\section{Introduction}\label{outline}
In 1977, Gieseker \cite{Gie} proved that the moduli space of surfaces of general type is a quasi-projective variety. Unlike the case for curves, it is not irreducible. Catanese and Manetti \cite{C1,C2,Ma} proved some results about the structure and number of components of moduli spaces of general type surfaces. However, still not much is known about such moduli spaces.
 In \cite{Tei2}, Teicher defined some new invariants of surfaces, stable on connected components of moduli space. The new invariants come from the cyclic structure of the fundamental group of the complement of a branch curve.

Let $X$ be an algebraic surface of degree $n$; one can consider it as a branched cover of the projective plane $\mathbb{CP}^2$ with respect to a generic projection $\pi : X\rightarrow \mathbb{CP}^2$. The branch curve $S$ is an irreducible cuspidal plane curve of even degree, namely,
$S$ admits only nodes and ordinary cusps as its singularities. Chisini's conjecture \cite{Chisini}, which was confirmed by Kulikov \cite{Ku99, Ku08}, states that: If $S$ is the branch curve of a generic projection $\pi : X\rightarrow \mathbb{CP}^2$, then $\pi$ is determined uniquely by $S$, except for the case when $X$ is the Veronese surface $V_2$ in $\mathbb{CP}^5$. Thus, one can reduce the classification of algebraic surfaces to that of cuspidal branch curves.

It is well known that the fundamental group of the complement of the branch curve $\pi_1(\mathbb{CP}^2-S)$ (we always denote the branch curve by $S$, and denote the group $\pi_1(\mathbb{CP}^2-S)$ by $G$, for convenience) does not change when the complex structure of $X$ changes continuously. Thus, we can use such an invariant to distinguish the connected components of the corresponding moduli space of surfaces. In fact, all surfaces in the same component of the moduli space have the same homotopy type and, therefore, the same fundamental group of the complement of the branch curves.

In \cite{MoTe87a} and \cite{MoTe87}, Moishezon-Teicher showed that $\pi_1(\mathbb{CP}^2-S)$ is connected with $\pi_1(X_{\text{Gal}})$, where the surface $X_{\text{Gal}}$ is the Galois cover of $X$. Thus, we can calculate fundamental groups of some surfaces of general type that usually may be very difficult to determine. Based on this idea, Moishezon-Teicher \cite{MoTe87} constructed a series of simply connected algebraic surfaces of general type, with positive and zero indices, thereby disproving the Bogomolov Watershed conjecture: An algebraic surface of general type, with a positive index, has an infinite fundamental group.

In recent years, some works were done in the study of $\pi_1(\mathbb{CP}^2-S)$ and $\pi_1(X_{\text{Gal}})$ -- for Cayley's singularities \cite{ADFT}; for different embeddings of $\mathbb{CP}^1\times \mathbb{CP}^1$ \cite{MoTe87}; for the Veronese surfaces $V_n, ~n\geq3$ \cite{19,20}, and for $V_2$ \cite{Za37}; for the Hirzebruch surfaces $F_1(a,b),~ F_2(2,2)$ \cite{AFT09,FT08a}; for $T\times T$ where $T$ is a complex torus \cite{Osaka,ATV08}; for $K3$ surfaces \cite{pillow}; for $\mathbb{CP}^1\times T$ \cite{AFT03,AGTV02,AG04}; for $\mathbb{CP}^1\times C_g$ where $C_g$ is a curve of genus $g$ \cite{FT12}; and for certain toric surfaces \cite{Ogata}. In \cite{ADKY}, one can also find a description of computations of braid monodromy and certain quotients of $\pi_1(\mathbb{CP}^2-S)$; the motivation came from the theory of symplectic 4-manifolds. In \cite{Li08}, Liedtke computed a quotient of $\pi_1(X_{\text{Gal}})$ that depends on $\pi_1(X)$ and data from the generic projection only, thereby simplifying the computations of Moishezon, Teicher, and others.
In \cite{A-R-T}, the authors computed the fundamental groups of Galois covers of surfaces of degree $\leq 4$.

In this paper, we study the fundamental groups of the complements of the branch curves and of the Galois covers of degree $5$ surfaces with nice planar degenerations (see Theorem \ref{thm-1}, \ref{thm-2}, \ref{thm-3}, \ref{thm-5}, \ref{thm-6}, \ref{thm-7}, \ref{thm-8}, \ref{thm-4}). As an application, we give a counter-example of the main unsolved question in \cite{Li08},  which asks:
Is $C^{\text{aff}}$ (see Section \ref{application} for the definition) trivial for every generic projection of degree $n\geq5$?

This paper is organized as follows:  In Section \ref{method}, we explain the main methods and give a fundamental and necessary background  that we use in this paper. Section \ref{small} is a complete calculation of the fundamental groups of the Galois covers of all the surfaces of degrees $5$ that degenerate to `nice' planar
arrangements, i.e., those in which no three planes meet in a line. We consider the degenerations, give the braid monodromy and the group $G$, and then compute the group $\pi_1(X_{\text{Gal}})$.
The surfaces that we consider are: the Hirzebruch surface $F_1(2,1)$ (Subsection \ref{sec1}), a union of the surface $\C\P^1 \times \C\P^1$ and a plane (Subsection \ref{cp1}), a union of the Veronese surface $V_2$ and a plane (Subsection \ref{sec2}), two cases of a union  of the Cayley surface and two planes (Subsection \ref{sec4}), a union of the quartic 4-point surface with a plane (Subsection \ref{extended-4}), the quintic 5-point surface (Subsection \ref{central}), a 4-point quintic degeneration (Subsection \ref{sec7}). In Section \ref{application}, we give a counter-example of the main unsolved question in \cite{Li08}.
In the Appendix, we analyze an example of degree $4$ that is missed in \cite{A-R-T}.

\paragraph{Acknowledgements:} We would like to thank Ciro Ciliberto for useful
discussions on degeneration of surfaces and for giving us the examples of the surfaces. We also wish to thank Robert Shwartz for helpful comments.

This work is supported by the Emmy Noether Research
Institute for Mathematics, the Minerva Foundation (Germany), NSFC and PAPD.

\section{Method and scientific background}\label{method}
In this section, we describe the main methods and the fundamental background used in this paper. The computations of the fundamental groups $\pi_1(\mathbb{CP}^2-S)$ and $\pi_1(X_{\text{Gal}})$ of surfaces $X$ of degree $5$,  with at worst isolated singularities and with nice degenerations, are explained here.

We start with an algebraic surface $X$ embedded in a projective space $\mathbb{CP}^n$. We project it onto the projective plane $\mathbb{CP}^2$ to get its branch curve $S$. Because it is not easy to describe $S$, we use a method called degeneration. The definition follows.

\begin{Def}
Let $D$ be the unit disc,
and let $X, Y$ be algebraic surfaces (or more generally, algebraic varieties).
Let $p: Y \rightarrow \mathbb{CP}^n$ and $p': X \rightarrow \mathbb{CP}^n$
be projective embeddings.
We say that $p'$ is a \emph{projective degeneration} of $p$
if there exists a flat family $\pi: V \rightarrow D$
and an embedding $F:V\rightarrow D \times \mathbb{CP}^n$,
such that $F$ composed with the first projection is $\pi$,
and:
\begin{itemize}
\item[(a)] $\pi^{-1}(0) \simeq X$;
\item[(b)] there is a $t_0 \neq 0$ in $D$ such that
$\pi^{-1}(t_0) \simeq Y$;
\item[(c)] the family $V-\pi^{-1}(0) \rightarrow D-{0}$
is smooth;
\item[(d)] restricted to $\pi^{-1}(0)$, $F = {0}\times p'$
under the identification of $\pi^{-1}(0)$ with $X$;
\item[(e)] restricted to $\pi^{-1}(t_0)$, $F = {t_0}\times p$
under the identification of $\pi^{-1}(t_0)$ with $Y$.
\end{itemize}
\end{Def}

We perform a sequence of projective degenerations $X: =X_r \leadsto X_{r-1} \leadsto \cdots X_{r-i} \leadsto
X_{r-(i+1)} \leadsto \cdots \leadsto X_0$, and we refer to each step along the way as a partial
degeneration (r is the number of partial degenerations).  The surface $X_0$ is a total
degeneration of $X$ if it is a union of linear spaces of dimension $n$.

In \cite[Sec.\,12]{CCFM}, there are
examples of surfaces that can projectively degenerate to a union of planes such that no 3 planes meet in a line.
In our paper we deal with degree $5$ surfaces; this means $n=5$. For more technical details see \cite{amte1}.

\begin{eg}\label{ex:equiv}
Consider the surface $\mathbb{CP}^1\times\mathbb{CP}^1$. Take $\ell_1=\mathbb{CP}^1\times pt$
and $\ell_2=pt\times\mathbb{CP}^1$. For $a,b\in\mathbb{N}$, consider the linear combination $a\ell_1+b\ell_2$.
We embed our surface into a projective space via the linear system $|a\ell_1+b\ell_2|$. For this
example, we take $a=1$ and $b=2$ as in Figure \ref{cp1cp1}.

We first degenerate the surface into two ``squares", such that each square is homeomorphic to
$\mathbb{CP}^1\times\mathbb{CP}^1$.  Then we degenerate each square into two planes to get a total degeneration,
which is a union of planes.

\begin{figure}[ht]
\begin{center}
\scalebox{0.60}{\includegraphics{cp1cp1.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{Degeneration of $\C\P^1 \times \C\P^1$}\label{cp1cp1}
\end{figure}

\end{eg}

Now, let us consider generic projections $\pi^{(i)} : X_{i} \rightarrow \mathbb{C}\mathbb{P}^2$ for $0 \leq i \leq r$. Let
$S_{i}$ be the branch curve of the generic projection for each $i$, and let $S_{i-1}$ be a degeneration of
$S_{i}$ for $1 \leq  i \leq r$. We regenerate $S_0$ to get the regenerated branch curve $S:=S_{r}$, using the regeneration Lemmas in \cite{BGT2}.

In the following diagram (Figure \ref{Degeneration-Regeneration}), we illustrate the connections between the
significant objects $X, X_0, S$, and $S_0$.

\begin{figure}[ht]
\begin{center}
\scalebox{0.90}{\includegraphics{a02.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{Diagram of degeneration-regeneration}\label{Degeneration-Regeneration}
\end{figure}

  Now we explain in general the regeneration process.
 Say that the degree of the degenerated branch curve $S_{0}$ is $m$.
Each of the $m$ lines of $S_{0}$ should be counted as a double line
in the scheme-theoretic branch locus, as each arises from a nodal line. Another way to see this is to note that
the regeneration of $X_0$ induces a regeneration of $S_{0}$ in such a way that each point on the
typical fiber, say $c$, is replaced by two nearby points $c, c'$. This means that a line, say $j$, regenerates to two parallel lines or to a conic \cite{amte1}, and it is replaced by $j$ and $j'$. The resulting branch curve $S$ is of degree $2m$.

In full generality, the curve $S_{0}$ may have $k$-point singularities for any value of $k$.  A 1-point regenerates to a conic $(j,j')$ with a branch point. A diagonal line in a 2-point regenerates to a conic that is tangent to the line \cite[Claim, p. 8]{19}.  A diagonal in a 3-point regenerates to a conic that is tangent to the other two lines \cite{pillow}. A 4-point regenerates to two conics and two pairs of parallel lines \cite[Figure 3]{19}. A 5-point regenerates as the 4-point regenerates but with an addition of the 5th line \cite[Figure 4]{FT08}.

In the next step, a tangent line regenerates into two parallel lines, and each tangency regenerates into three
cusps, following the regeneration rules given in \cite{19}. The curve $S$ is a cuspidal curve with nodes and
branch points. A branch point is topologically locally equivalent to $y^2+x=0$ or to $y^2-x=0$. A node (resp. a
cusp) is topologically locally equivalent to $y^2-x^2=0$ (resp. $y^2-x^3=0$).

We note that 1- and 2-points were considered in \cite{pillow,Ogata,MRT,MoTe87,18,19}, 3-points were considered in \cite{pillow,moshe,ADFT}, 4-points were considered in \cite{pillow,19}, and 5-points were considered in \cite{FT08}.
The regeneration process for large $k$ can be quite difficult, but work has been done for some specific values: see \cite{FT08}, \cite{amte1}, and \cite{agst} for 5-, 6-, and 8-points, respectively.

Now we explain how to derive the related braid monodromy for $S$ and  the fundamental group of its complement in $\mathbb{CP}^2$.
 We will follow the braid monodromy algorithm of Moishezon-Teicher \cite{BGT1,BGT2}. A detailed treatment can also be found in \cite{pillow,amte1}. Note that the braid group (and the braid monodromy) is very useful to study the projective plane \cite{GG04}.

