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Let $X$ be a hyperelliptic curve of genus $g$ (nonsingular, etc.) with a hyperelliptic cover $X \to \mathbb{P}^1$ corresponding to an invertible sheaf $\mathscr{L}$. The composition with the $(g-1)$ Veronese map $\mathbb{P}^1 \to \mathbb{P}^{g-1}$ is the morphism corresponding to the invertible sheaf $\mathscr{L}^{\otimes (g-1)}$. In Vakil's notes on algebraic geometry (20.5.6), he writes that the pullback homomorphism $\Gamma(\mathbb{P}^{g-1}, \mathscr{O}(1)) \to \Gamma(X, \mathscr{L}^{\otimes (g-1)})$ is injective because if there was a hyperplane $s \in \Gamma(X, \mathscr{O}_{\mathbb{P}^{g-1}}(1))$ that pulled back to 0 on $X$, then the image of $X$ would lie in that hyperplane, yet a rational normal curve cannot.

I am having trouble following this geometric argument; how would one show explicitly that this homomorphism is injective?

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up vote 2 down vote accepted

I think the geometric argument is better, what part do you have trouble with ?

An algebraic argument:denote by $\pi: X\to \mathbb P^1$ the cover. Then $\mathcal L=\pi^{*}O_{\mathbb P^1}(1)$ and $O_{\mathbb P^1}(1)\to \pi_*\mathcal L$ is injective. This implies that for any positive $n$, $O_{\mathbb P^1}(n)\to \pi_*(\mathcal L^{\otimes n})$ is injective. Hence $$ \Gamma(\mathbb P^1, O_{\mathbb P^1}(g-1))\to \Gamma(X, \mathcal L^{\otimes g-1})=\Gamma(\mathbb P^1, \pi_*(\mathcal L^{\otimes g-1}))$$ is injective. As $$ \Gamma(\mathbb P^{g-1}, O_{\mathbb P^{g-1}}(1))\to\Gamma(\mathbb P^1, O_{\mathbb P^1}(g-1)) $$ is injective (it takes $x_0, \dots x_{g-1}$ to $y_0^{g-1}, y_0^{g-2}y_1, \dots, y_1^{g-1}$ respectively), the composition $$ \Gamma(\mathbb P^{g-1}, O_{\mathbb P^{g-1}}(1))\to\Gamma(X, \mathcal L^{\otimes g-1})$$ is injective.

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