Let $Y:=\mathbb{P}^1_k$ and $X$ a hyperelliptic curve. We working with the definition 4.7 from Liu's "Algebraic Geometry and Arithmetic Curves" (page 288)
Namely there exist finite separable map $\pi:X \to Y$ of degree $2$.
I have a question about the tagged step in the proof of Lemma 7.4.8:
The rational point $y_0$ of $\mathbb{P}^1_k$ defines a Cartier divisor (resp the corresp invertible sheaf $O_Y(y_0)$) 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?
My idea was to use the fact that adjunction between $\pi^*$ and $\pi_*$ gives rise for morphism of $O_Y$ modules
$$O_Y(y_0) \to \pi_* \pi^* O_Y(y_0) =\pi_* O_X(D)$$
My question is if and why in this setting this map is injective? If this would be true then the left exact $H^0(-)$-functor would preserve the injectivity and we are done.
Or has Liu here a more simple argument in mind? since Liu hasn't mentioned that in the excerpt above that makes me to assume there is a more easier reason for this inclusion
