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)(c) 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
Comments (2)
Comment #8398 by ZL on
Comment #9008 by Stacks project on