The Stacks project

Proof. Let $X_ i$ be a scheme representing $F_ i$ and let $\xi _ i \in F_ i(X_ i) \subset F(X_ i)$ be the “universal family”. Because $F_ j \subset F$ is representable by open immersions, there exists an open $U_{ij} \subset X_ i$ such that $T \to X_ i$ factors through $U_{ij}$ if and only if $\xi _ i|_ T \in F_ j(T)$. In particular $\xi _ i|_{U_{ij}} \in F_ j(U_{ij})$ and therefore we obtain a canonical morphism $\varphi _{ij} : U_{ij} \to X_ j$ such that $\varphi _{ij}^*\xi _ j = \xi _ i|_{U_{ij}}$. By definition of $U_{ji}$ this implies that $\varphi _{ij}$ factors through $U_{ji}$. Since $(\varphi _{ij} \circ \varphi _{ji})^*\xi _ j =\varphi _{ji}^*(\varphi _{ij}^*\xi _ j) = \varphi _{ji}^*\xi _ i = \xi _ j$ we conclude that $\varphi _{ij} \circ \varphi _{ji} = \text{id}_{U_{ji}}$ because the pair $(X_ j, \xi _ j)$ represents $F_ j$. In particular the maps $\varphi _{ij} : U_{ij} \to U_{ji}$ are isomorphisms of schemes. Next we have to show that $\varphi _{ij}^{-1}(U_{ji} \cap U_{jk}) = U_{ij} \cap U_{ik}$. This is true because (a) $U_{ji} \cap U_{jk}$ is the largest open of $U_{ji}$ such that $\xi _ j$ restricts to an element of $F_ k$, (b) $U_{ij} \cap U_{ik}$ is the largest open of $U_{ij}$ such that $\xi _ i$ restricts to an element of $F_ k$, and (c) $\varphi _{ij}^*\xi _ j = \xi _ i$. Moreover, the cocycle condition in Section 26.14 follows because both $\varphi _{jk}|_{U_{ji} \cap U_{jk}} \circ \varphi _{ij}|_{U_{ij} \cap U_{ik}}$ and $\varphi _{ik}|_{U_{ij} \cap U_{ik}}$ pullback $\xi _ k$ to the element $\xi _ i$. Thus we may apply Lemma 26.14.2 to obtain a scheme $X$ with an open covering $X = \bigcup U_ i$ and isomorphisms $\varphi _ i : X_ i \to U_ i$ with properties as in Lemma 26.14.1. Let $\xi _ i' = (\varphi _ i^{-1})^* \xi _ i$. The conditions of Lemma 26.14.1 imply that $\xi _ i'|_{U_ i \cap U_ j} = \xi _ j'|_{U_ i \cap U_ j}$. Therefore, by the condition that $F$ satisfies the sheaf condition in the Zariski topology we see that there exists an element $\xi ' \in F(X)$ such that $\xi _ i = \varphi _ i^*\xi '|_{U_ i}$ for all $i$. Since $\varphi _ i$ is an isomorphism we also get that $(U_ i, \xi '|_{U_ i})$ represents the functor $F_ i$.

We claim that the pair $(X, \xi ')$ represents the functor $F$. To show this, let $T$ be a scheme and let $\xi \in F(T)$. We will construct a unique morphism $g : T \to X$ such that $g^*\xi ' = \xi$. Namely, by the condition that the subfunctors $F_ i$ cover $F$ there exists an open covering $T = \bigcup V_ i$ such that for each $i$ the restriction $\xi |_{V_ i} \in F_ i(V_ i)$. Moreover, since each of the inclusions $F_ i \subset F$ are representable by open immersions we may assume that each $V_ i \subset T$ is maximal open with this property. Because, $(U_ i, \xi '|_{U_ i})$ represents the functor $F_ i$ we get a unique morphism $g_ i : V_ i \to U_ i$ such that $g_ i^*\xi '|_{U_ i} = \xi |_{V_ i}$. On the overlaps $V_ i \cap V_ j$ the morphisms $g_ i$ and $g_ j$ agree, for example because they both pull back $\xi '|_{U_ i \cap U_ j} \in F_ i(U_ i \cap U_ j)$ to the same element. Thus the morphisms $g_ i$ glue to a unique morphism from $T \to X$ as desired. $\square$