\theta_{n-1}$. Each point of $S^n$ has coordinates
$\left.\right.$
$\begin{pmatrix}
\cos\varphi\sin\theta_1\dots\sin\theta_{n-1}\\
\sin\varphi\sin\theta_1\dots\sin\theta_{n-1}\\
\vdots\\
\cos\theta_{n-2}\sin\theta_{n-1}\\
\cos\theta_{n-1}
\end{pmatrix}$
$\left.\right.$
where $\varphi\in[0,2\pi]$ and $\theta_i\in[0,\pi]$. More precisely, this is a quotient map from $[0,2\pi]\times[0,\pi]\times...\times[0,\pi]$ onto the sphere. For the points on the northern hemisphere, the last angle $\theta_{n-1}$ is smaller than $\pi/2$. This hemisphere is the image of the restriction of this quotient map to the product where the last interval is only $[0,\pi/2]$. Seeing it as a quotient space helps to prove rigorously that the "doubling"-map is induced by a continuous map.

*Finally, show that this map identifies two points if and only if they are both on the boundary of $D^n$.
A: $$D^n/S^{n-1}\cong S^n$$Since we can consider the map: $$q:D^n\to S^n$$Which sends: $$\Bbb R^n\supset D^n\ni x=(x_1,x_2,\cdots,x_n)\mapsto(2\sqrt{\|x\|(1-\|x\|)}\cdot x,2\|x\|-1)\in S^n\subset\Bbb R^n\times\Bbb R$$
Geometrically, $q$ is wrapping the disk around the sphere by pinching the boundary onto the North pole $(0,0,\cdots,1)$.
Because $D^n$ is compact, $S^n$ is Hausdorff and $q$ is a continuous surjection, $q$ is a quotient map. It's easy to check that $q(x)=q(y)$ if and only if $x=y$ or $x,y\in S^{n-1}\subset D^n$. Therefore $q$ witnesses $D^n/S^{n-1}\cong S^n$.
