The Stacks project

Proof. In this proof we write $\mathcal{C} = \mathcal{X}_{\acute{e}tale}$ (resp. $\mathcal{C} = \mathcal{X}_{fppf}$) and we denote $\mathcal{C}' = \mathcal{X}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{C}' = \mathcal{X}_{flat, fppf}$). Then $\mathcal{C}'$ is a full subcategory of $\mathcal{C}$. In this proof we will think of objects $V$ of $\mathcal{C}$ as schemes over $\mathcal{X}$ and objects $U$ of $\mathcal{C}'$ as schemes smooth (resp. flat) over $\mathcal{X}$. Finally, we write $\mathcal{O} = \mathcal{O}_\mathcal {X}$ and $\mathcal{O}' = \mathcal{O}_{\mathcal{X}_{lisse,{\acute{e}tale}}}$ (resp. $\mathcal{O}' = \mathcal{O}_{\mathcal{X}_{flat,fppf}}$). In the notation above we have $\mathcal{O}(V) = \Gamma (V, \mathcal{O}_ V)$ and $\mathcal{O}'(U) = \Gamma (U, \mathcal{O}_ U)$. Consider the $\mathcal{O}$-module homomorphism $g_!\mathcal{O}' \to \mathcal{O}$ adjoint to the identification $\mathcal{O}' = g^{-1}\mathcal{O}$.

Recall that $g_!\mathcal{O}'$ is the sheaf associated to the presheaf $g_{p!}\mathcal{O}'$ given by the rule

where the colimit is taken in the category of abelian groups (Modules on Sites, Definition 18.16.1). Below we will use frequently that if

are morphisms and if $f' \in \mathcal{O}'(U')$ restricts to $f \in \mathcal{O}'(U)$, then $(V \to U, f)$ and $(V \to U', f')$ define the same element of the colimit. Also, $g_!\mathcal{O}' \to \mathcal{O}$ maps the element $(V \to U, f)$ simply to the pullback of $f$ to $V$.

Let us prove that $g_!\mathcal{O}' \to \mathcal{O}$ is surjective. Let $h \in \mathcal{O}(V)$ for some object $V$ of $\mathcal{C}$. It suffices to show that $h$ is locally in the image. Choose an object $U$ of $\mathcal{C}'$ corresponding to a surjective smooth morphism $U \to \mathcal{X}$. Since $U \times _\mathcal {X} V \to V$ is surjective smooth, after replacing $V$ by the members of an étale covering of $V$ we may assume there exists a morphism $V \to U$, see Topologies on Spaces, Lemma 73.4.4. Using $h$ we obtain a morphism $V \to U \times \mathbf{A}^1$ such that writing $\mathbf{A}^1 = \mathop{\mathrm{Spec}}(\mathbf{Z}[t])$ the element $t \in \mathcal{O}(U \times \mathbf{A}^1)$ pulls back to $h$. Since $U \times \mathbf{A}^1$ is an object of $\mathcal{C}'$ we see that $(V \to U \times \mathbf{A}^1, t)$ is an element of the colimit above which maps to $h \in \mathcal{O}(V)$ as desired.

Suppose that $s \in g_!\mathcal{O}'(V)$ is a section mapping to zero in $\mathcal{O}(V)$. To finish the proof we have to show that $s$ is zero. After replacing $V$ by the members of a covering we may assume $s$ is an element of the colimit

Say $s = \sum (\varphi _ i, s_ i)$ is a finite sum with $\varphi _ i : V \to U_ i$, $U_ i$ smooth (resp. flat) over $\mathcal{X}$, and $s_ i \in \Gamma (U_ i, \mathcal{O}_{U_ i})$. Choose a scheme $W$ surjective étale over the algebraic space $U = U_1 \times _\mathcal {X} \ldots \times _\mathcal {X} U_ n$. Note that $W$ is still smooth (resp. flat) over $\mathcal{X}$, i.e., defines an object of $\mathcal{C}'$. The fibre product

is surjective étale over $V$, hence it suffices to show that $s$ maps to zero in $g_!\mathcal{O}'(V')$. Note that the restriction $\sum (\varphi _ i, s_ i)|_{V'}$ corresponds to the sum of the pullbacks of the functions $s_ i$ to $W$. In other words, we have reduced to the case of $(\varphi , s)$ where $\varphi : V \to U$ is a morphism with $U$ in $\mathcal{C}'$ and $s \in \mathcal{O}'(U)$ restricts to zero in $\mathcal{O}(V)$. By the commutative diagram

we see that $((\varphi , 0) : V \to U \times \mathbf{A}^1, \text{pr}_2^*x)$ represents zero in the colimit above. Hence we may replace $U$ by $U \times \mathbf{A}^1$, $\varphi$ by $(\varphi , 0)$ and $s$ by $\text{pr}_1^*s + \text{pr}_2^*x$. Thus we may assume that the vanishing locus $Z : s = 0$ in $U$ of $s$ is smooth (resp. flat) over $\mathcal{X}$. Then we see that $(V \to Z, 0)$ and $(\varphi , s)$ have the same value in the colimit, i.e., we see that the element $s$ is zero as desired. $\square$