The Stacks project

Proof. Denote $\omega _{Y/X}$ the coherent $\mathcal{O}_ Y$-module such that there is an isomorphism

of $\pi _*\mathcal{O}_ Y$-modules, see Morphisms of Spaces, Lemma 67.20.10 and Descent on Spaces, Lemma 74.6.6. The canonical map $\mathcal{O}_ X \to \pi _*\mathcal{O}_ Y$ produces a canonical map

Since $V$ is Noetherian affine we may choose sections

generating the coherent module $\pi ^*\mathcal{F} \otimes _{\mathcal{O}_ X} \omega _{Y/X}$ over $V$. By Cohomology of Spaces, Lemma 69.13.4 we can choose integers $n_ i \geq 0$ such that $s_ i$ extends to a map $s_ i' : \mathcal{I}^{n_ i} \to \pi ^*\mathcal{F} \otimes _{\mathcal{O}_ Y} \omega _{Y/X}$. Pushing to $X$ we obtain maps

where the equality sign is Cohomology of Spaces, Lemma 69.4.3. To finish the proof we will show that the sum of these maps is surjective.

Let $x \in |X|$ be a point of $X$. Let $v \in |V|$ be a point mapping to $x$. We may choose an étale neighbourhood $(U, u) \to (X, x)$ such that

(disjoint union of algebraic spaces) such that $W \to U$ is an isomorphism and such that the unique point $w \in W$ lying over $u$ maps to $v$ in $V \subset Y$. To see this is true use Lemma 76.33.2 and Étale Morphisms, Lemma 41.18.1. After shrinking $U$ further if necessary we may assume $W$ maps into $V \subset Y$ by the projection. Since the formation of $\omega _{Y/X}$ commutes with étale localization we see that

We have $(\pi |_ W)_*\omega _{W/U} = \mathcal{O}_ U$ and this isomorphism is given by the trace map $\text{Tr}_\pi |_ U$ restricted to the first summand in the decomposition above. Since $W$ maps into $V$ we see that $\mathcal{I}^{n_ i}|_ W = \mathcal{O}_ W$. Hence

Chasing diagrams the reader sees (details omitted) that $\sigma _ i|_ U$ on the summand $\mathcal{O}_ U$ is the map $\mathcal{O}_ U \to \mathcal{F}$ corresponding to the section

Since the sections $s_ i$ generate the module $\pi ^*\mathcal{F} \otimes _{\mathcal{O}_ Y} \omega _{Y/X}$ over $V$ and since $W$ maps into $V$ we conclude that the restriction of $\bigoplus \sigma _ i$ to $U$ is surjective. Since $x$ was an arbitrary point the proof is complete. $\square$