Prove that a quotient of an n-dimensional ball by an equivalence relation is homeomorphic to an n-dimensional sphere. Question
Prove that a quotient of an n-dimensional ball by an equivalence relation, whose only non-trivial equivalence class is the n-1 dimensional sphere, is homeomorphic to an n-dimensional sphere.

Let $B_n = \{x \in \mathbb{R^n} : \|x\| \le 1\}$
and $S^n = \{x \in \mathbb{R^{n+1}} : \|x\| = 1\}$.
Define
$B_n$ as n dimensional ball.
$S^n$ as n dimensional sphere.
$\|x\|=(x_1^2+ \dots+x_n^2)^{1/2}$
I have defined a equivalence relation of $B_n$ by $x \sim y \iff \|x\|=\|y\|$ so that $[x]=\{y \in \mathbb{R^n} : \|y\|=\|x\|\}$. Then we have equivalence class as n-1 dimensional sphere.
And I was thinking of using a theorem:
Let $X$ be a topological space with an equivalence relation ~. Let $f :X \rightarrow Y$ be a continuous map with the properties


*

*$f(a)=f(b)$ iff $a$~$b$

*$f$ is onto

*$U$ is open in $Y$ if $f^{-1}(U)$ is open in $X$.


Then the unique map $g:X/\sim \rightarrow Y$ is homeomorphism

We can find the homeomorphism explicitly. Note that for all points $\mathbf x=(x_1,\ldots,x_n)$ in $B\setminus\partial B$, we have $0\le \|\mathbf x\|<1$, and of course $\|\mathbf x\|=1$ for the points on $\partial B$. We may try to map all points of the boundary to the north pole - and the interior to the rest; symmetry suggests to map the origin to the south pole. Now how to find a point $\mathbf y=(y_0,y_1,\ldots ,y_n)$ with $\|\mathbf y\|=1$ from a given point $\mathbf x\in B$ such that  $\|\mathbf x\|=1$ implies $y_0=1$ and $\|\mathbf x\|=0$ implies $y_0=-1$? This suggests $y_0=2\|\mathbf x\|-1$ and $y_k$ a multiple of $x_k$ such that $\|y\|$ becomes $1$.
