The map from the open disc to $\mathbb R^n$ is injective. To show that the open disc is homeomorphic to $\mathbb R^n$, the map usually used is $f:\mathbb R^n\to  D^n$ such that 
$f(x)=\frac{1}{1+|x|}x$. The step of showing that $f$ is injective is somewhat not clear. Take $x$ and $y$ on the disc such that 
$$\frac{1}{1+|x|}x=\frac{1}{1+|y|}y$$ how to proceed to show that $x=y$ ?
Another question, why taking only the interior of the disc, the map seems to be well defined on elements with $|x|=1$ is there any reason for that ?  thank you for your help!!
 A: First, $f$ is actually a homeomorphism $\mathbb{R}^n \to D^n$. You're right that it's defined on the boundary of the disc; it's even defined on the whole Euclidean space! But on the other hand $|f(x)| < 1$ for all $x \in \mathbb{R}^n$, and so $f(x)$ is always in the interior of $D^n$.
Now let's prove it's injective. Let $g : \mathbb{R} \to \mathbb{R}$ be given by $g(t) = t / (1+t)$. Then $g'(t) = 1/(1+t)^2 > 0$, thus $g$ is strictly increasing and injective. If $f(x) = f(y)$, then $$|f(x)| = |f(y)| \implies \frac{|x|}{1+|x|} = \frac{|y|}{1+|y|} \implies g(|x|) = g(|y|) \implies |x| = |y|.$$
Therefore $$\frac{1}{1+|x|} x = \frac{1}{1+|y|} y \implies x = y,$$ since the norms in the denominators are equal, and thus $f$ is injective.
Note that you're still a long way from proving that $f$ is a homeomorphism: you still need to prove that its surjective, and that its inverse is continuous (equivalently, that $f$ is open).
A: If $$\frac{1}{1+ \vert x \vert} x = \frac{1}{1 + \vert y \vert} y$$
then taking moduli, we obtain $$\frac{1}{1+\vert x \vert} \vert x \vert = \frac{1}{1+\vert y \vert } \vert y \vert$$
You can solve this to obtain $|x| = |y|$, from which the result follows:
$$\frac{1}{1+|x|} x = \frac{1}{1+|y|} y \Rightarrow \frac{1}{1+|x|} x = \frac{1}{1+|x|} y \Rightarrow x=y$$
