The Stacks project

Proof. The map $\mathcal{F}(-1) \to \mathcal{F}$ comes from Constructions, Equation (27.10.1.2) with $n = 1$, $m = -1$ and the section $s$ of $\mathcal{O}(1)$. Let's work out what this map looks like if we restrict it to $D_{+}(T_0)$. Write $D_{+}(T_0) = \mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n])$ with $x_ i = T_ i/T_0$. Identify $\mathcal{O}(1)|_{D_{+}(T_0)}$ with $\mathcal{O}$ using the section $T_0$. Hence if $s = \sum a_ iT_ i$ then $s|_{D_{+}(T_0)} = a_0 + \sum a_ ix_ i$ with the identification chosen above. Furthermore, suppose $\mathcal{F}|_{D_{+}(T_0)}$ corresponds to the finite $k[x_1, \ldots , x_ n]$-module $M$. Via the identification $\mathcal{F}(-1) = \mathcal{F} \otimes \mathcal{O}(-1)$ and our chosen trivialization of $\mathcal{O}(1)$ we see that $\mathcal{F}(-1)$ corresponds to $M$ as well. Thus restricting $\mathcal{F}(-1) \to \mathcal{F}$ to $D_{+}(T_0)$ gives the map

To see that the arrow is injective, it suffices to pick $a_0 + \sum a_ ix_ i$ outside any of the associated primes of $M$, see Algebra, Lemma 10.63.9. By Algebra, Lemma 10.63.5 the set $\text{Ass}(M)$ of associated primes of $M$ is finite. Note that for $\mathfrak p \in \text{Ass}(M)$ the intersection $\mathfrak p \cap \{ a_0 + \sum a_ i x_ i\} $ is a proper $k$-subvector space. We conclude that there is a finite family of proper sub vector spaces $V_1, \ldots , V_ m \subset \Gamma (\mathbf{P}^ n_ k, \mathcal{O}(1))$ such that if we take $s$ outside of $\bigcup V_ i$, then multiplication by $s$ is injective over $D_{+}(T_0)$. Similarly for the restriction to $D_{+}(T_ j)$ for $j = 1, \ldots , n$. Since $k$ is infinite, a finite union of proper sub vector spaces is never equal to the whole space, hence we may choose $s$ such that the map is injective. The cokernel of $\mathcal{F}(-1) \to \mathcal{F}$ is annihilated by $\mathop{\mathrm{Im}}(s : \mathcal{O}(-1) \to \mathcal{O})$ which is the ideal sheaf of $H$ by Divisors, Definition 31.14.8. Hence we obtain $\mathcal{G}$ on $H$ using Cohomology of Schemes, Lemma 30.9.8. $\square$