Consider the setting (Figure \ref{setup}). $S$ is an algebraic curve in $\mathbb{C}^2$, with $p =
\deg(S)$. Let $\pi: \mathbb{C}^2 \rightarrow \mathbb{C}$ be a generic projection onto the first coordinate.
Define the fiber $K(x) = \{y \mid (x,y) \in S\}$ in $S$ over a fixed point $x$, projected to the $y$-axis.
Define $N = \{x \mid \# K(x) < p \}$ and $M' = \{ s \in S \mid \pi_{\mid s} \mbox{ is not \'{e}tale at } s \}$;
note that $\pi (M') = N$. Let $\{A_j\}^q_{j=1}$ be the set of points of $M'$ and let $N = \{x_j\}^q_{j=1}$ be the
projection of $\{A_j\}^q_{j=1}$ on the $x$-axis. Recall that $\pi$ is generic, so we assume that $\# (\pi^{-1}(x) \cap M') =1$ for
every $x \in N$. Let $E$ (resp. $D$) be a closed disk on the $x$-axis (resp. the $y$-axis), such that $M'
\subset E \times D$ and $N \subset \mbox{Int}(E)$. We choose $u \in \partial E$, a real point far enough from the
set $N$, so that $x << u$ for every $x \in N$. Define $\mathbb{C}_u = \pi^{-1}(u)$ and number the points of
$K=\mathbb{C}_u\cap S$ as $\{1 , \dots , p\}$.

\begin{figure}[ht]
\begin{center}
\scalebox{0.80}{\includegraphics{setup.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{General setting}\label{setup}
\end{figure}

We now construct a g-base for the fundamental group $\pi_1(E - N, u)$. Take a set of paths
$\{\gamma_j\}^q_{j=1}$ that connect $u$ with the points $\{x_j\}^q_{j=1}$ of $N$. Now encircle each $x_j$ with
a small circle, $c_j$, oriented counterclockwise. Denote the path segment from $u$ to the boundary of this circle
by $\gamma'_j$. We define an element (a loop) in the g-base as $\delta_j = {\gamma'}_j c_j {\gamma'}^{-1}_j$.
Let $B_p[D,K]$ be the braid group, and let $H_1, \dots , H_{p-1}$ be its frame (for complete definitions, see
\cite[Section III.2]{BGT1}). The braid monodromy of $S$  is a map $\varphi: \pi_1(E - N, u)
\rightarrow B_p[D,K]$ defined as follows (see \cite{artin} in detail): every loop in $E - N$ starting at $u$ has liftings to a system of $p$
paths in $(E - N) \times D$ starting at each point of $K = 1,\ldots,p$. Projecting them to $D$, we obtain $p$
paths in $D$ defining a motion $\{1(t), \dots , p(t)\}$ (for $0 \leq t \leq 1$) of $p$ points in $D$ starting
and ending at $K$. This motion defines a braid in $B_p [D , K]$.

By the Artin Theorem \cite{BGT2}, for $j=1, \dots, q$, there exists a half-twist $Z_j \in B_p[D , K]$ and
$\varepsilon_j \in \mathbb{Z}$, such that $\varphi(\delta _j) = Z_j^{\varepsilon_j}$, where  $Z_j$ is a
half-twist.
  The following proposition allows us to determine the \emph{associated exponent} $\varepsilon$;

\begin{prop}
\label{prop:half-twist} (Moishezon-Teicher) \cite[p. 487 Proposition-Example VI.1.1.]{BGT1} Let
$E=\{x\in\mathbb{C}|\; |x|\leq 1\}$, $D=\{y\in\mathbb{C}|\; y\leq R\}$, $R>>1$, and $C$ be the curve
$y^2=x^\varepsilon$. Denote by $\varphi:\pi_1(E-N,1)\rightarrow B_2[D,\{1, -1\}]$ the braid monodromy of $C$.
Let $\gamma \in \pi_1(E-N,1)$ be a loop. Then $\varphi(\gamma)=h^\varepsilon$, where $h$ is the positive
half-twist defined by $[-1,1]$.
\end{prop}

We now explain how to get the braid monodromy around each singularity in $S$, following the notation of
Moishezon-Teicher. Let $A_j$ be a singularity in $S$ and $x_j$ its projection by $\pi$ to the $x$-axis. We
choose a point $x'_j$ next to $x_j$, such that \ $\pi^{-1}(x'_j)$ is a typical fiber.  We encircle $A_j$ with a
very small circle in such a way that the typical fiber $\pi^{-1}(x'_j)$ intersects the circle at two points, say
$a ,b$. We fix a skeleton $\xi_{x'_j}$ that connects $a$ and $b$, and denote it as $\langle a,b \rangle$. The
Lefschetz diffeomorphism $\Psi$ (see \cite{BGT1}) defines the corresponding skeleton $(\xi_{x'_j}) \Psi$ in the
typical fiber $\mathbb{C}_u$. This one defines a motion of its two endpoints, which induces a half-twist
$Z_j = \Delta \langle (\xi_{x'_j}) \Psi \rangle$. As above, $\varphi(\delta_j) = \Delta \langle (\xi_{x'_j})
\Psi \rangle^{\epsilon_j}$. The braid monodromy factorization associated to $S$ is
$\Delta^2_{p} = \prod\limits^q_{j=1} \varphi(\delta_j)$.

Using the braid monodromy factorization, we compute the fundamental group of the complement of $S$. By the van
Kampen Theorem \cite{vk}, there is a "good" geometric base $\{\G_{j}\}$ of $\pi_{1}(\C_{u} - S \cap \C_{u}, *)$,
such that the fundamental group $\pi_{1}(\mathbb{CP}^2 - S)$  of the complement of $S$ in $\mathbb{CP}^2$  is generated by the
images of $\{ \G_{j} \}$ with the relations $\varphi(\delta_{i})\G_{j} = \G_{j} \hspace{0.2cm}
\forall\, i,j$.

For our purposes, we take the curve $S$ to be the branch curve of a smooth surface $X$, which is a
cuspidal curve with nodes and branch points. Consider a small circle around a singularity. Denote by $a$ and $b$
the intersection points of the two branches with this small circle. Note that the branches meet at the singularity. Let $\G_{a},
\G_{b}$ be two non-intersecting loops in $\pi_{1}(\C_{u} - S \cap \C_{u}, *)$ around the intersection points of
the branches with the fiber $\C_{u}$ (constructed by cutting each of the paths and creating two loops that
proceed along the two parts and encircle $a$ and $b$); see \cite[Proposition-Example VI.1.1]{BGT1}. Then by the
van Kampen Theorem, we have the relations $\langle \G_{a},\G_{b} \rangle = \G_{a} \G_{b} \G_{a} \G_{b}^{-1}
\G_{a}^{-1} \G_{b}^{-1} = 1$ for a cusp, $[\G_{a},\G_{b}] = \G_{a} \G_{b}\G_{a}^{-1} \G_{b}^{-1} = 1$ for a
node, and $\G_a = \G_b$ for a branch point. These relations, with the addition of the projective relation,  generate the group $\pi_1(\mathbb{CP}^2-S)$ completely.
 In this manner, we construct a presentation for the group $\pi_1(\mathbb{CP}^2-S)$ by means of generators and
relations.

Recall now that $S$ is of degree $2m$ (after the regeneration process). Therefore the braids related to $S$ can be:
\begin{enumerate}
\item $Z_{j \;j'}$ for a branch point,
\item $Z^2_{i, j \;j'}$ is a product of two braids $Z^2_{i \; j}, Z^2_{i \;j'}$ for nodes,
\item $Z^3_{i, j \;j'}$ is a product of three braids $Z^3_{i\; j}$, $(Z^3_{i\; j})^{Z_{j\;j'}}$, $(Z^3_{i \; j})^{Z^{-1}_{j \; j'}}$ for cusps.
\end{enumerate}
We denote the generators of the group
$\pi_1(\mathbb{CP}^2-S)$ as $\G_1, \G_{1}', \dots, \G_{2m}, \G_{2m}'$. By the van Kampen
Theorem \cite{vk} we can get a presentation of $\pi_1(\mathbb{CP}^2-S)$ by means of generators $\{\G_{j}, \G_{j}'\}$ and relations of the types:
\begin{enumerate}
\item $\G_{j} = \G_{j}'$ for a branch point in a conic,
\item $[\G_{i},\G_{j}]=[\G_{i},\G_{j}']=e$ for  nodes,
\item $\langle\G_{i},\G_{j}\rangle=\langle\G_{i},\G_{j}'\rangle=\langle\G_{i},\G_{j}^{-1}\G_{j}'\G_{j}\rangle=e$ for cusps.
\end{enumerate}
To each list of relations we add the projective relation $\prod\limits_{j=m}^1 \G_{j}'\G_{j}=e$. Moreover, in some cases in the paper, we have parasitic intersections that contribute commutation relations. These intersections come from lines in $X_0$ that do not intersect, but when projecting $X_0$ onto $\C\P^2$, they will intersect. See details in \cite{MoTe87}.

Our techniques also allow us to compute fundamental groups of Galois covers. We recall from \cite{MoTe87} that if
 $f : X\rightarrow \mathbb{CP}^2$ is a generic projection of degree $n$, then ${X}_{\text{Gal}}$, the Galois cover, is defined as follows:
  $${X}_{\text{Gal}}=\overline{(X \times_{\mathbb{CP}^{2}} \ldots \times_{\mathbb{CP}^{2}} X)-\triangle},$$
where the product is taken $n$ times, and $\triangle$ is the diagonal. We apply the theorem of Moishezon-Teicher. There is an exact sequence
\begin{equation}\label{M-T}
0 \rightarrow K \rightarrow \pi_1(\mathbb{CP}^2-S) \rightarrow S_n \rightarrow 0,
\end{equation}
where the second map takes the generators of $\pi_1(\mathbb{CP}^2-S)$ to transpositions in the symmetric group $S_n$ according to the order of the lines in the degenerated surface. The fundamental group $\pi_1(X_{\text{Gal}})$ is the quotient of $K$ by the relations $\{\G_{j}^2=e,\G_{j}'^2=e\}$.
Then we obtain a presentation of the fundamental group of the Galois cover, and simplify the relations to
produce a canonical presentation that identifies with $\pi_1(X_{\text{Gal}})$, using the theory of Coxeter covers of the symmetric
groups.

\section{Calculations of the fundamental group}\label{small}
In this section, we consider the Hirzebruch surface $F_{1\,(2,1)}$ (Subsection \ref{sec1}), a union of the surface $\C\P^1 \times \C\P^1$ and a plane (Subsection \ref{cp1}), a union of the Veronese surface $V_2$ and a plane (Subsection \ref{sec2}), two cases of a union of the Cayley surface and two planes (Subsection \ref{sec4}), a union of the quartic 4-point surface with a plane (Subsection \ref{extended-4}), the quintic 5-point surface (Subsection \ref{central}), and the 4-point quintic degeneration (Subsection \ref{sec7}).

\begin{lemma} There are $8$ possible quintic degenerations, corresponding to Figures \ref{case1}, \ref{case3}, \ref{case2}, \ref{case4}, \ref{case7}, \ref{case6}, \ref{case5}, and \ref{case8}.
\end{lemma}

\begin{proof} We construct the degenerations combinatorially by gluing triangles. We add one triangle to quartic degenerations.
From the graph theory we get eight possible cases.  Figure \ref{case1} is from the Hirzebruch surface $F_{1(2,1)}$. Figure \ref{case3} is the degeneration of $\mathbb{CP}^1 \times \mathbb{CP}^1$ with a plane. Figure \ref{case2} is the degeneration of the Veronese surface with a plane. Figures \ref{case4} and \ref{case7} are the degenerations of the Cayley surface and two other planes. Figure \ref{case6} is the degeneration of the 4-point surface with a plane. Figure \ref{case5} is the degeneration of a 5-point quintic surface. Figure \ref{case8} is the degeneration of a 4-point quintic surface.

Every case is different.
It is not possible to glue them any further, because in
degree $5$ this would force us to have three planes meeting in a line.
\end{proof}

\subsection{The Hirzebruch surface $F_1(2,1)$}\label{sec1}
Let $F_1$ be the 1-th Hirzebruch surface, i.e., the projection of the vector bundle $\mathcal {O}_{\mathbb{CP}^1}(1)\bigoplus \mathcal {O}_{\mathbb{CP}^1}$.
Denote by $s$ the holomorphic section of $\mathcal {O}_{\mathbb{CP}^1}(1)$, and by $E_0\subset F_1$ the image of the section $(s,1)$ of $\mathcal {O}_{\mathbb{CP}^1}(1)\bigoplus \mathcal {O}_{\mathbb{CP}^1}$. The Picard group is always generated by the fiber $C$ and $E_0$. Note that $2C+E_0$ is very ample and thus defines an embedding $f_{\mid2C+E_0\mid}: F_1 \to \mathbb{CP}^N$. Let $F_1(2,1)=f_{\mid2C+E_0\mid}(F_1)$. By the constructions in \cite{MRT},
$F_1(2,1)$ degenerates to a union of five planes, as depicted in Figure \ref{case1}.
\vspace{0.2cm}

\begin{figure}[ht]
\begin{center}
\scalebox{0.60}{\includegraphics{winfig1.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{Degeneration of the Hirzebruch surface $F_1(2,1)$}\label{case1}
\end{figure}

\begin{thm}\label{thm-1}
The fundamental group of the Galois cover $\pi_1(X_{\text{Gal}})$ of the Hirzebruch surface $F_1(2,1)$ is trivial.
\end{thm}

\begin{proof}
See \cite[Theorem\,0.1]{MRT}

\end{proof}

\subsection{The union of $\mathbb{CP}^1 \times \mathbb{CP}^1$ degeneration and a plane}\label{cp1}
In this subsection we investigate the surface whose degeneration is depicted in Figure \ref{case3}, i.e., the union of the $\C\P^1 \times \C\P^1$ type degeneration with a plane.
In the degeneration, one can see the common edge, numbered as $1$.

\begin{figure}[H]
\begin{center}
\scalebox{0.60}{\includegraphics{winfig3.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{The union of $\C\P^1 \times \C\P^1$ degeneration and a plane}\label{case3}
\end{figure}

\begin{thm}\label{thm-2}
$\pi_1(X_{\text{Gal}})$  of surfaces with the degeneration as in Figure \ref{case3} is trivial.
\end{thm}

\begin{proof}
The branch curve $S$ in $\mathbb{CP}^2$ is an arrangement of 4 lines. We regenerate each vertex in turn and compute group $G$.

