Prove that the following map is a smooth, surjective submersion:
\begin{align*} \pi:\mathbb{S}^{2n+1}&\to\mathbb{CP}^n\\ (x_0,y_0,...,x_n,y_n)&\mapsto [x_0+iy_0:...:x_n+iy_n] \end{align*}
The surjectivity is immediate. I've also verified smoothness by considering the map $\mathbb{R}^{2n+2}\setminus\{0\}\to \mathbb{CP}^n$ with the same law and using the fact that $\mathbb{S}^{2n+1}\subset \mathbb{R}^{2n+2}\setminus\{0\}$ is an embbededing.
Now to prove $\pi$ is a submersion, I could use the standard approach of calculating the local formula for $d\pi$ and prove it has maximal rank. But that is obviously not the most pleasant thing to do.
I wonder if there is a more clever way or some kind of shorcut for this.
$S^{2n+1}\subset \mathbb{C}^{n+1}$. Let $p:\mathbb{C}^{n+1}-\{0\}$ be the quotient map, it is a submersion. For every $z\in \mathbb{C}^{n+1}-\{0\}$, $u\in T_z(\mathbb{C}^{n+1}-\{0\})$, $dp_zu=0$ if and only if $u=cz$. Suppose now $z\in S^{2n+1}$, $T_zS^{2n+1}=\{u:\langle z,u\rangle=0\}$. This implies that $ker dp_z\cap T_zS^{2n+1}=Vect_{\mathbb{R}}\{iz\}$ and the dimension of the image of $dp_z(TS^{2n+1})$ is $2n$.