Show using this map that $D^n / S^{n-1}$ is homeomorphic to $S^n$. 
Let $f : D^n \to S^n$, where $D^n = \{ x \in \mathbb{R}^n : \|x\| \leq 1 \}$ be defined as $x \mapsto (2\sqrt{1-\|x\|^2}x, 2\|x\|^2 -1).$ Show using this map that $D^n / S^{n-1}$ is homeomorphic to $S^n$.

Do I need to somehow show that the quotient map $q: D^n/S^{n-1} \to S^n$ is a homeomorphism here? I'm very confused about where to start. $f$ seems to be bijective, but I'm not sure if $f$ is homeomorphic. Do we even need $f$ to be homeomorphic in order for $q$ to be?
 A: Assuming $D^n$ and $S^n$ have the usual topologies, $f:D^n\to S^n$ is continuous
As you've said in the comments $f(S^{n-1}) = (0,...,0,1)$ and so we have a continuous quotient map $q:D^n/S^{n-1}\to S^n$.
Injectivity of $q$: Suppose $q(x_1)=q(x_2)$. Then we have
$2\|x_1\|^2-1=2\|x_2\|^2-1$
and
$2\sqrt{1-\|x_1\|^2}x_1=2\sqrt{1-\|x_2\|^2}x_2$.
The first equation implies $\|x_1\|=\|x_2\|$. If these norms are $1$ then they both correspond to the same identified point. If they are less than $1$ we can divide by $2\sqrt{1-\|x_2\|^2}$ in the second equation to obtain
$x_1=x_2$.
Surjectivity of $f$: Let $(x_1,\dots, x_{n+1})\in S^n$ such that $x_{n+1}\neq 1$. Then we must pick $x\in D^n$ such that $x_{n+1} = 2\|x\|^2-1 $. so $\|x\| = \sqrt{\frac{1+x_{n+1}}{2}}$. And the first $n$ components tell us
$(x_1,\dots,x_n) = 2\sqrt{1-\|x\|^2}x$.
Solving for $x$ we have
Let $x =  \frac{1}{2\sqrt{1-\|x\|^2}}(x_1,\dots,x_n)=\frac{1}{\sqrt{4-4\|x\|^2}}(x_1,\dots,x_n)=\frac{1}{\sqrt{2-2x_{n+1}}}(x_1,\dots,x_n)$.
You can separately check this $x$ works. If $x_{n+1}=1$ then just pick $x$ to be any point on the boundary of $D^n$.
Since $f$ is surjective, $q$ is surjective as well.
Then there is a well known theorem that says a continuous bijection from a compact space to a Hausdorff space is a homeomorphism . $D^n$ is compact and so the quotient $D^n/S^{n-1}$ is also compact. Moreover $S^n$ is Hausdorff, so we can apply this result to finish the proof.