Vertices 1, 3, and 4 are 1-points; therefore, they give rise to the braids  $Z_{1\; 1^{'}}$, $Z_{4\; 4^{'}}$, and $Z_{2\; 2^{'}}$, respectively, and hence to the following relations in $G$:
\begin{equation}\label{4-1}
\Gamma_1=\Gamma_1',~~~\Gamma_4=\Gamma_4',~~~\Gamma_2=\Gamma_2'.
\end{equation}

Vertex 5  is a 2-point that gives rise to the braid monodromy factors
 $Z^3_{3\; 3^{'},4}$, $(Z_{4\; 4^{'}})^{Z^2_{3\; 3^{'},4}}$ and to the following relations:

\begin{equation}\label{4-2}
\langle\Gamma_3,\Gamma_4\rangle=\langle\Gamma_3',\Gamma_4\rangle=\langle\Gamma_3^{-1}\Gamma_3'\Gamma_3,\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{4-3}
\Gamma_4'=\Gamma_4\Gamma_3'\Gamma_3\Gamma_4\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_4^{-1}.
\end{equation}

Vertex 2 is a 3-point that regenerates to 2 lines 1, 3 tangent to a conic $(2,2')$. Its braid monodromy factors are:
\begin{eqnarray}
\widetilde{\Delta}_2 =& {(Z_{1 \; 3}^2)}^{Z^{2}_{11', 2}} \cdot (Z_{1' \; 3}^2)^{Z^{2}_{11', 2}}
\cdot {(Z_{1 \; 3'}^2)}^{Z^{2}_{11', 2}} \cdot (Z_{1' \; 3'}^2)^{Z^{2}_{11', 2}}\\
&\cdot {(Z_{2 \; 2'})}^{Z^{2}_{1 \; 1', 2} {Z}^2_{2', 3 \; 3'}} \cdot Z^3_{2', 3 \; 3'} \cdot  Z^3_{1 \; 1', 2}. \nonumber
\end{eqnarray}

$\widetilde{\Delta}_2$ gives rise the following relations in $G$:
\begin{equation}\label{4-5}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_1',\Gamma_2\rangle=\langle\Gamma_1^{-1}\Gamma_1'\Gamma_1,\Gamma_2\rangle=e,
\end{equation}
\begin{equation}\label{4-6}
\langle\Gamma_2',\Gamma_3\rangle=\langle\Gamma_2',\Gamma_3'\rangle=\langle\Gamma_2',\Gamma_3^{-1}\Gamma_3'\Gamma_3\rangle=e,
\end{equation}
\begin{equation}\label{4-7}
\Gamma_3'\Gamma_3\Gamma_2'\Gamma_3^{-1}{\Gamma_3'}^{-1}=\Gamma_2\Gamma_1'\Gamma_1\Gamma_2\Gamma_1^{-1}\Gamma_1'^{-1}\Gamma_2^{-1},
\end{equation}
\begin{equation}\label{4-8}
[\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_3]=[\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_3']=
[\Gamma_2\Gamma_1'\Gamma_2^{-1},\Gamma_3]=[\Gamma_2\Gamma_1'\Gamma_2^{-1},\Gamma_3']=e.
\end{equation}

We also have the following parasitic and projective relations:

\begin{equation}\label{4-9}
[\Gamma_1,\Gamma_4]=[\Gamma_1,\Gamma_4']=[\Gamma_1',\Gamma_4]=[\Gamma_1',\Gamma_4']=e,
\end{equation}
\begin{equation}\label{4-10}
[\Gamma_2,\Gamma_4]=[\Gamma_2,\Gamma_4']=[\Gamma_2',\Gamma_4]=[\Gamma_2',\Gamma_4']=e,
\end{equation}
\begin{equation}\label{4-11}
\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1'\Gamma_1=e.
\end{equation}

After simplification,  we have the following relations in $G$:
$$\Gamma_1=\Gamma_1',\ \Gamma_2=\Gamma_2',\ \Gamma_4=\Gamma_4',\  \Gamma_3'=\Gamma_4^{-2}\Gamma_3\Gamma_4^2,$$
$$\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_2,\Gamma_3\rangle=\langle\Gamma_3,\Gamma_4\rangle=e,$$
$$[\Gamma_1,\Gamma_4]=[\Gamma_2,\Gamma_4]=[\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_3]=e,$$
$$\Gamma_3\Gamma_4^2\Gamma_3\Gamma_2^2\Gamma_1=e.$$

 It is easy to see that $[\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_4]=e$ and $\langle\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_2\rangle=e$. Let $\{\Gamma_2\Gamma_1\Gamma_2^{-1}, \Gamma_2,\Gamma_3,\Gamma_4 \}$ be the generators of $G/\langle\Gamma_1^2,\Gamma_2^2,\Gamma_3^2,\Gamma_4^2\rangle$, then $G/\langle\Gamma_1^2,\Gamma_2^2,\Gamma_3^2,\Gamma_4^2\rangle=\{\Gamma_2\Gamma_1\Gamma_2^{-1}, \Gamma_2,\Gamma_3,\Gamma_4 | ~~S_5~~ \text{type relations}\}\cong S_5$, hence $\pi_1(X_{\text{Gal}})$ is trivial.

This completes the proof.
\end{proof}

\subsection{The union of the Veronese $V_2$ degeneration and a plane}\label{sec2}
We denote by $V_2$ the Veronese surface of order $2$.
In this subsection we introduce the degeneration of a union of the Veronese surface $V_2$ and a plane. The degeneration of this surface is a union of five planes, where $V_2$ and the plane are united along the edge, numbered as 4, see Figure \ref{case2}.

We note that the surface $V_2$ is atypical in different algebraic-geometrical theories, for example, it is the exception case for Chisini's conjecture \cite{Chisini}.

\begin{figure}[H]
\begin{center}
\scalebox{0.60}{\includegraphics{winfig2.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{The union of $V_2$ degeneration and a plane}\label{case2}
\end{figure}

\begin{thm}\label{thm-3}
$\pi_1(X_{\text{Gal}})$ of surfaces with the degeneration as in Figure \ref{case2} is
 $\mathbb{Z}^8$.
\end{thm}

\begin{proof}
The branch curve $S$ in $\mathbb{CP}^2$ is an arrangement of 4 lines. We regenerate each vertex in turn and compute group $G$.

Vertex 4 is a 1-point that gives rise to braid  $Z_{4 \; 4^{'}}$, and derives the following relation in $G$:
\begin{equation}\label{3-1}
\Gamma_4=\Gamma_4'.
\end{equation}

Vertex 1 (resp. 3) is a 2-point that gives rise to the braid monodromy factors
$Z^3_{1\; 1^{'},2}$ and $(Z_{2\; 2^{'}})^{Z^2_{1\;1^{'},2}}$ (resp. $Z^3_{2^{'},3\;3^{'}}$ and $(Z_{2\;2^{'}})^{Z^2_{2^{'},3\;3^{'}}}$).
 These braids yield the following relations:

\begin{equation}\label{3-2}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_1',\Gamma_2\rangle=\langle\Gamma_1^{-1}\Gamma_1'\Gamma_1,\Gamma_2\rangle=e,
\end{equation}
\begin{equation}\label{3-3}
\Gamma_2'=\Gamma_2\Gamma_1'\Gamma_1\Gamma_2\Gamma_1^{-1}\Gamma_1'^{-1}\Gamma_2^{-1},
\end{equation}
\begin{equation}\label{3-4}
\langle\Gamma_2',\Gamma_3\rangle=\langle\Gamma_2',\Gamma_3'\rangle=\langle\Gamma_2',\Gamma_3^{-1}\Gamma_3'\Gamma_3\rangle=e,
\end{equation}
\begin{equation}\label{3-5}
\Gamma_2=\Gamma_3'\Gamma_3\Gamma_2'\Gamma_3^{-1}{\Gamma_3'}^{-1}.
\end{equation}

Vertex 2 is a 3-point (the intersection of the lines 1, 3, 4) that regenerates to a line 3 tangent to 2 conics $(1,1'), (4,4')$. The braid monodromy corresponding to this 3-point is:

\begin{eqnarray} \label{3-6}
{\widetilde{\Delta}} = Z_{1 \, 4}^2 \cdot Z_{1' \, 4}^2
\cdot Z_{1 \, 4'}^2 \cdot Z_{1' \, 4'}^2\cdot Z_{1',3\,3'}^3\cdot(Z_{1\,1'})^{Z_{1',3\,3'}^2}\cdot(Z_{3\,3',4}^3)^{Z_{1',3\,3'}^2}
\cdot(Z_{4\,4'})^{Z_{3\,3',4}^2Z_{1',3\,3'}^2}. \nonumber
\end{eqnarray}

${\widetilde{\Delta}}$ thus gives rise to the following relations:

\begin{equation}\label{3-7}
\langle\Gamma_1',\Gamma_3\rangle=\langle\Gamma_1',\Gamma_3'\rangle=\langle\Gamma_1',\Gamma_3^{-1}\Gamma_3'\Gamma_3\rangle=e,
\end{equation}
\begin{equation}\label{3-8}
\Gamma_1=\Gamma_3'\Gamma_3\Gamma_1'\Gamma_3^{-1}{\Gamma_3'}^{-1},
\end{equation}
\begin{equation}\label{3-9}
\langle\Gamma_4,\Gamma_3'\Gamma_3\Gamma_1'\Gamma_3{\Gamma_1'}^{-1}\Gamma_3^{-1}{\Gamma_3'}^{-1}\rangle=
\langle\Gamma_4,\Gamma_3'\Gamma_3\Gamma_1'\Gamma_3'{\Gamma_1'}^{-1}\Gamma_3^{-1}{\Gamma_3'}^{-1}\rangle=
\langle\Gamma_4,\Gamma_3'\Gamma_3\Gamma_1'\Gamma_3^{-1}\Gamma_3'\Gamma_3{\Gamma_1'}^{-1}\Gamma_3^{-1}{\Gamma_3'}^{-1}\rangle=e,
\end{equation}
\begin{equation}\label{3-10}
\Gamma_4'=\Gamma_4\Gamma_3'\Gamma_3\Gamma_1'\Gamma_3'\Gamma_3{\Gamma_1'}^{-1}\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_4
\Gamma_3'\Gamma_3\Gamma_1'\Gamma_3^{-1}{\Gamma_3'}^{-1}{\Gamma_1'}^{-1}\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_4^{-1},
\end{equation}
\begin{equation}\label{3-11}
[\Gamma_1,\Gamma_4]=[\Gamma_1,\Gamma_4']=[\Gamma_1',\Gamma_4]=[\Gamma_1',\Gamma_4']=e.
\end{equation}
We also have the following parasitic and projective relations:
\begin{equation}\label{3-12}
[\Gamma_2,\Gamma_4]=[\Gamma_2,\Gamma_4']=[\Gamma_2',\Gamma_4]=[\Gamma_2',\Gamma_4']=e,
\end{equation}
\begin{equation}\label{3-13}
\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1'\Gamma_1=e.
\end{equation}

By \eqref{3-1}, \eqref{3-8}, and \eqref{3-10}, we have
$$\Gamma_4=\Gamma_1\Gamma_3'\Gamma_3\Gamma_1^{-1}\Gamma_4\Gamma_1\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_1^{-1}.$$
As $[\Gamma_1,\Gamma_4]=e$, we obtain $[\Gamma_3'\Gamma_3,\Gamma_4]=e$.

By \eqref{3-1}, \eqref{3-3}, and \eqref{3-8}, we can choose $\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_3',\Gamma_4$ to be the generators of
$G$.

