$B^n/S^{n-1}$ is homeomorphic to $S^n$

Question

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 $S^n$.

Answer 1

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

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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.

Answer 2

$B^n/{S^{n-1}}$ 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).