If $f : U \rightarrow V$ is a one-to-one holomorphic map of open sets in $\mathbb{C}^n$ then the jacobian is not zero.

Question

I am just reading Griffiths and Harris Principles of algebraic geometry. I am confused on a slight issue of this argument page 20. Why f maps the locus $(z_1^{\prime} = \ldots = z_k^{\prime} = 0)$ one to one to the locus $(w_1 = \ldots = w_k = 0)$ ? I don't understand why the Jacobian matrix of f vanishes identically wherever its determinant is zero ?Here is also the start of the argument We prove this by induction on n; the case n = 1 is clear. Let $z = (z_1,\ldots,z_n)$ and $w = (w_1,\ldots,w_n)$ be coordinates on U and V, respectively, and suppose that $J(f)$ has rank k at $0 \in U$; we may assume that the matrix