After simplification, we have the following relations in $G$:
\begin{equation}\label{3-14}
\langle \Gamma_1,\Gamma_2\rangle=\langle \Gamma_2,\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_1\Gamma_3'\Gamma_3\rangle=
\langle \Gamma_2,\Gamma_3'\Gamma_3\Gamma_1\Gamma_3^{-1}{\Gamma_3'}^{-1}\rangle=e,
\end{equation}
\begin{equation}
\langle \Gamma_2,\Gamma_3'\rangle=\langle \Gamma_2,\Gamma_3'\Gamma_3{\Gamma_3'}^{-1}\rangle=\langle \Gamma_2,\Gamma_3'\Gamma_3\Gamma_3'\Gamma_3^{-1}{\Gamma_3'}^{-1}\rangle=e,
\end{equation}
\begin{equation}\label{3-15}
\langle \Gamma_1,\Gamma_3'\rangle=\langle \Gamma_1,\Gamma_3'\Gamma_3{\Gamma_3'}^{-1}\rangle=\langle \Gamma_1,\Gamma_3'\Gamma_3\Gamma_3'\Gamma_3^{-1}{\Gamma_3'}^{-1}\rangle=e,
\end{equation}
\begin{equation}\label{3-16}
\langle \Gamma_4,\Gamma_3'\rangle=\langle  \Gamma_4,\Gamma_3'\Gamma_3{\Gamma_3'}^{-1}\rangle=\langle \Gamma_4,\Gamma_3'\Gamma_3\Gamma_3'\Gamma_3^{-1}{\Gamma_3'}^{-1}\rangle=e,
\end{equation}
\begin{equation}\label{3-17}
[\Gamma_1,\Gamma_4]=[\Gamma_2,\Gamma_4]=[\Gamma_3'\Gamma_3,\Gamma_4]=e,
\end{equation}
\begin{equation}
\Gamma_2\Gamma_3^{-1}\Gamma_3'^{-1}\Gamma_1\Gamma_3'\Gamma_3\Gamma_1\Gamma_2\Gamma_1^{-1}\Gamma_3^{-1}\Gamma_3'^{-1}
\Gamma_1^{-1}\Gamma_3'\Gamma_3\Gamma_2^{-1}=\Gamma_3^{-1}\Gamma_3'^{-1}\Gamma_2\Gamma_3'\Gamma_3,
\end{equation}
\begin{equation}
\Gamma_4^2\Gamma_2\Gamma_3'\Gamma_3\Gamma_2\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_1\Gamma_3'\Gamma_3\Gamma_1=e.
\end{equation}

We define the surjective map $\varphi: G/\langle\Gamma_i^2,\Gamma_i'^{2}\rangle\to S_5$ by
$$\Gamma_1 \mapsto (1\,3),~\Gamma_2 \mapsto (2\,3),~\Gamma_3 \mapsto (3\,4),~\Gamma_3' \mapsto (3\,4)~\Gamma_4 \mapsto (4\,5).$$

\begin{figure}[H]
\begin{center}
\scalebox{0.55}{\includegraphics{winfig-cox7.eps},~~~~~~~~~~~~~~~~~~\includegraphics{winfig-cox8.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{Diagram for relations in $G$}\label{nosur}
\end{figure}

By \cite[pp.\,141]{RTV}, the generators of the kernel are  $(\Gamma_3\Gamma_1\Gamma_2\Gamma_1)^2,(\Gamma_3'\Gamma_1\Gamma_2\Gamma_1)^2$ and their conjugates.
Because $S_5$ is generated by 4 transpositions, we need to take 4 conjugates of $(\Gamma_3\Gamma_1\Gamma_2\Gamma_1)^2,(\Gamma_3'\Gamma_1\Gamma_2\Gamma_1)^2$.

Hence, $\pi_1(X_{\text{Gal}})\cong\ker \varphi\cong \mathbb{Z}^8$.

This completes the proof.
\end{proof}

\subsection{The union of the Cayley degeneration and two planes}\label{sec4}
The classification of singular cubic surfaces in $\C\P^3$ was done
in the 1860s, by Schl\"{a}fli \cite{Sch} and Cayley \cite{Cay}.
Surface XVI in Cayley's classification is now called the Cayley
cubic, and when embedded in $\C\P^3$, it is defined by the following
equation:
$$
4 (X^3 + Y^3 + Z^3 + W^3) - (X + Y + Z + W)^3=0.
$$
It has four singularities, which are ordinary double points.
Cayley noticed that this surface is a unique cubic surface having four
ordinary double points, which is the maximal possible number of
double points for a cubic surface (see, for example, Salmon's book
\cite{Sal}).

In this subsection we introduce two cases of a degeneration that is a union of the Cayley surface and two planes. We call them Type I (Figure \ref{case4}) and Type II (Figure \ref{case7}).

\subsubsection{Type I}
 In this subsection we consider the union of the Cayley surface and two planes (Type I).
 The degeneration of this surface is a union of five planes, where the common edges of the Cayley degeneration and the two planes are 4 and 5, see Figure \ref{case4}.

 The Hilbert scheme of del Pezzo surfaces of degree 5 contains configurations of planes as in Figure \ref{case4} (degenerations of Type I). Indeed, the rational normal scroll $F$ degenerates to the 4 ``top'' planes in Figure \ref{case4} (this is a toric degeneration, corresponding to the subdivision of the $(1,2)$--rectangle as Figure \ref{de4}). Then the ``bottom'' length 2 polyline $34$ in Figure \ref{case4} is a conic $C$, which degenerates here in two lines corresponding to the segments (edges) $2$ and $3$. The plane $P$ of $C$ becomes the plane spanned by edges $2$ and $3$, which is just the ``bottom plane'' in Figure \ref{case4}.

\begin{figure}[H]
\begin{center}
\scalebox{0.60}{\includegraphics{winfig4.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{Degeneration of Type I}\label{case4}
\end{figure}

The Cayley surface, its degeneration, and the related braid monodromy of the 3-points appearing there, were computed in
\cite{ADFT, A-R-T}. Here we have the same computations for the 3-points with modification of numbers, therefore we give the relations in group $G$, without the list of braids.

\begin{thm}\label{thm-5}
$\pi_1(X_{\text{Gal}})$ of surfaces with the degeneration as in Figure \ref{case4} is
 $(\mathbb{Z}_2^2\ltimes \mathbb{Z}^8)\bigoplus(\mathbb{Z}_2^2\ltimes \mathbb{Z}^8)$.
\end{thm}

\begin{proof}
The branch curve $S$ in $\mathbb{CP}^2$ is an arrangement of 5 lines. We regenerate each vertex in turn and compute group $G$.

Vertex 4 (resp. 3) is a 2-point. The braid
monodromy factors are $Z_{3',4\,4'}^3$ and $Z_{3\,3'}^{Z_{3',4\,4'}^2}$ (resp. $Z_{2',5\,5'}^3$ and $Z_{2\,2'}^{Z_{2',5\,5'}^2}$),
giving rise to the following relations in $G$:

\begin{equation}\label{5-11}
\langle\Gamma_3',\Gamma_4\rangle=\langle\Gamma_3',\Gamma_4'\rangle=\langle\Gamma_3',\Gamma_4^{-1}\Gamma_4'\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{5-12}
\Gamma_3=\Gamma_4'\Gamma_4\Gamma_3'\Gamma_4^{-1}{\Gamma_4'}^{-1},
\end{equation}
\begin{equation}\label{5-13}
\langle\Gamma_2',\Gamma_5\rangle=\langle\Gamma_2',\Gamma_5'\rangle=\langle\Gamma_2',\Gamma_5^{-1}\Gamma_5'\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{5-14}
\Gamma_2=\Gamma_5'\Gamma_5\Gamma_2'\Gamma_5^{-1}{\Gamma_5'}^{-1}.
\end{equation}

Vertex 1 is a 3-point that regenerates to a line 1 tangent to two conics $(4,4'), (5,5')$. The braid
monodromy factors are the following:
\begin{eqnarray}
\tilde{\Delta}_1 =  Z_{4 \; 5}^2 \cdot Z_{4' \; 5}^2
\cdot {\bar{Z}}_{4' \; 5'}^2 \cdot {(Z_{4 \; 5'}^2)}^{Z^{-2}_{55'}}
 \cdot Z^3_{1 \; 1', 5}\cdot {(Z_{5 \; 5'})}^{Z^{2}_{1 \; 1', 5}}\cdot
{Z}^3_{1\;1', 4} \cdot {(Z_{4 \; 4'})}^{Z^{2}_{1 \; 1', 4}}. \nonumber
\end{eqnarray}

 $\tilde{\Delta}_1$ gives rise to the following relations in $G$:

\begin{equation}\label{5-1}
[\Gamma_4,\Gamma_5]=[\Gamma_4',\Gamma_5]=[\Gamma_4,\Gamma_5^{-1}\Gamma_5'\Gamma_5]=[\Gamma_4',\Gamma_5^{-1}\Gamma_5'\Gamma_5]=e,
\end{equation}
\begin{equation}\label{5-2}
\langle\Gamma_1,\Gamma_5\rangle=\langle\Gamma_1',\Gamma_5\rangle=\langle\Gamma_1^{-1}\Gamma_1'\Gamma_1,\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{5-3}
\Gamma_5'=\Gamma_5\Gamma_1'\Gamma_1\Gamma_5\Gamma_1^{-1}\Gamma_1^{'-1}\Gamma_5^{-1},
\end{equation}
\begin{equation}\label{5-4}
\langle\Gamma_1,\Gamma_4\rangle=\langle\Gamma_1',\Gamma_4\rangle=\langle\Gamma_1^{-1}\Gamma_1'\Gamma_1,\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{5-5}
\Gamma_4'=\Gamma_4\Gamma_1'\Gamma_1\Gamma_4\Gamma_1^{-1}\Gamma_1'^{-1}\Gamma_4^{-1}.
\end{equation}

Vertex 2 is also a 3-point of Cayley type, the braid monodromy factors were computed in \cite{ADFT}, and give rise to the following relations:
\begin{equation}\label{5-10}
\Gamma_2=\Gamma_2',
\end{equation}
\begin{equation}\label{5-6}
[\Gamma_1',{\Gamma_2'}^{-1}\Gamma_3\Gamma_2']=[\Gamma_1',{\Gamma_2'}^{-1}\Gamma_3^{-1}\Gamma_3'\Gamma_3\Gamma_2']
=[\Gamma_1,{\Gamma_2'}^{-1}\Gamma_3\Gamma_2']=[\Gamma_1,{\Gamma_2'}^{-1}\Gamma_3^{-1}\Gamma_3'\Gamma_3\Gamma_2']=e,
\end{equation}
\begin{equation}\label{5-7}
\langle\Gamma_1,\Gamma_2'\rangle=\langle\Gamma_1',\Gamma_2'\rangle=\langle{\Gamma_1'}^{-1}\Gamma_1\Gamma_1',\Gamma_2'\rangle=e,
\end{equation}
\begin{equation}\label{5-8}
\Gamma_2'=\Gamma_1^{-1}\Gamma_1'^{-1}{\Gamma_2'}^{-1}\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2{\Gamma_2'}^{-1}\Gamma_3^{-1}{\Gamma_3'}^{-1}
\Gamma_2'\Gamma_1'\Gamma_1,
\end{equation}
\begin{equation}\label{5-9}
\langle\Gamma_2'\Gamma_2{\Gamma_2'}^{-1},\Gamma_3\rangle=\langle\Gamma_2'\Gamma_2{\Gamma_2'}^{-1},\Gamma_3'\rangle
=\langle\Gamma_2'\Gamma_2{\Gamma_2'}^{-1},\Gamma_3'\Gamma_3{\Gamma_3'}^{-1}\rangle=e.
\end{equation}

We also have the following parasitic and projective relations:
\begin{equation}\label{5-15}
[\Gamma_2,\Gamma_4]=[\Gamma_2,\Gamma_4']=[\Gamma_2',\Gamma_4]=[\Gamma_2',\Gamma_4']=e,
\end{equation}
\begin{equation}\label{5-16}
[\Gamma_3,\Gamma_5]=[\Gamma_3,\Gamma_5']=[\Gamma_3',\Gamma_5]=[\Gamma_3',\Gamma_5']=e,
\end{equation}
\begin{equation}\label{5-17}
\Gamma_5'\Gamma_5\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1'\Gamma_1=e.
\end{equation}

By \eqref{5-11} and \eqref{5-12} we get $\langle\Gamma_3,\Gamma_4\rangle=e$, and by \eqref{5-13} and \eqref{5-14}
we get $\langle\Gamma_2,\Gamma_5\rangle=e$.

By \eqref{5-9} and \eqref{5-10} we get
\begin{equation}
\langle\Gamma_2,\Gamma_3\rangle=e.
\end{equation}

It is easy to see that in $G$, we can choose $\{\Gamma_1,\Gamma_1',\Gamma_2,\Gamma_3,\Gamma_4,\Gamma_5\}$
to be the generators. Then $G/\langle\Gamma_i^2\rangle$  ($i=1,1',\dots,5,5'$) is generated by $\{\Gamma_1,\Gamma_1',\Gamma_2,\Gamma_3,\Gamma_4,\Gamma_5\}$
with the following relations:
\begin{equation}\label{5-18}
\Gamma_1^2=\Gamma_1'^{2}=\Gamma_2^2=\Gamma_3^2=\Gamma_4^2=\Gamma_5^2=e,
\end{equation}
\begin{equation}\label{5-19}
[\Gamma_2,\Gamma_4]=[\Gamma_3,\Gamma_5]=[\Gamma_4,\Gamma_5]=e,
\end{equation}
\begin{eqnarray}\label{5-20}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_1,\Gamma_4\rangle=\langle\Gamma_1,\Gamma_5\rangle=e,\\
\langle\Gamma_1',\Gamma_2\rangle=\langle\Gamma_1',\Gamma_4\rangle=\langle\Gamma_1',\Gamma_5\rangle=e,\nonumber
\end{eqnarray}
\begin{equation}\label{5-21}
\langle\Gamma_2,\Gamma_3\rangle=\langle\Gamma_2,\Gamma_5\rangle=\langle\Gamma_3,\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{5-22}
[\Gamma_1,{\Gamma_2}^{-1}\Gamma_3\Gamma_2]=[\Gamma_1',{\Gamma_2}^{-1}\Gamma_3\Gamma_2]=e.
\end{equation}
We define the surjective map $\rho: G/\langle\Gamma_i^2,\Gamma_i^{'2}\rangle\to S_5$ according to Figure \ref{case4} by

$$\Gamma_1 \mapsto (1\,2),\,\Gamma_1' \mapsto (1\,2),\,\Gamma_2 \mapsto (1\,3),\,\Gamma_3 \mapsto (2\,3),~\Gamma_4 \mapsto (2\,4),~\Gamma_5 \mapsto (1\,5).$$

Then the diagrams for the above map $\rho$ are as shown in Figure \ref{cox1}.

\begin{figure}[H]
\begin{center}
\scalebox{0.55}{\includegraphics{winfig-cox1.eps},~~~~~~~~~~~~~~~~~~\includegraphics{winfig-cox2.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{Diagrams for $\rho$ }\label{cox1}
\end{figure}

By \cite[pp.\,141]{RTV}, the subgroup of $\text{ker} \rho$ contributed by the left part of Figure \ref{cox1} is generated by $\Gamma_1\Gamma_2\Gamma_3\Gamma_2,\,\Gamma_2\Gamma_3\Gamma_1\Gamma_3,\,\Gamma_3\Gamma_1\Gamma_2\Gamma_1$, arising from the cycle
$\Gamma_1,\,\Gamma_2,\,\Gamma_3$, and by the conjugates of $(\Gamma_2\Gamma_1\Gamma_5\Gamma_1)^2$ and $(\Gamma_3\Gamma_1\Gamma_4\Gamma_1)^2$,
arising from the relations involving three edges meeting at a vertex.

