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Since I'm used to working with algebraic $\pi_1$'s, which don't work well with surfaces, I find myself lacking geometric intuition when I attempt to do these types of purely geometric arguments. I'm hoping someone can point out a fact or two that will help me gain some insight.

Let's say we have a complex surface, $X$, and we are given a map $X \rightarrow \mathbb{A}^1_{\mathbb{C}}$ such that the fibers are all $\mathbb{P}^1_{\mathbb{C}}$'s. Let's say we are given a prescribed branch divisor, $B$, made up of the prime divisors $B_1, ..., B_r$ (I want them to be horizontal - meaning that each $B_i$ considered over any $a$ in $\mathbb{A}^1_{\mathbb{C}}$ is just a point). Assume for the sake of simplicity that $B_1$ meets $B_2$ once and no two other branch points meet.

Let's fix a complex $t$ (over which $B_1$ and $B_2$ don't meet), and pick our basepoint to be some point over $t$ that doesn't meet the branch points. A $G$-Galois cover $Y \rightarrow X$ branched at most at $B$ gives a $G$-Galois cover of $t \times_{\mathbb{A}^1_{\mathbb{C}}} X \cong \mathbb{P}^1_{\mathbb{C}}$. This cover can now be described as $\gamma_i \mapsto g_i$ where $\gamma_i$ is a loop around $B_i \times_{\mathbb{A}^1_{\mathbb{C}}} t$ (meaning the fiber of $B_i$ over $t$). This would imply in the cover of the curve $\langle g_1 \rangle$ is the inertia of some point over $B_1 \times_{\mathbb{A}^1_{\mathbb{C}}} t$ and $\langle g_2 \rangle$ is the inertia of some point over $B_2 \times_{\mathbb{A}^1_{\mathbb{C}}} t$. In fact, we know what point it is, if you fix a basepoint above. Fix such a basepoint.

It seems that this data alone determines the cover over the entire surface. Is that right? By this I mean: take $\pi_1(X \setminus B, basept)$ and map $\gamma_i$, which is a loop in our curve which lies in our surface, $X$, to $g_i$. And if it is true -- can we somehow use this to know what the inertia groups are for the prime divisors over $B_1$ and $B_2$ that meet? (after all, they don't all meet. all we know is there is some prime divisor over $B_1$ that meets some prime divisor over $B_2$.) Finding the inertia groups of those divisors in terms of whatever can I can get my hands on through curves (which means including the cover over $t$) is really my ultimate goal.

3 clarification; added 8 characters in body

Since I'm used to working with algebraic $\pi_1$'s, which don't work well with surfaces, I find myself lacking geometric intuition when I attempt to do these types of purely geometric arguments. I'm hoping someone can point out a fact or two that will help me gain some insight.

Let's say we have a complex surface, $X$, and we are given a map $X \rightarrow \mathbb{A}^1_{\mathbb{C}}$ such that the fibers are all $\mathbb{P}^1_{\mathbb{C}}$'s. Let's say we are given a prescribed branch divisor, $B$, made up of the prime divisors $B_1, ..., B_r$ (I want them to be horizontal - meaning that each comes from a fiber of $X$). B_i$considered over any$a$in$\mathbb{A}^1_{\mathbb{C}}$is just a point). Assume for the sake of simplicity that$B_1$meets$B_2$once and no two other branch points meet. Let's fix a complex$t$(over which$B_1$and$B_2$don't meet), and pick our basepoint to be some point over$t$that doesn't meet the branch points. A$G$-Galois cover$Y \rightarrow X$branched at most at$B$gives a$G$-Galois cover of$t \times_{\mathbb{A}^1_{\mathbb{C}}} X \cong \mathbb{P}^1_{\mathbb{C}}$. This cover can now be described as$\gamma_i \mapsto g_i$where$\gamma_i$is a loop around$B_i \times_{\mathbb{A}^1_{\mathbb{C}}} t$(meaning the fiber of$B_i$over$t$). This would imply in the cover of the curve$\langle g_1 \rangle$is the inertia of some point over$B_1 \times_{\mathbb{A}^1_{\mathbb{C}}} t$and$\langle g_2 \rangle$is the inertia of some point over$B_2 \times_{\mathbb{A}^1_{\mathbb{C}}} t$. In fact, we know what point it is, if you fix a basepoint above. Fix such a basepoint. It seems that this data alone determines the cover over the entire surface. Is that right? By this I mean: take$\pi_1(X \setminus B, basept)$and map$\gamma_i$, which is a loop in our curve which lies in our surface,$X$, to$g_i$. And if it is true -- can we somehow use this to know what the inertia groups are for the prime divisors over$B_1$and$B_2$that meet? (after all, they don't all meet. all we know is there is some prime divisor over$B_1$that meets some prime divisor over$B_2$.) Finding the inertia groups of those divisors in terms of whatever can I get my hands on through curves (which means including the cover over$t\$) is really my ultimate goal.

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