Classifying a Branched Covering Space This question comes from the proof of proposition 2.2 in  Henry Laufer's 'Normal Two-Dimensional Singularities" text. I am excerpting the part I don't understand, and I think it's a self-contained question.
We have an irreducible analytic branched cover $W$ with the branch locus given by $\zeta_1\zeta_2=0$ in a polydisk neighborhood $\Delta$ of the origin. This cover is finite sheeted, and comes from finitely-many quadradic transformations, but that is not important to us at the moment. The covering map is given by $\rho$. First we extend $\Delta-\{\zeta_1\zeta_2=0\}$ to $\mathbb{C}^2-\{\zeta_1\zeta_2=0\}$, and notice that there is a deformation retract from this space to $|\zeta_1|=|\zeta_2|=1$, a torus. Therefore the fundamental group of our cover is
$\pi_1(W)=\pi_1(\mathbb{C}^2-\{\zeta_1\zeta_2=0\})=\mathbb{Z}\oplus\mathbb{Z}$
Now any covering space should correspond to a subgroup of the fundamental group.
Q1: The way I see this is that given the topology of the complement of the branch locus we can construct the covering space given a subgroup of $\pi_1$ because the subgroup will tell us 'how to connect the sheets'. Is that correct?
"$W$ is a finite sheeted covering space and thus corresponds to a subgroup $G\subset \pi_1$ of finite index. Thus $G\approx\mathbb{Z}\oplus\mathbb{Z}$ is determined by its two generators in $\pi_1$."
Q2: Why finite index? Does $G\approx\mathbb{Z}\oplus\mathbb{Z}$ mean $G=\pi_1$, and if so, why is that?
Now we can write the covering in a form
$\rho :\mathbb{C}^2-\{\xi\eta=0\}\to \mathbb{C}^2-\{\zeta_1\zeta_2=0\}$
with $\rho(\xi,\eta)=(\xi^\alpha\eta^\beta,\xi^\gamma\eta^\delta)$ such that
$\alpha\delta-\beta\gamma\neq 0$
Q3: This condition confuses me a little. I guess you want to make sure you map things not in the branch locus to things not in the ramification locus, but if $\xi\eta\neq 0$ and $\alpha\beta=\delta\gamma$ then
$\zeta_1\zeta_2=(\xi\eta)^{\alpha\beta}(\xi\eta)^{\alpha\beta}\neq 0$
so I obviously don't understand the condition.
 A: Q1.  That's correct.  Covering spaces of a space $X$ can be classified by first constructing the simply connected universal covering space of $X$, which is made up of homotopy equivalence classes of paths in $X$ starting at the base point.  Then, for any subgroup $G$ of the fundamental group $\pi_1(X)$ of $X$, you can identify paths $p$ and $q$ in the universal covering space if the loop $p q^{-1}$ is in $G$.  This gives you a covering space of $X$ with fundamental group $G$.
Q2.  The number of sheets of the covering equals the index of $G$ in $\pi_1=\pi_1(X)$.  So, if $W$ is finitely sheeted, $G$ must have finite index.  Since $\pi_1={\Bbb Z}\oplus {\Bbb Z}$, a subgroup of finite index of $\pi_1$ must be generated by two linearly independent integer vectors,
$$
G=<(\alpha,\gamma),(\beta,\delta)>, \qquad \alpha, \beta, \gamma, \delta \in {\Bbb Z}, \text{ say.}
$$
${\Bbb Z}\oplus {\Bbb Z}$ is then isomorphic to $G$ via the map sending $(1,0)\mapsto (\alpha,\gamma)$ and $(0,1)\mapsto(\beta,\delta)$.  Unless the covering is trivial, $G$ will not equal $\pi_1$.
Q3.   If $\alpha\delta=\beta\gamma$, then $(\alpha,\gamma)$ and $(\beta,\delta)$ will not be linearly independent, so $G$ will not have finite index.  