Because the cycle only involves three vertices and because $S_3$ is generated
by two transpositions, it is enough to take the preimages of two transpositions that
arise from the cyclic relation. In particular, all elements of the kernel that arise from the
cyclic relation are generated by $\Gamma_1\Gamma_2\Gamma_3\Gamma_2,\,\Gamma_2\Gamma_3\Gamma_1\Gamma_3$ and their conjugates.
It is easy to see that they have order 2 and commute. Then the subgroup generated by these 2 elements is $\mathbb{Z}_2^2$.

By \cite[pp.\,141]{RTV}, the generators of the kernel that arise from three edges meeting at a vertex are $(\Gamma_2\Gamma_1\Gamma_5\Gamma_1)^2, (\Gamma_3\Gamma_1\Gamma_4\Gamma_1)^2$ and their conjugates. Because $S_5$
is generated by 4 transpositions, we need to take 4 conjugates of $(\Gamma_2\Gamma_1\Gamma_5\Gamma_1)^2, (\Gamma_3\Gamma_1\Gamma_4\Gamma_1)^2$.

Hence, $\pi_1(X_{\text{Gal}})=\ker \rho\cong (\mathbb{Z}_2^2\ltimes \mathbb{Z}^8)\bigoplus(\mathbb{Z}_2^2\ltimes \mathbb{Z}^8)$.

This completes the proof.

\end{proof}

\subsubsection{Type II} \label{typeII}
In this subsection we consider the union of the Cayley surface and two planes (Type II).
 The degeneration of this surface is a union of five planes, where the common edges of the Cayley degeneration and one of the planes is 3, and two planes have a common edge that is 5, see Figure \ref{case7}.

The Hilbert scheme of del Pezzo surfaces of degree 5 contains configurations of planes as in Figure  \ref{case7} (degenerations of Type II). Indeed, the Veronese  $V$  degenerates to the 4 ''rightmost'' planes (except a plane with vertices 123) in Figure
 \ref{case7} (this is a toric degeneration).
Then in vetex 3, the length 2 line $13$ of $V$ corresponds to a conic $C$, which degenerates here in two lines corresponding to the segments (edges) $1$ and $2$. The plane $P$ of $C$ becomes the plane spanned by edges $1$ and $2$, which is just a plane with vertices 123 in Figure  \ref{case7}.

\begin{figure}[H]
\begin{center}
\scalebox{0.60}{\includegraphics{winfig7.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{Degeneration of Type II}\label{case7}
\end{figure}

\begin{thm}\label{thm-6}
$\pi_1(X_{\text{Gal}})$ of surfaces with the degeneration as in Figure \ref{case7} is $\mathbb{Z}_2^2\ltimes \mathbb{Z}^4$.
\end{thm}

\begin{proof}

The branch curve $S$ in $\mathbb{CP}^2$ is an arrangement of 5 lines. We regenerate each vertex in turn and compute group $G$.

Vertices 1 and 5 give rise to the braids  $Z_{1\, 1^{'}}$ and $Z_{5\, 5^{'}}$ respectively, and hence to the relations:
\begin{equation}\label{8-1}
\Gamma_1=\Gamma_1',~~~~~~\Gamma_5=\Gamma_5'.
\end{equation}

Vertex 3 is a 2-point with the braid monodromy factors being $Z_{2',3\,3'}^3$ and $(Z_{2\,2^{'}})^{Z^2_{2^{'},3\,3^{'}}}$. Then we get the following relations:
\begin{equation}\label{8-7}
\langle\Gamma_2',\Gamma_3\rangle=\langle\Gamma_2',\Gamma_3'\rangle=\langle\Gamma_2',\Gamma_3^{-1}\Gamma_3'\Gamma_3\rangle=e,
\end{equation}
\begin{equation}\label{8-8}
\Gamma_2=\Gamma_3'\Gamma_3\Gamma_2'\Gamma_3^{-1}{\Gamma_3'}^{-1}.
\end{equation}

Vertex 2  is also a 3-point of Cayley type, the braid monodromy factors were computed in \cite{ADFT}, and give rise to the following relations:
\begin{equation}\label{8-2}
\Gamma_2=\Gamma_2',
\end{equation}
\begin{equation}\label{8-3}
[\Gamma_1',{\Gamma_2'}^{-1}\Gamma_4\Gamma_2']=[\Gamma_1',{\Gamma_2'}^{-1}\Gamma_4^{-1}\Gamma_4'\Gamma_4\Gamma_2']
=[\Gamma_1,{\Gamma_2'}^{-1}\Gamma_4\Gamma_2']=[\Gamma_1,{\Gamma_2'}^{-1}\Gamma_4^{-1}\Gamma_4'\Gamma_4\Gamma_2']=e,
\end{equation}
\begin{equation}\label{8-4}
\langle\Gamma_1,\Gamma_2'\rangle=\langle\Gamma_1',\Gamma_2'\rangle=\langle{\Gamma_1'}^{-1}\Gamma_1\Gamma_1',\Gamma_2'\rangle=e,
\end{equation}
\begin{equation}\label{8-5}
\Gamma_2'=\Gamma_1^{-1}\Gamma_1'^{-1}{\Gamma_2'}^{-1}\Gamma_4'\Gamma_4\Gamma_2'\Gamma_2{\Gamma_2'}^{-1}\Gamma_4^{-1}{\Gamma_4'}^{-1}
\Gamma_2'\Gamma_1'\Gamma_1,
\end{equation}
\begin{equation}\label{8-6}
\langle\Gamma_2'\Gamma_2{\Gamma_2'}^{-1},\Gamma_4\rangle=\langle\Gamma_2'\Gamma_2{\Gamma_2'}^{-1},\Gamma_4'\rangle
=\langle\Gamma_2'\Gamma_2{\Gamma_2'}^{-1},\Gamma_4'\Gamma_4{\Gamma_4'}^{-1}\rangle=e.
\end{equation}

Vertex 4 is a 3-point regenerating to a line 3 tangent to two conics $(4,4'),~(5,5')$ (as line 4,\,5 are diagonals) with the braid monodromy factors giving rise to the following relations in $G$:
\begin{equation}\label{8-9}
[\Gamma_4,\Gamma_5]=[\Gamma_4',\Gamma_5]=[\Gamma_4,\Gamma_5^{-1}\Gamma_5'\Gamma_5]=[\Gamma_4',\Gamma_5^{-1}\Gamma_5'\Gamma_5]=e,
\end{equation}
\begin{equation}\label{8-10}
\langle\Gamma_3,\Gamma_5\rangle=\langle\Gamma_3',\Gamma_5\rangle=\langle\Gamma_3^{-1}\Gamma_3'\Gamma_3,\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{8-11}
\Gamma_5'=\Gamma_5\Gamma_3'\Gamma_3\Gamma_5\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_5^{-1},
\end{equation}
\begin{equation}\label{8-12}
\langle\Gamma_3,\Gamma_4\rangle=\langle\Gamma_3',\Gamma_4\rangle=\langle\Gamma_3^{-1}\Gamma_3'\Gamma_3,\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{8-13}
\Gamma_4'=\Gamma_4\Gamma_3'\Gamma_3\Gamma_4\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_4^{-1}.
\end{equation}

We also have the following parasitic and projective relations:
\begin{equation}\label{8-14}
[\Gamma_1,\Gamma_3]=[\Gamma_1',\Gamma_3]=[\Gamma_1,\Gamma_3']=[\Gamma_1',\Gamma_3']=e,
\end{equation}
\begin{equation}\label{8-15}
[\Gamma_1,\Gamma_5]=[\Gamma_1',\Gamma_5]=[\Gamma_1,\Gamma_5']=[\Gamma_1',\Gamma_5']=e,
\end{equation}
\begin{equation}\label{8-16}
[\Gamma_2,\Gamma_5]=[\Gamma_2',\Gamma_5]=[\Gamma_2,\Gamma_5']=[\Gamma_2',\Gamma_5']=e,
\end{equation}
\begin{equation}\label{8-17}
\Gamma_5'\Gamma_5\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1'\Gamma_1=e.
\end{equation}

By the relations above, we can choose $\Gamma_i,~i=1, \dots, 5$ to be generators of $G$. The relations in $G/\langle \Gamma_i^2\rangle$ are the following:
\begin{equation}\label{8-19}
[\Gamma_1,\Gamma_3]=[\Gamma_1,\Gamma_5]=[\Gamma_2,\Gamma_5]=[\Gamma_1,\Gamma_2\Gamma_4\Gamma_2]=[\Gamma_4,\Gamma_5]=e,
\end{equation}
\begin{equation}\label{8-20}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_2,\Gamma_3\rangle=\langle\Gamma_2,\Gamma_4\rangle
=\langle\Gamma_3,\Gamma_4\rangle=\langle\Gamma_3,\Gamma_5\rangle=e.
\end{equation}

We define the surjective map $\rho': G/\langle\Gamma_i^2\rangle\to S_5$ by
$$\Gamma_1 \mapsto (1\,2),~\Gamma_2 \mapsto (1\,3),~\Gamma_3 \mapsto (3\,4),~\Gamma_4 \mapsto (2\,3),~\Gamma_5 \mapsto (4\,5).$$

The diagram for the map $\rho'$ is Figure \ref{cox3}.
\begin{figure}[H]
\begin{center}
\scalebox{0.55}{\includegraphics{winfig-cox5.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{Diagram for $\rho'$}\label{cox3}
\end{figure}

By \cite[pp.\,141]{RTV}, $\text{ker} \rho'$ is generated by $\Gamma_1\Gamma_2\Gamma_4\Gamma_2,\,\Gamma_2\Gamma_4\Gamma_1\Gamma_4,\,\Gamma_4\Gamma_1\Gamma_2\Gamma_1$, arising from the cycle
$\Gamma_1,\,\Gamma_2,\,\Gamma_4$, and  by the conjugates of $(\Gamma_2\Gamma_3\Gamma_4\Gamma_3)^2$, arising from the relation involving three  edges meeting at a vertex.

Because the cycle only involves three vertices and because $S_3$ is generated
by two transpositions, it is enough to take the preimages of two transpositions that
arise from the cyclic relation. In particular, all elements of the kernel that arise from the
cyclic relation are generated by $\Gamma_1\Gamma_2\Gamma_4\Gamma_2,\,\Gamma_2\Gamma_4\Gamma_1\Gamma_4$ and their conjugates.
It is easy to see that they have order 2 and commute. Then the subgroup generated by these 2 elements is $\mathbb{Z}_2^2$.

By \cite[pp.\,141]{RTV}, the generators of the kernel that arise from three edges meeting at a vertex are    $(\Gamma_2\Gamma_3\Gamma_4\Gamma_3)^2$ and its conjugates. Because $S_5$ is generated by 4 transpositions, we need to take 4 conjugates of $(\Gamma_2\Gamma_3\Gamma_4\Gamma_3)^2$.

Hence, $\pi_1(X_{\text{Gal}})=\ker \rho'\cong\mathbb{Z}_2^2\ltimes \mathbb{Z}^4$.

This completes the proof.
\end{proof}

\subsection{A union of the 4-point quartic degeneration and a plane}\label{extended-4}
In this subsection, we take a degeneration of a quartic surface to a plane arrangement with a $4$-point.

The Hilbert scheme of del Pezzo surfaces of degree 5 contains reducible surfaces that consist in a general degree 4 complete intersection $F$ of type $(2,2)$ in $\mathbb P^ 4$ (which is itself a del Pezzo surface), plus a plane $P$ meeting $F$ along a line.

