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My answer does not intend to completely solve your problem, but I wish to write down a couple of examples that (I hope) could help you gain some geometrical insight.

I will take $G$ abelian, since in this case the map $\pi_1(X \setminus B) \to G$ factors through $H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take $X=\mathbb{P}^2$ (if you want a map with fibre $\mathbb{P}^1$, just blow-up a point).

EXAMPLE 1.

$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$,

$B_1$ and $B_2$ two distinct lines intersecting in a point $p$,

$B=B_1 + B_2$.

Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the two loops $\gamma_1$ and $\gamma_2$ (around $B_1$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2=0$, hence it is isomorphic to $\mathbb{Z}$. The unique $\mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_1$, and consequently $\gamma_2$, to the generator $g$. The inertia group (stabilizer) over $B_1 + B_2$ is of course isomorphic to $\mathbb{Z}_2$, but notice that the preimage of $p$ in $Y$ is $singular$. In fact, $Y$ is isomorphic to a quadric cone in $\mathbb{P}^3$.

EXAMPLE 2.

$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$,

$B_1$, $B_2$, $B_3$ three distinct lines intersecting pairwise in three distinct points; set $p_{ij}:=B_i \cap B_j$.

$B=B_1+B_2+B_3$.

Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the three loops $\gamma_1$, $\gamma_2$ and $\gamma_3$ (around $B_1$, $B_2$ and $B_3$, respectively) with the relation $\gamma_1+\gamma_2+\gamma_3=0$, hence it is isomorphic to $\mathbb{Z}^2$. Up to permutation of the indices, the unique $\mathbb{Z}_2 \times \mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_i$ to $g_i$, for all $i=1,2,3$.

The inertia group of a point over $B_1$, distinct from $p_{12}$ and $p_{13}$, is isomorphic to $\langle g_1 \rangle$ and similarly for the other two lines. But the inertia group over the three points $p_{ij}$ is the whole group $\mathbb{Z}_2 \times \mathbb{Z}_2$.

In this case $Y$ is smooth, in fact it is isomorphic to $\mathbb{P}^2$. However, there are three "intermediate covers" $Y_1$, $Y_2$, $Y_3$, corresponding to the three non-trivial subgroups of $G$; any each $Y_i$ is then isomorphic to a quadric cone.

A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ is developed in full details.

4 deleted 1 characters in body

My answer does not intend to completely solve your problem, but I wish to write down a couple of examples that (I hope) could help you gain some geometrical insight.

I will take $G$ abelian, since in this case the map $\pi_1(X \setminus B) \to G$ factors through $H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take $X=\mathbb{P}^2$ (if you want a map with fibre $\mathbb{P}^1$, just blow-up a point).

EXAMPLE 1.

$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$,

$B_1$ and $B_2$ two distinct lines intersecting in a point $p$,

$B=B_1 + B_2$.

Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the two loops $\gamma_1$ and $\gamma_2$ (around $B_1$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2=0$, hence it is isomorphic to $\mathbb{Z}$. The unique $\mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_1$, and consequently $\gamma_2$, to the generator $g$. The inertia group (stabilizer) over $B_1 + B_2$ is of course isomorphic to $\mathbb{Z}_2$, but notice that the preimage of $p$ in $Y$ is $singular$. In fact, $Y$ is isomorphic to a quadric cone in $\mathbb{P}^3$.

EXAMPLE 2.

$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$,

$B_1$, $B_2$, $B_3$ three distinct lines intersecting pairwise in three distinct points; set $p_{ij}:=B_i \cap B_j$.

$B=B_1+B_2+B_3$.

Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the three loops $\gamma_1$, $\gamma_2$ and $\gamma_3$ (around $B_1$, $B_2$ and $B_2$, B_3$, respectively) with the relation$\gamma_1+\gamma_2+\gamma_3=0$, hence it is isomorphic to$\mathbb{Z}^2$. Up to permutation of the indices, the unique$\mathbb{Z}_2 \times \mathbb{Z}_2$-cover$Y \to X$branched on$B$is obtained by sending$\gamma_i$to$g_i$, for all$i=1,2,3$. The inertia group of a point over$B_1$, distinct from$p_{12}$and$p_{13}$, is isomorphic to$\langle g_1 \rangle$and similarly for the other two lines. But the inertia group over the three points$p_{ij}$is the whole group$\mathbb{Z}_2 \times \mathbb{Z}_2$. In this case$Y$is smooth, in fact it is isomorphic to$\mathbb{P}^2$. However, there are three "intermediate coveringscovers"$Y_1$,$Y_2$,$Y_3$, corresponding to the three non-trivial subgroups of$G$; any$Y_i$is then isomorphic to a quadric cone. A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case$G=\mathbb{Z}_2 \times \mathbb{Z}_2$is developed in full details. 3 added 32 characters in body My answer does not intend to completely solve your problem, but I wish to write down a couple of examples that (I hope) could help you gain some geometrical insight. I will take$G$abelian, since in this case the map$\pi_1(X \setminus B) \to G$factors through$H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take$X=\mathbb{P}^2$(if you want a map with fibre$\mathbb{P}^1$, just blow-up a point). EXAMPLE 1.$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$,$B_1$and$B_2$two distinct lines intersecting in a point$p$,$B=B_1 + B_2$. Then$H_1(X \setminus B, \mathbb{Z})$is generated by the two loops$\gamma_1$and$\gamma_2$(around$B_1$and$B_2$, respectively) with the relation$\gamma_1+\gamma_2=0$, hence it is isomorphic to$\mathbb{Z}$. The unique$\mathbb{Z}_2$-cover$Y \to X$branched on$B$is obtained by sending$\gamma_1$, and consequently$ \gamma_2$, to the generator$g$. The inertia group (stabilizer) over$B_1 + B_2$is of course isomorphic to$\mathbb{Z}_2$, but notice that the preimage of$p$in$Y$is$singular$. In fact,$Y$is isomorphic to a quadric cone in$\mathbb{P}^3$. EXAMPLE 2.$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$,$B_1$,$B_2$,$B_3$three distinct lines intersecting pairwise in three distinct points; set$p_{ij}:=B_i \cap B_j$.$B=B_1+B_2+B_3$. Then$H_1(X \setminus B, \mathbb{Z})$is generated by the three loops$\gamma_1$,$\gamma_2$and$\gamma_3$(around$B_1$,$B_2$and$B_2$, respectively) with the relation$\gamma_1+\gamma_2+\gamma_3=0$, hence it is isomorphic to$\mathbb{Z}^2$. Up to permutation of the indices, the unique$\mathbb{Z}_2 \times \mathbb{Z}_2$-cover$Y \to X$branched on$B$is obtained by sending$\gamma_i$to$g_i$, for all$i=1,2,3$. The inertia group of a point over$B_1$, distinct from$p_{12}$and$p_{13}$, is isomorphic to$\langle g_1 \rangle$and similarly for the other two lines. But the inertia group over the three points$p_{ij}$is the whole group$\mathbb{Z}_2 \times \mathbb{Z}_2$. In this case$Y$is smooth, in fact it is isomorphic to$\mathbb{P}^2$. However, there are three "intermediate coverings"$Y_1$,$Y_2$,$Y_3$, corresponding to the three non-trivial subgroups of$G$; any$Y_i$is then isomorphic to a quadric cone. A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case$G=\mathbb{Z}_2 \times \mathbb{Z}_2\$ is developed in full details.

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