Lemma 79.3.11. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$. Let $F$ be a presheaf of sets on $(\mathit{Sch}/S)_{fppf}$. Assume

1. $F$ is a sheaf for the Zariski topology on $(\mathit{Sch}/S)_{fppf}$,

2. there exists an index set $I$ and subfunctors $F_ i \subset F$ such that

1. each $F_ i$ is an fppf sheaf,

2. each $F_ i \to F$ is representable by algebraic spaces,

3. $\coprod F_ i \to F$ becomes surjective after fppf sheafification.

Then $F$ is an fppf sheaf.

Proof. Let $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and let $s \in F(T)$. By (2)(c) there exists an fppf covering $\{ T_ j \to T\}$ such that $s|_{T_ j}$ is a section of $F_{\alpha (j)}$ for some $\alpha (j) \in I$. Let $W_ j \subset T$ be the image of $T_ j \to T$ which is an open subscheme Morphisms, Lemma 29.25.10. By (2)(b) we see $F_{\alpha (j)} \times _{F, s|_{W_ j}} W_ j \to W_ j$ is a monomorphism of algebraic spaces through which $T_ j$ factors. Since $\{ T_ j \to W_ j\}$ is an fppf covering, we conclude that $F_{\alpha (j)} \times _{F, s|_{W_ j}} W_ j = W_ j$, in other words $s|_{W_ j} \in F_{\alpha (j)}(W_ j)$. Hence we conclude that $\coprod F_ i \to F$ is surjective for the Zariski topology.

Let $\{ T_ j \to T\}$ be an fppf covering in $(\mathit{Sch}/S)_{fppf}$. Let $s, s' \in F(T)$ with $s|_{T_ j} = s'|_{T_ j}$ for all $j$. We want to show that $s, s'$ are equal. As $F$ is a Zariski sheaf by (1) we may work Zariski locally on $T$. By the result of the previous paragraph we may assume there exist $i$ such that $s \in F_ i(T)$. Then we see that $s'|_{T_ j}$ is a section of $F_ i$. By (2)(b) we see $F_{i} \times _{F, s'} T \to T$ is a monomorphism of algebraic spaces through which all of the $T_ j$ factor. Hence we conclude that $s' \in F_ i(T)$. Since $F_ i$ is a sheaf for the fppf topology we conclude that $s = s'$.

Let $\{ T_ j \to T\}$ be an fppf covering in $(\mathit{Sch}/S)_{fppf}$ and let $s_ j \in F(T_ j)$ such that $s_ j|_{T_ j \times _ T T_{j'}} = s_{j'}|_{T_ j \times _ T T_{j'}}$. By assumption (2)(b) we may refine the covering and assume that $s_ j \in F_{\alpha (j)}(T_ j)$ for some $\alpha (j) \in I$. Let $W_ j \subset T$ be the image of $T_ j \to T$ which is an open subscheme Morphisms, Lemma 29.25.10. Then $\{ T_ j \to W_ j\}$ is an fppf covering. Since $F_{\alpha (j)}$ is a sub presheaf of $F$ we see that the two restrictions of $s_ j$ to $T_ j \times _{W_ j} T_ j$ agree as elements of $F_{\alpha (j)}(T_ j \times _{W_ j} T_ j)$. Hence, the sheaf condition for $F_{\alpha (j)}$ implies there exists a $s'_ j \in F_{\alpha (j)}(W_ j)$ whose restriction to $T_ j$ is $s_ j$. For a pair of indices $j$ and $j'$ the sections $s'_ j|_{W_ j \cap W_{j'}}$ and $s'_{j'}|_{W_ j \cap W_{j'}}$ of $F$ agree by the result of the previous paragraph. This finishes the proof by the fact that $F$ is a Zariski sheaf. $\square$

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