Lemma 9.3.14 (Constant rank theorem)F802E1301C0F

Proof. After permuting the coordinates of the source and of the target, we may assume that the matrix $(\partial F_ i/\partial x_ j(p))_{i,j\leq r}$ is invertible. The map $\phi (x)=(F_1(x),\ldots ,F_ r(x),x_{r+1},\ldots ,x_ m)$ has invertible derivative at $p$, so by the holomorphic inverse function theorem (proposition 5.2.4(1)) it maps a neighborhood of $p$ biholomorphically onto a polydisc $Q'\times Q''$ centered at $\phi (p)$, on which the rank of $G=F\circ \phi ^{-1}$ is $r$. This map has the form $G(y)=(y',G''(y))$ with $y'=(y_1,\ldots ,y_ r)$ and $G''=(G_{r+1},\ldots ,G_ K)$. For $i,j>r$, the minor of the Jacobian matrix of $G$ formed by the rows $1,\ldots ,r,i$ and the columns $1,\ldots ,r,j$ equals $\partial G_ i/\partial y_ j$, because the first $r$ rows are $(I,0)$; since the rank is $r$, these derivatives vanish. As $Q''$ is convex, $G''(y',y'')=G''(y',c'')$, where $c''$ is the center of $Q''$. The map $\psi (w',w'')=(w',w''+G''(w',c''))$ is a biholomorphism of $Q'\times \mathbb C^{K-r}$ onto itself, with inverse $(w',w'')\mapsto (w',w''-G''(w',c''))$, and $G(y',y'')=\psi (y',0)$. The fibers of $G$ are the slices $\{ y'\} \times Q''$, and its image is $\psi (Q'\times \{ 0\} )$, a closed submanifold of $Q'\times \mathbb C^{K-r}$. $\square$