$D^2$ is not homeomorphic to $D^n/S^{n-1} $ The following example was left in exercise of my topology class. I think I would need help on how to prove the asserstion asked.
Let $D^n$ denote the unit ball in the n-th dimensional euclidean space and let $S^{n-1} $be the unit sphere.

Show that $D^2 $ is not homeomorphic to $D^n /S^{n-1}$.

These types of questions are proved by assuming that there exists an homeomorphism and then there is a fundamental property( upto homeomorphism) which is not satisfied in one of the sets but is satisfied in the other sets.

But I am not able to think of such a property in the current question.

 A: $S^n$ is not homeomorphic to $D^2$ for any $n$. That's because $D^2\setminus\{0\}$ is not simply connected (it retracts onto the boundary circle) but $S^n$ minus a point is... That uses the fact that $S^1$ isn't simply connected, which follows from perhaps one of the most known results of algebraic topology: $$\pi_1(S^1)\cong\Bbb Z$$And $S^n\setminus\{p\}$ is homeomorphic to $\Bbb R^n$ for any $p$, so it must be simply connected if $n\ge1$. If $n=0$, $S^0=\{-1,1\}$ (or $D^0/\emptyset$) is obviously not homeomorphic to $D^2$.
As for calculating $(n\ge1)$:
$$D^n/S^{n-1}\cong S^n$$Consider the map: $$D^n\to S^n$$Which sends: $$x=(x_1,x_2,\cdots,x_n)\mapsto(2\sqrt{\|x\|(1-\|x\|)}\cdot x,2\|x\|-1)$$Giving the left hand side coordinates of $\Bbb R^n$ and the right hand side coordinates of $\Bbb R^n\times\Bbb R$. This is a quotient map for compactness reasons, and the relation it induces is precisely the relation that kills $S^{n-1}=\partial D^n$.
A: As Randall already pointed out in the comments, $D^n/S^{n-1}$ is homeomorphic to a sphere, in this case $S^n$. This is proven here by Henno Brandsma using the one-point or Alexandroff compactification (See here). In short, the argument looks like this:
$$D^n/S^{n-1}
\cong(D^n\setminus S^{n-1})^*
\cong((D^n)^\circ)^*
\cong(\mathbb{R}^n)^*
\cong S^n.$$
The first step is the fifth point here, the second step is using $\partial D^n=S^{n-1}$, the third step is using $(D^n)^\circ\cong\mathbb{R}^n$ (See here.) and the forth step can be imagined as stereographic projection and adding the missing point.
To now solve your problem: $D^2$ and $D^n/S^{n-1}\cong S^n$ are not homeomorphic (and not even homotopy equivalent) since their $n$-th homotopy groups are not isomorphic:
$$\pi_n(D^2)
=1\not\cong\mathbb{Z}
\cong\pi_n(S^n)
\cong\pi_n(D^n/S^{n-1}).$$
You could also say, that $D^2$ is contractible (homotopy equivalent to a point, meaning all homotopy groups are trivial), while $D^n/S^{n-1}$ is not.
