Lemma 103.14.5. Let $\mathcal{X}$ be an algebraic stack. Notation as in Lemmas 103.14.2 and 103.14.4.

1. We have $g_!\mathcal{O}_{\mathcal{X}_{lisse,{\acute{e}tale}}} = \mathcal{O}_\mathcal {X}$.

2. We have $g_!\mathcal{O}_{\mathcal{X}_{flat, fppf}} = \mathcal{O}_\mathcal {X}$.

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

$V \longmapsto \mathop{\mathrm{colim}}\nolimits _{V \to U} \mathcal{O}'(U)$

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

$V \to U \to U'$

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

$\mathop{\mathrm{colim}}\nolimits _{V \to U} \mathcal{O}'(U)$

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

$V' = V \times _{(\varphi _1, \ldots , \varphi _ n), U} W$

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

$\xymatrix{ V \ar[rr]_-{(\varphi , 0)} \ar[rrd]_\varphi & & U \times \mathbf{A}^1 \\ & & U \ar[u]_{(\text{id}, 0)} }$

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. $\square$

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