ag.algebraic geometry - How to use automorphisms to produce isotrivial non trivial families

If $C=B$ with the same $Z/2$ action, and $g(C/Z_2) geq 2$, then you always get a non-trivial family:

Suppose it was trivial. Then it had another projection $q : D to C$. Since $C$ is an 'etale cover of $B'$, and both have genus at least two, $g(C) > g(B')$ by the Hurwitz-formula. That is, also by the Hurwitz formula, all the maps from $B'$ to $C$ are the constant maps. So, all sections $B' to D$ of $f$ would have to be contained in a fiber of $q$, i. e. they have to be one fiber of $q$. However there are two natural sections $E$ and $F$ of $p$: $[c] to [(c,c)]$ and $[c] to [(c,-c)]$. So both of these have to be contracted to a point by $q$. Given any $b in B$, a choice $c in C$ such that $[c]=b$ gives an isomorphism $a_c : p^{-1}(d) to C$, which sends $[(c,c')]$ to $c'$. Now one can construct automorphisms $varphi_{c'}$ of $C$ for any $c' in C$ by the following composition:

$
C to^{a_c^{-1}} p^{-1}([c]) to^q C to^s p^{-1}([c']) to^{a_{c'}} C
$

where

$
S=(q|_{p^{-1}([c'])})^{-1}
$

One specialty about $varphi_{c'}$ is that it takes $c$ and $-c$ to $c'$ and $-c'$, respectively. This follows from the fact, that $E$ and $F$ are contracted to a point by $q$. So, $C$ has infinitely many different automorphisms, which is a contradiction by the assumption that $g(C) geq 2$. That is our assumption is false, therefore $Y$ is not a product family.