a concept question about the zeros of the several variable holomorphic function There is a theorem in the tasty bits of several complex variables as follows (Proposition 6.3.2):
Proposition 6.3.2. Let $D \subset \mathbb{C}$ and $U^{\prime} \subset \mathbb{C}^{n-1}$ be domains, and $f \in \mathcal{O}\left(U^{\prime} \times D\right)$. Let $m \in \mathbb{N}$ be such that for each $z^{\prime} \in U^{\prime}$, the function $z_n \mapsto f\left(z^{\prime}, z_n\right)$ has precisely $m$ geometrically distinct zeros. Then locally near each point in $U^{\prime}$ there exist $m$ holomorphic functions $\alpha_1\left(z^{\prime}\right), \ldots, \alpha_m\left(z^{\prime}\right)$, positive integers $k_1, \ldots, k_m$, and a nonvanishing holomorphic function $u$ such that
$$
f\left(z^{\prime}, z_n\right)=u\left(z^{\prime}, z_n\right) \prod_{j=1}^m\left(z_n-\alpha_j\left(z^{\prime}\right)\right)^{k_j} .
$$
We can only define $\alpha_1$ through $\alpha_m$ locally (on a smaller domain) as we do not know how $\alpha_1$ through $\alpha_m$ are ordered, and the order could change as we move around $U^{\prime}$ if it is not simply connected. If $U^{\prime}$ is simply connected, then the functions can be defined globally by analytic continuation.

I can't understand why there is some global obstruction due to ordering of them?
 A: As there is no discriminant (all the roots are distinct), we can analytically continue the roots in $U'$.  However, $U'$ is not going to be simply connected in general, and analytic continuation only gives you a holomorphic function on a simply connected set.  So we can't find the functions globally.  If you locally take say $\alpha_1(z'_0)$ and continue it as a holomorphic function around a path in $U'$ it could happen that when you get back to $z'_0$ you actually have the value $\alpha_2(z'_0)$.
Consider the domain $U' = \mathbb{C} \setminus \{0 \}$.  Suppose your function is $f(z_1,z_2) = z_1-z_2^2$.  Then $\alpha_1$ and $\alpha_2$ are $\pm \sqrt{z_1}$.  But if you start say at $z_1=1$ and let $\alpha_1(1) = 1$ and $\alpha_2(1)=-1$, then as you travel once around the unit circle trying to continue $\alpha_1$, you'll get that you'd have to have $\alpha_1(1)=-1$, which is a contradiction.
So no matter how you order the $\alpha_j$s, when you go around some curve trying to continue it, you'll end up with a different order of roots.  The global issue is that of topology.
See the Monodromy theorem.  It is sufficient to just look at a one complex variable book for this, it is the same idea.