We can show that the configuration shown in Figure \ref{case6} is a limit of smooth del Pezzo surfaces of degree 5.
Indeed, the surface $F$ can degenerate to the union of 4 planes filling up the subdivided square in Figure \ref{case6}:
Simply degenerate the two quadrics cutting out $F$ in two general quadric cones with the same vertex (which is then the 4-tuple point $p$ that is common to the 4 planes). In this degeneration the lines contained in $F$, which are 16, are mapped to the following configuration of lines: Take a general quadric $Q$ in $\mathbb P^ 4$, it cuts each of the 4 lines through $p$ in two points; then on each of the 4 planes take the four lines pairwise joining the pairs of points not on the same line through $p$, these are the 16 limits in question. Then a general plane through a line on $F$ goes to a plane like the ``rightmost top'' plane in Figure \ref{case6}.

\begin{figure}[H]
\begin{center}
\scalebox{0.60}{\includegraphics{winfig6.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{A union of the 4-point quartic degeneration and a plane}\label{case6}
\end{figure}

\begin{thm}\label{thm-7}
$\pi_1(X_{\text{Gal}})$ of surfaces with the degeneration as in Figure \ref{case6} is trivial.
\end{thm}

\begin{proof}
The branch curve $S$ in $\mathbb{CP}^2$ is an arrangement of 5 lines. We regenerate each vertex in turn and compute group $G$.

Vertices 1 and 2 are 1-points, giving the braids  $Z_{1\, 1^{'}}$ and $Z_{2\, 2^{'}}$ respectively, and hence the following relations in $G$:
\begin{equation}\label{7-1}
\Gamma_1=\Gamma_1',~~~~~~\Gamma_2=\Gamma_2'.
\end{equation}

Vertex 3  is a 2-point and it gives the braid monodromy factors
 $Z^3_{4\,4^{'},5}$ and $(Z_{5\,5^{'}})^{Z^2_{4\,4^{'},5}}$. The relations in $G$ are:
\begin{equation}\label{7-2}
\langle\Gamma_4,\Gamma_5\rangle=\langle\Gamma_4',\Gamma_5\rangle=\langle\Gamma_4^{-1}\Gamma_4'\Gamma_4,\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{7-3}
\Gamma_5'=\Gamma_5\Gamma_4'\Gamma_4\Gamma_5\Gamma_4^{-1}{\Gamma_4'}^{-1}\Gamma_5^{-1}.
\end{equation}

Vertex 4 is also a 2-point and it gives the braid monodromy factors
 $Z^3_{3\,3^{'},5}$ and $(Z_{5\,5^{'}})^{Z^2_{3\,3^{'},5}}$. The relations in $G$ are:
\begin{equation}\label{7-4}
\langle\Gamma_3,\Gamma_5\rangle=\langle\Gamma_3',\Gamma_5\rangle=\langle\Gamma_3^{-1}\Gamma_3'\Gamma_3,\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{7-5}
\Gamma_5'=\Gamma_5\Gamma_3'\Gamma_3\Gamma_5\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_5^{-1}.
\end{equation}

The braid monodromy factors corresponding to the 4-point (vertex 5) were computed in \cite{Ogata}.
These braids give rise to the following relations in $G$:
\begin{equation}\label{7-6}
\langle\Gamma_1',\Gamma_2\rangle=\langle\Gamma_1',\Gamma_2'\rangle=\langle\Gamma_1',\Gamma_2^{-1}\Gamma_2'\Gamma_2\rangle=e,
\end{equation}
\begin{equation}\label{7-7}
\langle\Gamma_3,\Gamma_4\rangle=\langle\Gamma_3',\Gamma_4\rangle=\langle\Gamma_3^{-1}\Gamma_3'\Gamma_3,\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{7-8}
[\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2^{-1}{\Gamma_2'}^{-1},\Gamma_4] = e,
\end{equation}
\begin{equation}\label{7-9}
[\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2^{-1}{\Gamma_2'}^{-1},\Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_4^{-1}\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3] = e,
\end{equation}
\begin{equation}\label{7-10}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_1,\Gamma_2'\rangle=\langle\Gamma_1,\Gamma_2^{-1}\Gamma_2'\Gamma_2\rangle=e,
\end{equation}
\begin{equation}\label{7-11}
\langle\Gamma_3,\Gamma_4^{-1}\Gamma_4'\Gamma_4\rangle=\langle\Gamma_3',\Gamma_4^{-1}\Gamma_4'\Gamma_4\rangle=\langle\Gamma_3^{-1}\Gamma_3'\Gamma_3,\Gamma_4^{-1}\Gamma_4'\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{7-12}
[\Gamma_2'\Gamma_2\Gamma_1\Gamma_2^{-1}{\Gamma_2'}^{-1},\Gamma_4^{-1}\Gamma_4'\Gamma_4] = e,
\end{equation}
\begin{equation}\label{7-13}
[\Gamma_2'\Gamma_2\Gamma_1\Gamma_2^{-1}{\Gamma_2'}^{-1}, \Gamma_3^{-1}{\Gamma_3'}^{-1}\Gamma_4^{-1}{\Gamma_4'}^{-1}\Gamma_4\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3] = e,
\end{equation}
\begin{equation}\label{7-14}
\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2\Gamma_1'^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1} = \Gamma_4\Gamma_3'\Gamma_4^{-1},
\end{equation}
\begin{equation}\label{7-15}
\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2'\Gamma_1'^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1} = \Gamma_4\Gamma_3'\Gamma_3{\Gamma_3'}^{-1}\Gamma_4^{-1},
\end{equation}
\begin{equation}\label{7-16}
\Gamma_2'\Gamma_2\Gamma_1\Gamma_2\Gamma_1^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1} = \Gamma_4^{-1}\Gamma_4'\Gamma_4\Gamma_3'\Gamma_4^{-1}{\Gamma_4'}^{-1}\Gamma_4,
\end{equation}
\begin{equation}\label{7-17}
\Gamma_2'\Gamma_2\Gamma_1\Gamma_2'\Gamma_1^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1} = \Gamma_4^{-1}\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3\Gamma_3'^{-1}\Gamma_4^{-1}{\Gamma_4'}^{-1}\Gamma_4.
\end{equation}

We also have the following parasitic and projective relations:
\begin{equation}\label{7-18}
[\Gamma_1, \Gamma_5]=[\Gamma_1', \Gamma_5]=[\Gamma_1, \Gamma_5']=[\Gamma_1', \Gamma_5']=e,
\end{equation}
\begin{equation}\label{7-19}
[\Gamma_2, \Gamma_5]=[\Gamma_2', \Gamma_5]=[\Gamma_2, \Gamma_5']=[\Gamma_2', \Gamma_5']=e,
\end{equation}
\begin{equation}\label{7-20}
\Gamma_5'\Gamma_5\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1'\Gamma_1=e.
\end{equation}

By \eqref{7-14} and \eqref{7-15}, we have $\Gamma_3=\Gamma_3'$.

Combining it with \eqref{7-7}, \eqref{7-14}, and \eqref{7-15}, we get
\begin{equation}\label{7-21}
\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2\Gamma_1'^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1} = \Gamma_3^{-1}\Gamma_4\Gamma_3.
\end{equation}

By \eqref{7-11} and \eqref{7-16} we have
\begin{equation}\label{7-22}
\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2\Gamma_1'^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1} = \Gamma_3^{-1}\Gamma_4^{-1}\Gamma_4'\Gamma_4\Gamma_3.
\end{equation}

It follows that $\Gamma_4=\Gamma_4'$.

Thus, we get the following relations in $G$:
\begin{equation}\label{7-23}
\Gamma_1=\Gamma_1',~~~\Gamma_2=\Gamma_2',~~~\Gamma_3=\Gamma_3',~~~\Gamma_4
=\Gamma_4',~~~\Gamma_5'=\Gamma_4^{-2}\Gamma_5\Gamma_4^2,
\end{equation}
\begin{equation}\label{7-24}
\Gamma_2\Gamma_1\Gamma_2^{-1}=\Gamma_4\Gamma_3\Gamma_4^{-1},
\end{equation}
\begin{equation}\label{7-25}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_2,\Gamma_4\rangle=\langle\Gamma_3,\Gamma_4\rangle
=\langle\Gamma_3,\Gamma_5\rangle=\langle\Gamma_4,\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{7-26}
[\Gamma_1,\Gamma_5]=[\Gamma_2,\Gamma_3]=[\Gamma_2,\Gamma_5]=
[\Gamma_2^2\Gamma_1\Gamma_2^{-2},\Gamma_4]=
[\Gamma_2^2\Gamma_1\Gamma_2^{-2},\Gamma_3^{-2}\Gamma_4\Gamma_3^2]=e,
\end{equation}
\begin{equation}\label{7-27}
\Gamma_5\,\Gamma_4^2\,\Gamma_5\,\Gamma_3^2\,\Gamma_2^2\,\Gamma_1^2=e.
\end{equation}

In $G/\langle\Gamma_i^2\rangle,~i=1,2,3,4,5$, it is easy to see that the generators are $\{\Gamma_1,\Gamma_2,\Gamma_4,\Gamma_5\}$, and the relations are the following:
\begin{equation}\label{7-28}
\Gamma_1^2=\Gamma_2^2=\Gamma_3^2=\Gamma_4^2=\Gamma_5^2=e,
\end{equation}
\begin{equation}\label{7-29}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_2,\Gamma_4\rangle=\langle\Gamma_4,\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{7-30}
[\Gamma_1,\Gamma_4]=[\Gamma_1,\Gamma_5]=[\Gamma_2,\Gamma_5]=e.
\end{equation}

It is easy to see that $G\cong S_5$, thus $\pi_1(X_{\text{Gal}})$ is trivial.

This completes the proof.

\end{proof}

\subsection{A 5-point quintic degeneration}\label{central}
In this subsection we consider a quintic whose degeneration is depicted in Figure \ref{case5}. This degeneration gives a 5-point, in this case an intersection of five lines and also five planes.
According to \cite{CCFM}, the configuration in Figure \ref{case5} is a Zappatic surface of type $E_5$.
It is well known that a general del Pezzo $S$ of degree $n$ in $\mathbb P^ n$ can degenerate to a configuration of points of type $E_n, n=3,\ldots, 9$. Firstly, we degenerate $S$ \cite{CLM} to the cone over a general hyperplane section (elliptic curve) of $S$.
Secondly, we degenerate the hyperplane section to a cycle of lines.

\begin{figure}[htbp]
\begin{center}
\scalebox{0.68}{\includegraphics{winfig5.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.3cm}
\caption{A 5-point quintic degeneration}\label{case5}

\end{figure}

The regeneration and the related braid monodromy of the 5-points were done in \cite{FT08}.  We use the result from \cite[Corollary\,2.5]{FT08} to give the braid monodromy relating to the 5-point.

\begin{thm}\label{thm-8}
$\pi_1(X_{\text{Gal}})$ of surface with a 5-point quintic degeneration as in Figure \ref{case5} is trivial.
\end{thm}

\begin{proof}
The branch curve $S$ in $\mathbb{CP}^2$ is an arrangement of 5 lines. We regenerate each vertex in turn and compute group $G$.

Vertices 1, 2, 3, 4, and 5 are 1-points, therefore the related braid monodromy factors are  $Z_{1\,1^{'}}$, $Z_{2\,2^{'}}$, $Z_{4\,4^{'}}$, $Z_{5\,5^{'}}$, and $Z_{3\,3^{'}}$, respectively, and hence we have the following relations:
\begin{equation}\label{6-1}
\Gamma_1=\Gamma_1',~~~\Gamma_2=\Gamma_2',~~~\Gamma_3=\Gamma_3',~~~\Gamma_4=\Gamma_4',~~~\Gamma_5=\Gamma_5'.
\end{equation}

