Homeomorphism between $D^n$ and $[0,1]^n$ I know that $D^n=\{x\in \mathbb{R}^n:\|x\|\leq 1\}$ is homeomorphic to $[0,1]^n$, but how to write down homeomorphism? How to find explicit formula?
Thanks in advance.
 A: Let $\| v \|_{p}$ denotes the $p$-norm of $v \in \Bbb{R}^{n}$. In particular,
$$ \| v \|_{2} = [ v_{1}^{2} + \cdots + v_{n}^{2} ]^{1/2} \quad \text{and} \quad \| v \|_{\infty} = \max \{ |v_{1}|, \cdots, |v_{n}| \}. $$
Then the following map
$$ F : D^{n} \to [-1, 1]^{n} : v \mapsto \frac{\|v\|_{2}}{\|v\|_{\infty}} v $$
gives the homeomorphism with the inverse
$$ G : [-1, 1]^{n} \to D^{n} : w \mapsto \frac{\|w\|_{\infty}}{\|w\|_{2}} w. $$
(Of course, we set $F(0) = 0 = G(0)$.)
The idea is simple:
$$ D^{n} = \{ \| v \|_{2} \leq 1 \} \quad \text{and}  \quad [-1, 1]^{n} = \{ \| v\|_{\infty} \leq 1 \}. $$
So the map $F$ rescales the vector so that $\| F(v) \|_{\infty} = \| v \|_{2}$. Checking that both $F$ and $G$ are indeed continuous is not theoretically hard, though it may be somewhat cumbersome.
The following graph may help you what is actually going on in $n = 3$. The sphere (above) is mapped into the cube (below) by the mapping $F$. (The seams between the faces of the cube are software artifacts.)

A: First place $I^n$ in the center of your coordinate system using translation $T$. Then define $F:\mathbb R^n\rightarrow \mathbb R^n$ as fallows $F(x)=x/\|\partial T(I^n)\cap l(x)\|$, where $l(x)$ is a line passing through $0$ and $x$. Now morphism you looking for is $F\circ T$.
