# Quotient map of the complex projective space

It's not to hard to see that the quotient map $\pi\colon \mathrm{C}^{n+1}\backslash \{0\} \to \mathbb CP^n$ is smooth and surjective. Does that imply that it is a submersion as well?

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Generally, a surjective smooth map need not be a submersion. For instance $x\mapsto x^3$ is smooth surjective on the real line but isn't submersive over $0$. However, $\pi$ is submersive, and you can check this in local coordinates, i.e. using charts for the projective space. – Olivier Bégassat Oct 7 '12 at 19:06
Thanks for your comment! Could you please elaborate a little bit in terms of the charts? I understand that I have to prove that the differential is surjective, so the relation between these concepts is not clear to me. – user43014 Oct 7 '12 at 22:00

As Olivier Bégassat mentions in the comments, a smooth surjective map need not be a submersion. However, $\pi$ is a submersion.

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 f)(x_0, \dots, x_n) &= \varphi_0(f(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

$$\left[\begin{array}{ccccc}-\frac{x_1}{x_0^2} & \frac{1}{x_0} & 0 & \dots & 0\\ -\frac{x_1}{x_0^2} & 0 & \frac{1}{x_0} & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ -\frac{x_n}{x_0^2} & 0 & 0 & \dots & \frac{1}{x_0}\end{array}\right].$$

Note that the matrix has rank $n$, so the differential is surjective. Therefore, $\pi$ is a submersion.

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