Proof.
Let us construct the functor from right to left. Let $F \to X$ be a map of sheaves on $(\mathit{Sch}/S)_{fppf}$ such that each $V_ i = X_ i \times _ X F$ is an algebraic space. We have the projection $V_ i \to X_ i$. Then both $V_ i \times _ X X_ j$ and $X_ i \times _ X V_ j$ represent the sheaf $X_ i \times _ X F \times _ X X_ j$ and hence we obtain an isomorphism
\[ \varphi _{ii'} : V_ i \times _ X X_ j \to X_ i \times _ X V_ j \]
It is straightforward to see that the maps $\varphi _{ij}$ are morphisms over $X_ i \times _ X X_ j$ and satisfy the cocycle condition. The functor from right to left is given by this construction $F \mapsto (V_ i, \varphi _{ij})$.
Let us construct a functor from left to right. The isomorphisms $\varphi _{ij}$ give isomorphisms
\[ \varphi _{ij} : V_ i \times _ X X_ j \longrightarrow X_ i \times _ X V_ j \]
over $X_ i \times X_ j$. Set $F$ equal to the coequalizer in the following diagram
\[ \xymatrix{ \coprod _{i, i'} V_ i \times _ X X_ j \ar@<1ex>[rr]^-{\text{pr}_0} \ar@<-1ex>[rr]_-{\text{pr}_1 \circ \varphi _{ij}} & & \coprod _ i V_ i \ar[r] & F } \]
The cocycle condition guarantees that $F$ comes with a map $F \to X$ and that $X_ i \times _ X F$ is isomorphic to $V_ i$. The functor from left to right is given by this construction $(V_ i, \varphi _{ij}) \mapsto F$.
We omit the verification that these constructions are mutually quasi-inverse functors. The final statements (1) and (2) follow from the constructions.
$\square$
Comments (2)
Comment #978 by David Holmes on
Comment #998 by Johan on