Let $f:C\longrightarrow C'$ be a finite morphism of curves. Let $A'$ be a line bundle on $C'$ and let $A$ be it's pullback to $C$. If $A'$ is globally generated, is it true that $A$ is globally generated too?
Since $A'$ is globally generated, we have a surjection $H^0(C',A')\otimes \mathcal{O}_{C'}\longrightarrow A'\longrightarrow 0$.
Pullback operation is right exact, therefore we get $H^0(C',A')\otimes\mathcal{O}_C \longrightarrow A \longrightarrow 0$.
But since $f$ is a finite morphism, $h^0(C,A)$ could be greater than $h^0(C',A')$. Therefore we could compose with projection to get the required surjection. Is that right?
But $H^0(C',A')\otimes\mathcal{O}_C \longrightarrow A \longrightarrow 0$ --- this surjection shows that $A$ can be generated by fewer global sections isn't it? Thank you!
Given a section $\sigma'\in H^0(C',A')$ and its pullback $\sigma=f^*(\sigma') \in H^0(C,A)$, we have for every $c\in C$ the formula: $$\sigma(c)=\sigma'(f(c))\in A[c]=A'[f(c)]$$ So if $\sigma'(f(c))\neq 0$, which can always be achieved for a suitable choice of $\sigma '$ by the hypothesis of global generation of $A'$, we will have $\sigma(c)\neq0$, and this proves that $A$ is globally generated too.
Nota Bene
The above is valid for arbitrary morphisms (finite or not) between arbitrary varieties of arbitrary (maybe different) dimensions.