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Let $B^n$ be a n-dimensional disc (ball) with boundary $S^{n-1}$. Prove that $B^n/S^{n-1}$ is homeomorphic to $S^n$.

Could someone check my proving, please?

Let put the ball $B^n=\{x\in\mathbb R^n\mid |x|\leq 1\}$ into $\mathbb R^{n+1}$ using $x\mapsto (x,0)$. Then we can use homeomorphism $(x,0)\mapsto (x,\sqrt{1-|x|})$. We fall into the half of the sphere $S^n$. The boundary of the ball after this will turn just into points of the form $(*,0)$. If you pull it to the point then we will get suspension $S^{n-1}$ or $S^n$.

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Metso
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    The demonstration seems correct to me. Btw,I do not know how you prove that suspension of $S^n$ is $S^{n+1}$ but probably you know about onepoint compactification that is another way to solve this problem – Tommaso Scognamiglio Feb 06 '18 at 20:27
  • I am having trouble showing that $f$ is a quotient map. For this, it is enough to show that $f$ is surjective, continuous and open. $f$ is a composition of continuous functions, so $f$ is continuous. Composition of open functions is open, but the map $x\mapsto x^2$ is not open. – AgentSmith May 03 '20 at 15:27

2 Answers2

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This question has been answered by @SteveD in the comments, as reproduced below.

This seems fine, but you can also define the map explicitly. Namely, a function $f: \Bbb{B}^n \to \Bbb{S}^n$, given by: $$ f(b) = \left(2b\sqrt{1-|b|^2},2|b|^2 - 1\right) $$ This maps the boundary of $\Bbb{B}^n$ to the north pole of $\Bbb{S}^n$. It's not hard to show this is surjective, and injective on the interior of $\Bbb{B}^n$, so it's the quotient you want.

Paul Frost
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    On the interior of the ball, this map $f$ is the composite $h\circ g$ of the homeomorphisms $g$ and $h$, where $g: \mathrm{Int}(\mathbb{B}^n)\rightarrow \mathbb{R}^n$ is given by $b\mapsto\frac{b}{\sqrt{1-|b|^2}}$ and $h: \mathbb{R}^n\rightarrow S^n\backslash {p} $ is the inverse stereographic projection. – AgentSmith May 02 '20 at 20:42
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    After some time reverse engineering where these equations come from, I found another way to understand them: let $d=x/|x|$ be the direction vector and $r=|x|$, the homeomorphism $g: \text{Int}(\mathbb{B}^n)\rightarrow\mathbb{R}^n$ is given by $x \rightarrow \tan\sin^{-1}(r) \cdot d = \frac{x}{\sqrt{1-|x|^2}}$ and the quotient map $f: \mathbb{B}^n \rightarrow \mathbb{S}^n$ is given by $f(x)=(\sin(2\sin^{-1}r)\cdot d, \cos(2\sin^{-1}r))=\left(2x\sqrt{1-|x|^2}, 2|x|^2-1\right)$. Overall, just imagine $|x|$ as an "angle" sweeping out a circle. This is one intuitive way to arrive at these equations – cicolus Apr 21 '22 at 00:29
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$B^n/{S^{n-1}}$ is a compact Hausdorff space $X$ with a point $p$ (the class of the collapsed $S^{n-1}$) such that $X\setminus \{p\}$ is homeomorphic to $B^n \setminus S^{n-1}$, which is the open $n$-ball $U_n$, which is homeomorphic to $\mathbb{R^n}$ (via $f: \mathbb{R}^n \to U_n; f(x)=\frac{1}{1+\|x\|}\cdot x$ e.g.).

On the other hand $S^n$ is also a compact Hausdorff space $Y$ such that $Y\setminus \{q\}$ (where we can take $q$ the north pole, or any point, by homogeneity) is homeomorphic to $\mathbb{R}^n$ as well, by stereographic projection.

So both spaces are concrete instantiations of the Alexandroff extension( a.k.a. the one-point compactification) of $\mathbb{R}^n$ and thus homeomorphic to each other by this extension's unicity up to homeomorphism (with compact Hausdorff spaces).

Henno Brandsma
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