Extending a nonsingular holomorphic map to a biholomorphic map

Question

If $F:\mathbb C^k\to\mathbb C^n$ is a non-singular holomorphic function ($k<n$). How does that imply that there are vectors $u_{k+1},...,u_n\in \mathbb C^n$ such that together with the vectors $\partial F/\partial z_1(a),...,\partial F/\partial z_k(a)$ form a basis for $\mathbb C^n$. Moreover, the map $\tilde F(z_1,...,z_n):=F(z_1,..,z_k)+z_{k+1}u_{k+1}+...+z_nu_n$ is biholomorphic?

Answer 1

I assume by nonsingular you mean the derivative is of full rank (rank $k$) at all points. Now it is not true in general globally, that is, you cannot pick a fixed set of $u_{k+1}$ through $u_n$ such that $\tilde{F}$ is a biholomorphism globally: Think $k=1$ and $n=2$. Let $F(z) = (z+z^2,z^3)$. Then $F' = (1+2z,3z^2)$ and that is never zero. A nonzero vector $u = (u_1,u_2)$ will be linearly independent with $F'$ if $(1+2z)u_2 - (3z^2)u_1 \not= 0$. But write that as $(-3u_1)z^2 + (2u_2) z + u_2$ which is a nonconstant polynomial in $z$ if $u$ is nonzero and so it has a zero. Thus no single $u$ as above will do the job at all points.

However, locally it will work of course: Just complete the basis at $a$ and you have a biholomorphism. If the vectors are linearly independent at a point, they are linearly independent nearby simply by continuity.