According to \cite[Corollary\,2.5]{FT08},
the braid monodromy corresponding to the 5-point yields the following relations in $G$:
\begin{equation}\label{6-2}
[\Gamma_3,\Gamma_4]=[\Gamma_3',\Gamma_4]=e,
\end{equation}
\begin{equation}\label{6-3}
\langle\Gamma_4',\Gamma_5\rangle=\langle\Gamma_4',\Gamma_5'\rangle=\langle\Gamma_4',\Gamma_5^{-1}\Gamma_5'\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{6-4}
\langle\Gamma_2,\Gamma_4\rangle=\langle\Gamma_2',\Gamma_4\rangle=\langle\Gamma_2^{-1}\Gamma_2'\Gamma_2,\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{6-5}
[\Gamma_4\Gamma_3\Gamma_4^{-1},\Gamma_5'\Gamma_5\Gamma_4'\Gamma_5^{-1}{\Gamma_5'}^{-1}]=
[\Gamma_4\Gamma_3'\Gamma_4^{-1},\Gamma_5'\Gamma_5\Gamma_4'\Gamma_5^{-1}{\Gamma_5'}^{-1}]=e,
\end{equation}
\begin{equation}\label{6-6}
\Gamma_4\Gamma_2'\Gamma_2\Gamma_4\Gamma_2^{-1}{\Gamma_2'}^{-1}\Gamma_4^{-1}=\Gamma_5'\Gamma_5\Gamma_4'\Gamma_5^{-1}{\Gamma_5'}^{-1},
\end{equation}
\begin{equation}\label{6-7}
[\Gamma_1,\Gamma_4]=[\Gamma_1',\Gamma_4]=
[\Gamma_1,\Gamma_5'\Gamma_5\Gamma_4'\Gamma_5^{-1}{\Gamma_5'}^{-1}]=[\Gamma_1',\Gamma_5'\Gamma_5\Gamma_4'\Gamma_5^{-1}{\Gamma_5'}^{-1}]=e,
\end{equation}
\begin{equation}\label{6-8}
\langle\Gamma_1',\Gamma_2\rangle=\langle\Gamma_1',\Gamma_2'\rangle=\langle\Gamma_1',\Gamma_2^{-1}\Gamma_2'\Gamma_2\rangle=e,
\end{equation}
\begin{equation}\label{6-9}
\langle\Gamma_4\Gamma_3\Gamma_4^{-1},\Gamma_5\rangle=\langle\Gamma_4\Gamma_3'\Gamma_4^{-1},\Gamma_5\rangle=\langle\Gamma_4\Gamma_3^{-1}\Gamma_3'\Gamma_3\Gamma_4^{-1},\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{6-10}
\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2\Gamma_1'^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1}=\Gamma_4^{-1}\Gamma_5\Gamma_4\Gamma_3'\Gamma_4^{-1}\Gamma_5^{-1}\Gamma_4,
\end{equation}
\begin{equation}\label{6-11}
\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2'\Gamma_1'^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1}=
\Gamma_4^{-1}\Gamma_5\Gamma_4\Gamma_3'\Gamma_3\Gamma_3'^{-1}\Gamma_4^{-1}\Gamma_5^{-1}\Gamma_4,
\end{equation}
\begin{equation}\label{6-12}
[\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2^{-1}{\Gamma_2'}^{-1},\Gamma_4^{-1}\Gamma_5\Gamma_4]=e,
\end{equation}
\begin{equation}\label{6-13}
[\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1'\Gamma_2^{-1}{\Gamma_2'}^{-1}\Gamma_3^{-1}\Gamma_3'^{-1},\Gamma_4^{-1}\Gamma_5^{-1}\Gamma_5'\Gamma_5\Gamma_4]=e,
\end{equation}
\begin{equation}\label{6-14}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_1,\Gamma_2'\rangle=\langle\Gamma_1,\Gamma_2^{-1}\Gamma_2'\Gamma_2\rangle=e,
\end{equation}
\begin{equation} \label{6-15}
\begin{split}
\langle\Gamma_4\Gamma_3\Gamma_4^{-1},\Gamma_5^{-1}\Gamma_5'\Gamma_5\rangle&=\langle\Gamma_4\Gamma_3'\Gamma_4^{-1},
\Gamma_5^{-1}\Gamma_5'\Gamma_5\rangle\\
&=\langle\Gamma_4\Gamma_3^{-1}\Gamma_3'\Gamma_3\Gamma_4^{-1},\Gamma_5^{-1}\Gamma_5'\Gamma_5\rangle=e,
\end{split}
\end{equation}
\begin{equation}\label{6-16}
\Gamma_2'\Gamma_2\Gamma_1\Gamma_2\Gamma_1^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1}=
\Gamma_4^{-1}\Gamma_5^{-1}\Gamma_5'\Gamma_5\Gamma_4\Gamma_3'\Gamma_4^{-1}\Gamma_5^{-1}{\Gamma_5'}^{-1}\Gamma_5\Gamma_4,
\end{equation}
\begin{equation}\label{6-17}
\Gamma_2'\Gamma_2\Gamma_1\Gamma_2'\Gamma_1^{-1}\Gamma_2^{-1}{\Gamma_2'}^{-1}=
\Gamma_4^{-1}\Gamma_5^{-1}\Gamma_5'\Gamma_5\Gamma_4\Gamma_3'\Gamma_3\Gamma_3'^{-1}\Gamma_4^{-1}\Gamma_5^{-1}{\Gamma_5'}^{-1}\Gamma_5\Gamma_4,
\end{equation}
\begin{equation}\label{6-18}
[\Gamma_2'\Gamma_2\Gamma_1\Gamma_2^{-1}{\Gamma_2'}^{-1},\Gamma_4^{-1}\Gamma_5^{-1}\Gamma_5'\Gamma_5\Gamma_4]=e,
\end{equation}
\begin{equation}\label{6-19}
[\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1\Gamma_2^{-1}{\Gamma_2'}^{-1}\Gamma_3^{-1}\Gamma_3'^{-1},\Gamma_4^{-1}\Gamma_5^{-1}
{\Gamma_5'}^{-1}\Gamma_5\Gamma_5'\Gamma_5\Gamma_4]=e.
\end{equation}

We also have the following projective relation:
\begin{equation}\label{6-20}
\Gamma_5'\Gamma_5\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1'\Gamma_1=e.
\end{equation}

After we simplify the relations in $G $, the generators of  $G$ are $\{\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4,\Gamma_5\}$.
The relations are the following:
\begin{equation}\label{6-21}
[\Gamma_1,\Gamma_4]=[\Gamma_1,\Gamma_5]=[\Gamma_2,\Gamma_5]=[\Gamma_3,\Gamma_4]=[\Gamma_4,\Gamma_2^2\Gamma_5^2]=e,
\end{equation}
\begin{equation}\label{6-22}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_2,\Gamma_4\rangle=\langle\Gamma_3,\Gamma_5\rangle=\langle\Gamma_4,\Gamma_5\rangle=e,
\end{equation}
\begin{equation}\label{6-23}
\Gamma_2\Gamma_1\Gamma_2^{-1}=\Gamma_3^{-1}\Gamma_4^{-1}\Gamma_5\Gamma_4\Gamma_3,
\end{equation}
\begin{equation}\label{6-24}
\Gamma_5^2\,\Gamma_4^2\,\Gamma_3^2\,\Gamma_2^2\,\Gamma_1^2=e.
\end{equation}

Note that by \eqref{6-23}, $\Gamma_2=\Gamma_1\Gamma_3^{-1}\Gamma_4^{-1}\Gamma_5\Gamma_4\Gamma_3\Gamma_1^{-1}$. We substitute this expression in  $\langle\Gamma_1,\Gamma_2\rangle=e$ and  obtain $\langle\Gamma_1,\Gamma_3\rangle=e$. We substitute this expression in the other relations that include $\Gamma_2$, and they become redundant. Now we can say that $G$ is generated by $\{\Gamma_1,\Gamma_3,\Gamma_4,\Gamma_5\}$, and considering the quotient $G/\langle\Gamma_i^2\rangle$, we get
\begin{equation}\label{6-25}
[\Gamma_1,\Gamma_4]=[\Gamma_1,\Gamma_5]=[\Gamma_3,\Gamma_4]=e,
\end{equation}
\begin{equation}\label{6-26}
\langle\Gamma_1,\Gamma_3\rangle=\langle\Gamma_3,\Gamma_5\rangle=\langle\Gamma_4,\Gamma_5\rangle=e.
\end{equation}

Thus $G/\langle\Gamma_1^2,\Gamma_3^2,\Gamma_4^2,\Gamma_5^2\rangle\cong S_5$, and it follows that $\pi_1(X_{\text{Gal}})$ is trivial.

This completes the proof.
\end{proof}

\subsection{A 4-point quintic degeneration}\label{sec7}
In this subsection we consider a quintic that degenerates to a union of planes as shown in Figure \ref{case8}. This degeneration gives a special 4-point, in this case an intersection of four edges.

A rational normal scroll $F$ of degree 5 in $\mathbb P^ 6$ can degenerate to the cone over a hyperplane section of it, which is a rational normal curve $C$ of degree 5 in $\mathbb P^ 5$. Then $C$ can be degenerated to a chain of lines which is well known. This yields $F$ degenerates to 5 planes as in Figure \ref{case8}.

\begin{figure}[H]
\begin{center}
\scalebox{0.60}{\includegraphics{winfig8.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{A 4-point quintic degeneration}\label{case8}
\end{figure}

\begin{thm}\label{thm-4}
$\pi_1(X_{\text{Gal}})$  of surface with a 4-point quintic degeneration as in Figure \ref{case8} is trivial.
\end{thm}

\begin{proof}
The branch curve $S$ in $\mathbb{CP}^2$ is an arrangement of 4 lines. We regenerate each vertex in turn and compute group $G$.

Vertices 1, 2, 3, and 4 are 1-points, which give rise to the braids  $Z_{1 \; 1^{'}}$, $Z_{2 \; 2^{'}}$, $Z_{3\; 3^{'}}$, and $Z_{4 \; 4^{'}}$, respectively, and hence to the following relations in $G$:
\begin{equation}\label{9-1}
\Gamma_1=\Gamma_1',~~~\Gamma_2=\Gamma_2',~~~\Gamma_3=\Gamma_3',~~~\Gamma_4=\Gamma_4'.
\end{equation}

We have to compute the regeneration of the remaining 4-point (vertex 5) which is not the usual 4-point as in Figure \ref{case6}. Regenerating it, we get the curve presented in Figure \ref{4pointregeneration}.
\begin{figure}[H]
\begin{center}
\scalebox{0.60}{\includegraphics{winfig4point.eps}}
\end{center}
\setlength{\abovecaptionskip}{-0.15cm}
\caption{4-point regeneration}\label{4pointregeneration}
\end{figure}

Then the braid monodromy table of the curve in Figure \ref{4pointregeneration} is:
\vspace{0.2cm}
\begin{center}
  \begin{tabular}{|l|l|l|l|}

  \hline
  $j$ & $x_j$ & $\epsilon_{x_j}$ & $\delta_{x_j}$ \\\hline
  1 & $\langle 3,4\rangle$ & 2 & $\Delta\langle 3,4\rangle$ \\
  2 & $\langle 2,3\rangle$ & 2 & $\Delta\langle 2,3\rangle$ \\
  3 & $\langle 1,2\rangle$ & 2 & $\Delta\langle 1,2\rangle$ \\
  4 & $\langle 2,3\rangle$ & 4 & $\Delta^2\langle 2,3\rangle$ \\
  5 & $\langle 5,6\rangle$ & 4 & $\Delta^2\langle 5,6\rangle$ \\
  6 & $\langle 4,5\rangle$ & 2 & $\Delta\langle 4,5\rangle$ \\\hline

\end{tabular}\quad\quad\quad\quad
\begin{tabular}{|l|l|l|l|}

  \hline
 $j$ & $x_j$ & $\epsilon_{x_j}$ & $\delta_{x_j}$ \\\hline
  7 & $\langle 3,4\rangle$ & 2 & $\Delta\langle 3,4\rangle$ \\
  8 & $\langle 2,3\rangle$ & 2 & $\Delta\langle 2,3\rangle$ \\
  9 & $\langle 5,6\rangle$ & 4 & $\Delta^2\langle 5,6\rangle$ \\
  10 & $\langle 4,5\rangle$ & 1 & $\Delta^{1/2}_{I_2\mathbb{R}}\langle 4\rangle$ \\
  11 & $\langle 1,2\rangle$ & 1 & $\Delta^{1/2}_{I_4I_2}\langle 1\rangle$\\
  12 & $\langle 1,2\rangle$ &   &  \\\hline
\end{tabular}
\end{center}

\vspace{0.2cm}
By Moishezon-Teicher's algorithm \cite{BGT1,BGT2}, we get
the braid monodromy factors corresponding to the 4-point (vertex 5):
\begin{align*}
\widetilde{\Delta}_5=&Z_{2'\;3}^2\cdot Z_{2\;3}^2\cdot Z_{1\,1', 3}^2\cdot Z_{1\,1', 2}^3\cdot Z_{3', 4\,4'}^3\cdot (\bar{Z}_{2'\;3'}^2)^{Z_{3', 44'}^2}\cdot (Z_{2\;3'}^2)^{Z_{3', 4\,4'}^2Z_{23}^2Z_{11', 2}^2}\\
&\cdot(Z_{1\,1',3'}^2)^{Z_{3', 44'}^2Z_{11', 3}^2Z_{11', 2}^2}
\cdot Z_{2', 4\,4'}^3\cdot (Z_{2\;2'})^{Z_{11', 2}^2Z_{2', 44'}^2Z_{3, 44'}^{-2}}\cdot (Z_{3\;3'})^{Z_{3', 44'}^2}\cdot(Z_{11', 44'}^2)^{Z_{11',3}^2Z_{11', 2}^2}.
\end{align*}

