Diffeomorphism: Unit Ball vs. Euclidean Space

In my differential geometry class we are being asked to prove that the open unit ball

$B^n$ = { $x$ $\in$ $\mathbb{R}$$^n$ such that |$x$| < $1$}

is diffeomorphic to $\mathbb{R}$$^n$

I am having a hard time with this as I am brand new not only to differential geometry, but also topology.

I know that I need to construct a smooth, differentiable bijection between the two with a differentiable inverse, but beyond that, I am unsure of where to start. Some guidance in the right direction would be greatly appreciated.

Or you can try $f(x)=x/\sqrt{1-|x|^2}$ for $x\in B^n$.

Let $\phi \colon [0,1) \to [0, \infty)$ a diffeomorphism with inverse $\psi$. Some possible choices: $t \mapsto \frac{t}{1-t}$, $t \mapsto \tan (\frac{\pi}{2}\cdot t)$.

The map $$x \mapsto \phi(||x||) \cdot \frac{x}{||x||}$$

is a diffeomorphism from $B^n$ to $ \mathbb{R}^n$ with inverse $$y \mapsto \psi(||y||) \cdot \frac{y}{||y||}$$

$\bf{Added:}$ It turns out that the choice of the diffeomorphism from $[0,1)$ to $[0,\infty)$ matters a lot, since $x \mapsto ||x||$ is not smooth at $0$. This was brought to my attention by @Freeze_S and I thank him a lot! One can check that the map obtained for $\phi(t) = \frac{t}{1-t}$ is only $C^1$ at $0$... However, we can use the map so kindly suggested by @Jesus RS: ( big thanks! ) $\phi(t) = \frac{t}{\sqrt{1-t^2}}$ with inverse $\psi(s) = \frac{s}{\sqrt{1+s^2}}$ and it will work just fine. The diffeomorphisms are, as written by @Jesus RS: $$x \mapsto \frac{x}{\sqrt{1-||x||^2}} \\ y \mapsto \frac{y}{\sqrt{1+||y||^2}}$$

In fact, as long as $\phi(t)$ is an odd function of $t$ things will work OK. So, another example is

$$x \mapsto \frac{\tan (\frac{\pi}{2} \cdot ||x|| )}{||x||} \cdot x $$

Standard Approach

Consider the identification: $$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{\sqrt{1-\|x\|_E^2}}$$ Its inverse is given explicitely by: $$\Psi:\mathbb{R}^n\to\mathbb{B}^n:\Psi(y):=\frac{y}{\sqrt{1+\|y\|_E^2}}$$ The argument of the roots never vanish: $$1-\|x\|_E^2\neq0\quad1+\|y\|_E^2\neq0$$ So they're both differentiable.

Alternative Approach

Consider the identification: $$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{1-\|x\|_E^2}$$ By the inverse function theorem: $$\mathcal{N}d\Phi(x)\equiv(0)\implies\mathrm{d}\Psi(y)=\mathrm{d}\Phi(\Psi(y))\in\mathcal{L}(\mathbb{R}^n)$$ But this holds globally as the identification is a bijection.

Problematic Approach

Consider the identification: $$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{\sqrt{1-\|x\|_\infty^2}}$$ This is most likely not differentiable.

Not every smooth function induces a smooth map: $$\Phi:\mathbb{B}\to\mathbb{R}:\quad\varphi(|x|):=\frac{1}{1-|x|}$$ Just have a careful look at its diagram:

(Note that it is not even differentiable at zero!)

The problem is that the norm is not smooth in general: $$\|\cdot\|:V\to[0,\infty)$$

One can master this almost only by patching a bump on top: $$\varphi(r):=e^{-\frac{1}{r^2}}\cdot\frac{1}{1-r}$$

For a mere diffeomorphism still the whole story highly depends on the chosen norm!