I'm solving an exercise about the complex projective space, and during a step of the solution I'm asked to find a surjective map $D^{2n} \to \mathbb{CP}^n$. I defined the map in this way $$ (z_0,\dots, z_{n-1}) \mapsto [z_0;\dots;z_{n-1};\sqrt{1-\|z\|^2}]$$ where $z:=(z_0,\dots, z_{n-1})$.
Obviously the $z_i$'s are complex variables. Intuitively it's seems surjective, but I can't prove it formally. My attempt is to prove that this system always admit a $\lambda \in \mathbb{C}$ and $y=(y_0,\dots, y_{n-1}) \in D^{2n} \subset \mathbb{C}^n$ as a solution of it:
$$\begin{cases} \lambda z_0 = y_0 \\ \lambda z_1 = y_1 \\ \vdots \\ \lambda z_n = \sqrt{ 1-y_0\bar{y_0}-y_1\bar{y_1}-\cdots-y_{n-1}\bar{y_{n-1}} } \end{cases}$$
This will mean that the maps is surjective thanks to that $\lambda$.
Any suggestion? Thanks in advance!
You are basically there. You just need to determine $\lambda$. Given an arbitrary $[z_0,\dotsc,z_n] \in \mathbb{CP}^n$, you want to find $\lambda$ such that
$$\sum_{k=0}^n \lvert \lambda z_k\rvert^2 = 1,$$
and $\lambda\cdot z_n \geqslant 0$. These conditions are easy to satisfy.