$\widetilde{\Delta}_5$ gives rise to the following relations in $G$:
\begin{equation}\label{9-2}
[\Gamma_2,\Gamma_3] =[\Gamma_2',\Gamma_3] = [\Gamma_1,\Gamma_3] =[\Gamma_1',\Gamma_3] =e,
\end{equation}
\begin{equation}\label{9-3}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_1',\Gamma_2\rangle=\langle\Gamma_1^{-1}\Gamma_1'\Gamma_1,\Gamma_2\rangle=e,
\end{equation}
\begin{equation}\label{9-4}
\langle\Gamma_3',\Gamma_4\rangle=\langle\Gamma_3',\Gamma_4'\rangle=\langle\Gamma_3',\Gamma_4^{-1}\Gamma_4'\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{9-5}
[\Gamma_3\Gamma_2'\Gamma_3^{-1},\Gamma_4'\Gamma_4\Gamma_3'\Gamma_4^{-1}{\Gamma_4'}^{-1}] =
[\Gamma_2\Gamma_1'\Gamma_1\Gamma_2\Gamma_1^{-1}{\Gamma_1'}^{-1}\Gamma_2^{-1},
\Gamma_3^{-1}\Gamma_4'\Gamma_4\Gamma_3'\Gamma_4^{-1}{\Gamma_4'}^{-1}\Gamma_3] = e,
\end{equation}
\begin{equation}\label{9-6}
[\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_3^{-1}\Gamma_4'\Gamma_4\Gamma_3'\Gamma_4^{-1}{\Gamma_4'}^{-1}\Gamma_3] = e,
\end{equation}
\begin{equation}\label{9-7}
\langle\Gamma_3^{-1}\Gamma_2'\Gamma_3,\Gamma_4\rangle=\langle\Gamma_3^{-1}\Gamma_2'\Gamma_3,\Gamma_4'\rangle=
\langle\Gamma_3^{-1}\Gamma_2'\Gamma_3,\Gamma_4^{-1}\Gamma_4'\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{9-8}
\Gamma_2\Gamma_1'\Gamma_1\Gamma_2\Gamma_1^{-1}{\Gamma_1'}^{-1}\Gamma_2^{-1}=
\Gamma_3^{-1}\Gamma_4'\Gamma_4\Gamma_3\Gamma_2'\Gamma_3^{-1}\Gamma_4^{-1}{\Gamma_4'}^{-1}\Gamma_3,
\end{equation}
\begin{equation}\label{9-9}
\Gamma_3=
\Gamma_4'\Gamma_4\Gamma_3'\Gamma_4^{-1}{\Gamma_4'}^{-1},
\end{equation}
\begin{equation}\label{9-10}
[\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_3^{-1}\Gamma_4\Gamma_3] = [\Gamma_2\Gamma_1'\Gamma_2^{-1},\Gamma_3^{-1}\Gamma_4\Gamma_3]=
[\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_3^{-1}\Gamma_4'\Gamma_3] = [\Gamma_2\Gamma_1'\Gamma_2^{-1},\Gamma_3^{-1}\Gamma_4'\Gamma_3]=e.
\end{equation}

We also have the following projective relation:
\begin{equation}\label{9-11}
\Gamma_4'\Gamma_4\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1'\Gamma_1=e.
\end{equation}

Now, we simplify the group $G$ by using \eqref{9-1}, \eqref{9-2}, \eqref{9-3} and \eqref{9-4}; then we have
\begin{equation}\label{9-12}
[\Gamma_1,\Gamma_3] = [\Gamma_2,\Gamma_3] =[\Gamma_2,\Gamma_4^2\Gamma_3\Gamma_4^{-2}] =e,
\end{equation}
\begin{equation}\label{9-13}
[\Gamma_1^{-2}\Gamma_2\Gamma_1^2,\Gamma_4^2\Gamma_3\Gamma_4^{-2}] = [\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_4^2\Gamma_3\Gamma_4^{-2}] =
[\Gamma_2\Gamma_1\Gamma_2^{-1},\Gamma_4]=e,
\end{equation}
\begin{equation}\label{9-14}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_2,\Gamma_4\rangle=\langle\Gamma_3,\Gamma_4\rangle=e,
\end{equation}
\begin{equation}\label{9-15}
\Gamma_3=
\Gamma_4^2\Gamma_3\Gamma_4^{-2},
\end{equation}
\begin{equation}\label{9-16}
\Gamma_1^{-2}\Gamma_2\Gamma_1^2=
\Gamma_4^2\Gamma_2\Gamma_4^{-2},
\end{equation}
\begin{equation}\label{9-17}
\Gamma_4^2\Gamma_3^2\Gamma_2^2\Gamma_1^2=e.
\end{equation}

Thus in $G/\langle\Gamma_i^2\rangle$, we have the following relations:
 \begin{equation}\label{9-18}
[\Gamma_1,\Gamma_3] = [\Gamma_2,\Gamma_3] =[\Gamma_2\Gamma_1\Gamma_2,\Gamma_4] =e,
\end{equation}
\begin{equation}\label{9-19}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_2,\Gamma_4\rangle=\langle\Gamma_3,\Gamma_4\rangle=e.
\end{equation}
It is easy to see that
 \begin{equation}\label{9-20}
[\Gamma_2\Gamma_1\Gamma_2,\Gamma_3] =\langle\Gamma_2\Gamma_1\Gamma_2,\Gamma_2\rangle=e.
\end{equation}

We can then choose $\{\Gamma_2\Gamma_1\Gamma_2,\Gamma_2,\Gamma_3,\Gamma_4\}$ to be the generators of $G/\langle\Gamma_i^2\rangle$. It follows that $G/\langle\Gamma_i^2\rangle\cong S_5$. Thus $\pi_1(X_{\text{Gal}})$  is trivial.

This completes the proof.
\end{proof}

\section{On a question of Liedtke} \label{application}
In order to compute the fundamental group of the affine piece of Galois cover of surfaces, we introduce the following notions: $S^{\text{Aff}}$ is the affine part of $S$ in $\mathbb{C}^2$, $x_0$ is a point in  $\mathbb{C}^2-S^{\text{Aff}}$, and  $\pi_1(\mathbb{C}^2-S^{\text{Aff}}, x_0)$ is the fundamental group of the complement of $S^{\text{Aff}}$ in $\mathbb{C}^2$. This group is in fact $G$, without the projective relation.

We have the following exact sequence (\cite[Section 3]{Li08}, \cite[Chapter 0.3]{MoTe87}):
\begin{equation}\label{exact}
 1\rightarrow \pi_1(X_{\text{Gal}}^{\text{Aff}})\rightarrow
\frac{\pi_1(\mathbb{C}^2-S^{\text{Aff}}, x_0)}{\langle \Gamma_i^2\rangle}\xrightarrow{\psi}S_n \rightarrow1.
\end{equation}

Liedtke introduced a normal subgroup $C^{\text{Aff}}$ of $\pi_1(\mathbb{C}^2-S^{\text{Aff}}, x_0)$:
\begin{Def} \cite[Definition 3.3]{Li08}
$C^{\text{Aff}}$ is the subgroup normally generated by the
following elements inside $\pi_1(\mathbb{C}^2-S^{\text{Aff}}, x_0)$:
  \begin{itemize}
  \item[] $[\gamma\Gamma_i\gamma,\Gamma_j^{-1}]$, if $\psi(\gamma\Gamma_i\gamma^{-1})$ and $\psi(\Gamma_j)$ are disjoint transpositions,

  \item[] $\langle \gamma\Gamma_i\gamma,\Gamma_j^{-1} \rangle$, if $\psi(\gamma\Gamma_i\gamma^{-1})$ and $\psi(\Gamma_j)$
      have precisely one letter in common,
  \end{itemize}
where $\gamma$ runs through $\pi_1(\mathbb{C}^2-S^{\text{Aff}}, x_0)$.
\end{Def}

$C^{\text{Aff}}$ is trivial if the commutator and triple commutator relations hold in the above definition.
After taking the quotient of the short exact sequence \eqref{exact} by $C^{\text{Aff}}$, the resulting short exact sequence splits in a very nice way. Then he simplifies the computation of $\pi_1(\mathbb{C}^2-S^{\text{Aff}}, x_0)$ and $\pi_1(\mathbb{CP}^2-S)$ (cf: \cite{Li08}).
 He then states the following question, which is the main unsolved problem:
\begin{Qst}
Is $C^{\text{Aff}}$ trivial for every generic projection of degree $n\geq5$?
\end{Qst}

This is true in all known examples (such as Hirzebruch surface, $\mathbb{CP}^1\times \mathbb{CP}^1$, $\mathbb{CP}^1\times C_g, \cdots$).
Here, we give a counter-example of degree $5$.
\begin{thm}
$C^{\text{Aff}}$ of surfaces with the degeneration as in Figure \ref{case7} is non-trivial.
\end{thm}
\begin{proof}
In the proof of Theorem \ref{thm-6}, we simplify the relations without using the projective relation, thus $\{\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4,\Gamma_5\}$
are generators of $\pi_1(\mathbb{C}^2-S^{\text{Aff}}, x_0)/\langle \Gamma_i^2\rangle$, and we have the following relations:
\begin{equation*}
[\Gamma_1,\Gamma_3]=[\Gamma_1,\Gamma_5]=[\Gamma_2,\Gamma_5]=[\Gamma_1,\Gamma_2\Gamma_4\Gamma_2]=[\Gamma_4,\Gamma_5]=e,
\end{equation*}
\begin{equation*}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_2,\Gamma_3\rangle=\langle\Gamma_2,\Gamma_4\rangle
=\langle\Gamma_3,\Gamma_4\rangle=\langle\Gamma_3,\Gamma_5\rangle=e.
\end{equation*}

The projection $\psi:\frac{\pi_1(\mathbb{C}^2-S^{\text{Aff}}, x_0)}{\langle \Gamma_i^2\rangle}\rightarrow S_n$
is defined as follows:
$$\Gamma_1 \mapsto (1\,2),~\Gamma_2 \mapsto (1\,3),~\Gamma_3 \mapsto (3\,4),~\Gamma_4 \mapsto (2\,3),~\Gamma_5 \mapsto (4\,5).$$

\begin{figure}[H]
\begin{center}
\scalebox{0.55}{\includegraphics{winfig-cox5.eps}}
\end{center}
\end{figure}
It is easy to see that $\psi(\Gamma_3\Gamma_4\Gamma_3^{-1})$ and $\psi(\Gamma_2)$  are disjoint transpositions.
By \cite[pp.\,141]{RTV}, $[\Gamma_3\Gamma_4\Gamma_3,\Gamma_2^{-1}]$ is a generator of $\text{Ker}\psi$ which is non trivial. Thus,
$C^{\text{Aff}}$ is non-trivial.

This completes the proof.
\end{proof}

\section{Appendix: A missing case in \cite{A-R-T}}\label{Appendix}
 In \cite{A-R-T}, the authors classified fundamental groups of Galois covers of surfaces of degree $4$ degenerating to nice plane arrangements, but did not specifically include the following case (Figure \ref{de4}).
Indeed, the rational normal scroll $F$ degenerates to a union of planes in Figure \ref{de4} (this is a toric degeneration, corresponding to the subdivision of the $(1,2)$--rectangle).

\begin{figure}[H]
\begin{center}
\scalebox{0.75}{\includegraphics{winfig9.eps}}
\end{center}
\setlength{\abovecaptionskip}{-1.2cm}
\caption{Degree $4$}\label{de4}
\end{figure}

\begin{thm}\label{thm-9}
$\pi_1(X_{\text{Gal}})$ of surfaces with the degeneration as in Figure \ref{de4} is trivial.
\end{thm}

\begin{proof}
Vertices 1, 2, and 3 are 1-points, therefore the related braids are  $Z_{1\,1^{'}}$, $Z_{2\,2^{'}}$, and $Z_{3\,3^{'}}$ respectively, and hence the relations in $G$ are the following:
\begin{equation}\label{de-1}
\Gamma_1=\Gamma_1',~~~\Gamma_2=\Gamma_2',~~~\Gamma_3=\Gamma_3'.
\end{equation}

Vertex 4 is a 3-point (line 1 and 3 are diagonals) that regenerates to a line 2 tangent to  two conics, $(1,1')$ and $(3,3')$.

We give the  following relations in $G$:
\begin{equation}\label{de-2}
[\Gamma_1,\Gamma_3]=[\Gamma_1',\Gamma_3]=[\Gamma_1,\Gamma_3^{-1}\Gamma_3'\Gamma_3]
=[\Gamma_1',\Gamma_3^{-1}\Gamma_3'\Gamma_3]=e,
\end{equation}
\begin{equation}\label{de-3}
\langle\Gamma_2,\Gamma_3\rangle=\langle\Gamma_2',\Gamma_3\rangle
=\langle\Gamma_2^{-1}\Gamma_2'\Gamma_2,\Gamma_3\rangle=e,
\end{equation}
\begin{equation}\label{de-4}
\Gamma_3'=\Gamma_3\Gamma_2'\Gamma_2\Gamma_3\Gamma_2^{-1}{\Gamma_2'}^{-1}\Gamma_3^{-1},
\end{equation}
\begin{equation}\label{de-5}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_1,\Gamma_2'\rangle=
\langle\Gamma_1,\Gamma_2^{-1}\Gamma_2'\Gamma_2\rangle=e,
\end{equation}
\begin{equation}\label{de-6}
\Gamma_1'=\Gamma_1\Gamma_2'\Gamma_2\Gamma_1\Gamma_2^{-1}{\Gamma_2'}^{-1}\Gamma_1^{-1}.
\end{equation}
\begin{equation}
\Gamma_3'\Gamma_3\Gamma_2'\Gamma_2\Gamma_1'\Gamma_1=e.
\end{equation}

After an easy simplification, we get the following relations (combining projective relation):
\begin{equation}\label{de-7}
[\Gamma_1,\Gamma_3]=e,
\end{equation}
\begin{equation}\label{de-8}
\langle\Gamma_1,\Gamma_2\rangle=\langle\Gamma_2,\Gamma_3\rangle=e,
\end{equation}
\begin{equation}\label{de-9}
\Gamma_3=\Gamma_2^{-2}\Gamma_3\Gamma_2^2,\,\ \ \ \Gamma_1=\Gamma_2^{-2}\Gamma_1\Gamma_2^2,
\end{equation}
\begin{equation}\label{de-10}
\Gamma_3^2\Gamma_2^2\Gamma_1^2=e.
\end{equation}

Thus $G/\langle\Gamma_1^2,\Gamma_2^2,\Gamma_3^2\rangle\cong S_4$, hence $\pi_1(X_{\text{Gal}})$ is trivial.

This completes the proof.
\end{proof}