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):

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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?

  • $\begingroup$ 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 $f \in L(D) = H^0(Y,O_Y(D)) \iff f\circ \pi \in L(\pi^*(D))=H^0(X,O_X(\pi^* (D)))$ $\endgroup$ – reuns Aug 24 '19 at 18:28
  • $\begingroup$ I meant $\implies$ and it is $\iff$ when the morphism is surjective $\endgroup$ – reuns Aug 24 '19 at 18:36
  • $\begingroup$ 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 $F:=O_X(y_o)$. $\endgroup$ – user698176 Aug 24 '19 at 18:46
  • $\begingroup$ I don't understand what you wrote. It is obvious $f \mapsto f \circ \pi$ is injective for $\pi$ non-constant. $\endgroup$ – reuns Aug 24 '19 at 18:53
  • $\begingroup$ yes of course, I see now. here non constant = surjective. $\endgroup$ – user698176 Aug 24 '19 at 18:57

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