Let $D_{k}^n$ be the closed disk of radius $k$ centered at the origin in $\mathbb{R}^n$. Let $A$ be $D_{1}^n$ and let $B$ be $D_{2}^n$. Then we claim that $B/A$ is homeomorphic to a closed disk. In particular, I claim $B/A\cong D_1^n$. We want to find an explicit homeomorphism, and demonstrate that it is continuous and bijective. We need not prove that the inverse is continuous since $B/A$ is compact and $D_1^n$ is Hausdorff.
I am having a hard time coming up with the specific map. What would be the way to proceed?
Best Answer
Define $$f : D^n_2 \to D^n_1, h(x) = \begin{cases} 0 & \lVert x \rVert \le 1 \\ \frac{\lVert x \rVert - 1}{\lVert x \rVert } x & \lVert x \rVert \ge 1 \end{cases} $$ This is a continuous map such that $f(D^n_1) = \{0\}$, hence it induces of continuous map $f' : D^n_2/D^n_1 \to D^n_1$ (universal property of the quotient).
$f$ is surjective: For $y \ne 0$ define $x(y) = \frac{\lVert y \rVert + 1}{\lVert y \rVert } y$. Then $\lVert x(y) \rVert = \lVert y \rVert + 1 \in (1,2]$, hence $x(y) \in D^n_2$. Inserting yields $f(x(y)) = y$.
When do we have $f(x) = f(x')$? Certainly if $x, x' \in D^n_1$. If only one of $x, x'$ is contained in $D^n_1$, then it is impossible because $f(z) \ne 0$ for $z \notin D^n_1$. If both $x,x' \in D^n_2 \setminus D^n_1$, then from $\lVert f(x) \rVert = \lVert f(x') \rVert$ we infer $\lVert x \rVert - 1 = \lVert x' \rVert - 1$, i.e. $\lVert x \rVert = \lVert x' \rVert$. This implies $x = x'$.
The above considerations show that $f'$ is a bijection.