Homeomorphism $T(x)=(\mathbb{R}^{n}_{p} ,\mapsto B(0,1), ||.||_p)$ Let $w_0\in \mathbb{R}^n_{p}$ and let $T(x)=(\mathbb{R}^n_{p} ,\mapsto B(0,1), ||.||_p)$, show that 

$T(x)=\frac{x-w_0}{1+||x-w_0||_p}$

Prove T(x) is an homeomorphism
I'm kinda lost cause i already know that $f(x)=d(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is continuous but the thing is about i get confused by the ball $B(0,1)$ and also that we're working with norms but also i know a metric is induced by a norm but still i can figure out how this is an homeomorphism.

After composing with translations (which are homeomorphisms), this problem reduces to the case where $w_0 = 0$. To prove $T$ is a homeomorphism in this case, show that $T: x \mapsto \frac{x}{1 + ||x||_p}$ is continuous and has continuous inverse.
The hardest thing to do here is to write down $T^{-1}$. Two observations are helpful here. First, $T$ maps lines through $0$ to themselves, so its inverse should as well. We can also figure out what the modulus of $T^{-1}(y)$ would need to be. For this, if $y = \frac{x}{1 + ||x||_p}$, then $||y||_p = \frac{||x||_p}{1 + ||x||_p}$, and doing some algebraic manipulation gives us that $$ ||x||_p = \frac{||y||_p}{1 - ||y||_p}.$$ With this information in mind, we put $S : y \mapsto  \frac{y}{1 - ||y||_p}$, and sure enough we can verify that $S$ and $T$ are inverse to each other. Both maps can be seen to be continuous by the continuity of $||\cdot ||_p$, completing the proof.
