# Question on Sections of Hyperelliptic Curve

I have a question about a step in the proof of Lemma 7.4.8 from Liu's "Algebraic Geometry and Arithmetic Curves" (page 288):

If we assume that $$X$$ is hiperelliptic and we have our finite separable map $$\pi:X \to \mathbb{P}^1_k$$ of degree $$2$$.

A rational point $$y_0$$ of $$\mathbb{P}^1_k$$ defines a Cartier divisor and let $$D:= \pi^*y_0 \in Div(X)$$ it's pullback under $$\pi$$.

Denote by $$O_X(D)$$ the corresponding invertible sheaf to $$D$$ and the take into account that the author uses the notation $$L(D)=H^0(X,O_X(D))$$.

My question is why under giving setting we have an inclusion

$$H^0(\mathbb{P}^1_k, O_{\mathbb{P}^1_k}(y_0)) \subset H^0(X,O_X(D))$$

as stated in the excerpt?

• If you have a morphism $\pi : X \to Y= \Bbb{P}^1$, for $f([u:v])\in k(u/v)$ rational on $Y$, then $\pi^*$ is the unique homomorphism $Div(Y) \to Div(X)$ such that $\pi^*(Div(f)) = Div(f\circ \pi)$. At the unramified points $\pi^*(p)=\sum_{q \in \pi^{-1}(p)} q$. If $Div(f) \ge D$ then $Div(f\circ \pi) \ge \pi^* (D)$ so that $f \in L(D) = H^0(Y,O_Y(D)) \iff f\circ \pi \in L(\pi^*(D))=H^0(X,O_X(\pi^* (D)))$ – reuns Aug 24 '19 at 18:28
• I meant $\implies$ and it is $\iff$ when the morphism is surjective – reuns Aug 24 '19 at 18:36
• I still not see how your observation that $Div(f\circ \pi) \ge \pi^* (D)$ implies that the map $H^0(Y,O_Y(D)) \to H^0(X,O_X(\pi^*D)), f \mapsto f \circ \pi$ is injective. by the way: does this map you described above exactly the same as that one we obtain using adjunction between $\pi^*$ and $\pi_*$ and then applying global section functor $H^0()$ to canonical $F \to \pi_* \pi^* F$ with $F:=O_X(y_o)$. – user698176 Aug 24 '19 at 18:46
• I don't understand what you wrote. It is obvious $f \mapsto f \circ \pi$ is injective for $\pi$ non-constant. – reuns Aug 24 '19 at 18:53
• yes of course, I see now. here non constant = surjective. – user698176 Aug 24 '19 at 18:57