# Pullback of globally generated line bundle is globally generated?

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.
• Thanks Georges! But doesn't it mean that $A$ can be generated by a subspace $H^0(C',A')$ of $H^0(C,A)$? That is the stalks of $A$ are generated by fewer sections. But will this not mean that the complement say $\mathcal{O}^r$ is in the kernel of the surjection $H^0(C,A)\otimes\mathcal{O}_C\longrightarrow A\longrightarrow 0$? – gradstudent Sep 9 '15 at 11:42
• Yes, sections coming from $H^0(C',A')$ suffice to generate $A$. – Georges Elencwajg Sep 9 '15 at 12:20
• But the kernel $K$ of $H^0(C,A)\otimes \mathcal{O}_C\longrightarrow A$ has no global sections right? Then how does it contain copies of $\mathcal{O}$ – gradstudent Sep 9 '15 at 12:43