Quotient map $\pi\colon \mathbb{C}^{n+1}\backslash \{0\} \to \mathbb{CP}^n$ is a submersion It's not to hard to see that the quotient map $\pi\colon \mathbb{C}^{n+1}\backslash \{0\} \to \mathbb{CP}^n$ is smooth and surjective. Does that imply that it is a submersion as well?
 A: As Olivier Bégassat mentions in the comments, a smooth surjective map need not be a submersion. However, it is true that $\pi$ is a submersion. To see this, let $U_i = \{[x_0, \dots, x_n] \in \mathbb{CP}^n \mid x_i \neq 0\}$ and define 
\begin{align*}
\varphi : U_i &\to \mathbb{C}^n\\
[x_0, \dots, x_n] &\mapsto \left(\frac{x_0}{x_i}, \dots, \frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i}, \dots, \frac{x_n}{x_i}\right).
\end{align*}
The pairs $(U_i, \varphi_i)$ are standard charts on $\mathbb{CP}^n$.
Let $p \in \mathbb{C}^{n+1}\setminus\{0\}$, then $\pi(p) \in U_i$ for some $i = 0, \dots, n$. Without loss of generality, suppose $\pi(p) \in U_0$. If $p = (x_0, \dots, x_n)$ then 
\begin{align*}
(\varphi_0\circ \pi)(x_0, \dots, x_n) &= \varphi_0(\pi(x_0, \dots, x_n))\\
&=\varphi_0([x_0, \dots, x_n])\\
&=\left(\frac{x_1}{x_0}, \dots, \frac{x_n}{x_0}\right).
\end{align*}
The differential has standard matrix (of size $n\times(n+1)$) given by 
$$\left[\begin{array}{ccccc}-\frac{x_1}{x_0^2} & \frac{1}{x_0} & 0 & \dots & 0\\
-\frac{x_2}{x_0^2} & 0 & \frac{1}{x_0} & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ -\frac{x_n}{x_0^2} & 0 & 0 & \dots & \frac{1}{x_0}\end{array}\right].$$
Note that the matrix has rank $n$, so the differential is surjective. Therefore, $\pi$ is a submersion.
A: I will basically replicate Michael Albanese's answer in terms of real coordinates.
Let $U_i=\pi(\{z_i\neq 0\})\subset\mathbb CP^n$ be the affine opens that cover $\mathbb CP^n$ along with their standard charts. We can restrict $\pi:\mathbb C^{n+1}-\{0\})\to\mathbb CP^n$ to map
$$
\pi:\{z_i\neq 0\}\to U_i,
$$
and if we show smoothness on each restriction, then we've shown smoothness on $\pi$. Assume WLOG that $i=0$, so we are looking at
$$
\pi:\{z_0\neq 0\}\to U_0,
$$
which looks in (complex) coordinates as
$$
(z_0,\dots,z_n)\mapsto (z_1/z_0,\dots,z_n/z_0).
$$
We use the coordinates $x+iy\leftrightarrow(x,y)$ on $\mathbb C$, and write $z_k\in\mathbb C$ as $z_k^{(1)}+iz_k^{(2)}$. Note that
$$
z_k/z_0=\frac{(z_k^{(1)}+iz_k^{(2)})\cdot(z_0^{(1)}-iz_0^{(2)})}{\vert z_0\vert^2}=\frac{z_k^{(1)}z_0^{(1)}+z_k^{(2)}z_0^{(2)}}{\vert z_0\vert^2}+i\frac{z_k^{(2)}z_0^{(1)}-z_k^{(1)}z_0^{(2)}}{\vert z_0\vert^2}.
$$
Therefore $\pi$ looks in real coordinates as
$$
(z_0^{(1)},z_0^{(2)},\dots,z_n^{(1)},z_n^{(1)})\mapsto\left(\frac{z_k^{(1)}z_0^{(1)}+z_k^{(2)}z_0^{(2)}}{\vert z_0\vert^2}, \frac{z_k^{(2)}z_0^{(1)}-z_k^{(1)}z_0^{(2)}}{\vert z_0\vert^2}\right)_k.
$$
Computing the derivative w.r.t $z_k^{(1)}$ and $z_k^{(2)}$ in the $k$-th (complex) component, we find that the corresponding $2\times 2$ block of the Jacobian is given by
$$
\pmatrix{
\frac{z_0^{(1)}}{\vert z_0\vert^2} & \frac{z_0^{(2)}}{\vert z_0\vert^2}\\
\frac{-z_0^{(2)}}{\vert z_0\vert^2} & \frac{z_0^{(1)}}{\vert z_0\vert^2},
}
$$
which is non-singular because $z_0\neq 0$. If you replace the entries $1/z_0$ in the $k$-th row of is Michael's complex Jacobian by the $2\times 2$ blocks above (and also replace the first column by the appropriate partial derivatives, whose values don't matter to argue that we have full rank), then we see that the real Jacobian has full rank. This shows that $\pi$ is a smooth submersion.
