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

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 $S^n$.

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.