I try to figure out the following statement in Griffiths & Harris, Principles in Algebraic Geometry, page $9$.
The zero locus of an analytic function $f(z_1,...,z_n,w)$, not vanishing identically on the $w$-axis, projects locally onto the hyperplane $w=0$ as a finite-sheeted cover branched over the zero locus of an analytic function.
The authors argue that by the Weierstrass representation theorem, in a nbh of the origin (in $\mathbb{C}^n$) $f$ can be uniquely represented as $g.h$ in which $g$ is a Weierstrass polynomial (say of degree $d$) and $h(0) \neq 0$. Now consider the roots of $g$, $b_1(z),...,b_d(z)$ we then have $$f(z,w) = g.h = (w - b_1(z))...(w- b_d(z))h(z,w)$$
The roots $b_i(z)$ of the polynomial $g(z,.)$ are, away from those values of $z$ for which $g(z,.)$ has a multiple root, locally-singed value holomorphic functions in $z$. Since the discriminant of $g(z,.)$ is an analytic function in $z$.
I really don't understant the argument above, as far as I can explain, locally, the projection $$\pi: ((z,w) \mid f= 0) \to \mathbb{C}^{n},(z,w) \mapsto (z,0)$$ has inverse image at $(z,0)$ is $\pi^{-1}(z,0) = \bigcup_{i=1}^d ((z,b_i(z))$. Therefore, as long as the discriminant is nonzero, this projection (again, locally) is a $d$-sheet covering and because of the regularity of the discriminant we can always choose a nbd of $z$ for which it (the discriminant) is nonzero? Is this explanation right? I would be appreciate if someone can ellaborate it, thanks in advance.
Everything here is local. Suppose $f$ is the function which is holomorphic in a neighborhood of the origin in $\mathbb{C}^n$ (with coordinates $(z_1, ..., z_{n-1},w)$) and is not identically zero on the $w$-axis. The Weierstrass preparation theorem tells us that $f$ vanishes like a polynomial, i.e., $$f = g \cdot h,$$ where $g$ is the polynomial and $h$ does not vanish in the neighborhood.
The branching of the cover will clearly occur when $g$ has a multiple root (it wont look like the zero-set of an analytic function with non-vanishing gradient locally, i.e., it will not look like a manifold here). Away from these points, however, the function looks likes a $\deg(g)$-sheeted cover over $f=0$.
I'm not sure what you're asking.