How can I show that
the open unit cube $(-1,1)^n \subset \mathbb{R}^n$ and
the open unit ball $B = \{x \in \mathbb{R}^n \mid \|x\| < 1\}$
are diffeomorphic?
I know that one can proof this by showing that those two sets are diffeomorphic to the whole space $\mathbb{R}^n$. But is there a direct way (not over the $\mathbb{R}^n$ ) to proof this? Thus, is there a smooth, differentiable bijection between the two sets with a differentiable inverse?
$\newcommand{\Reals}{\mathbf{R}}$Let $f:(-1, 1) \to \Reals$ be your favorite smooth, increasing bijection, such as $$ f(x) = \tanh^{-1} x \quad\text{or}\quad f(x) = \frac{x}{1 - x^{2}} \quad\text{or}\quad f(x) = \tan \tfrac{\pi}{2} x. $$ The mapping $$ \phi(x_{1}, \dots, x_{n}) = \frac{\bigl(f(x_{1}), \dots, f(x_{n})\bigr)}{\sqrt{1 + f(x_{1})^{2} + \dots + f(x_{n})^2}} $$ is obviously a diffeomorphism from the open cube $C$ to the open unit ball $B$.
Oh, that fact isn't obvious...?
But look: If we define $\psi_{1}:C \to \Reals^{n}$ and $\psi_{2}:\Reals^{n} \to B$ by $$ \psi_{1}(x_{1}, \dots, x_{n}) = \bigl(f(x_{1}), \dots, f(x_{n})\bigr) \quad\text{and}\quad \psi_{2}(y) = \frac{y}{\sqrt{1 + \|y\|^{2}}}, $$ it's obvious $\psi_{1}$ and $\psi_{2}$ are diffeomorphisms, and $\phi = \psi_{2} \circ \psi_{1}$.
The deeper point of this imaginary dialogue is, constructing a diffeomorphism from $C$ to $B$ by "going through $\Reals^{n}$" is more elegant than trying to find a "direct" mapping: In fancy terms, using $\Reals^{n}$ as "intermediary" exploits the common one-point compactification of $C$ and $B$.
By contrast, a "direct" mapping (presumably, a diffeomorphism from $B$ to $C$ that extends to the respective boundaries) cannot be smooth at the boundary of $B$. One finds oneself visiting the land of unnecessary vexation.(1)