The Stacks project

Proof. To prove this we will use Schemes, Lemma 26.15.4. First we define a contravariant functor $F$ from the category of schemes to the category of sets. Namely, for a scheme $T$ we set

The restriction mapping $F(T) \to F(T')$ given a morphism $T' \to T$ is just gotten by composition. For any $W \in \mathcal{B}$ we consider the subfunctor $F_ W \subset F$ consisting of those systems $(g, \{ h_ U\} )$ such that $g(T) \subset W$.

First we show $F$ satisfies the sheaf property for the Zariski topology. Suppose that $T$ is a scheme, $T = \bigcup V_ i$ is an open covering, and $\xi _ i \in F(V_ i)$ is an element such that $\xi _ i|_{V_ i \cap V_ j} = \xi _ j|_{V_ i \cap V_ j}$. Say $\xi _ i = (g_ i, \{ h_{i, U}\} )$. Then we immediately see that the morphisms $g_ i$ glue to a unique global morphism $g : T \to S$. Moreover, it is clear that $g^{-1}(U) = \bigcup g_ i^{-1}(U)$. Hence the morphisms $h_{i, U} : g_ i^{-1}(U) \to X_ U$ glue to a unique morphism $h_ U : g^{-1}(U) \to X_ U$. It is easy to verify that the system $(g, \{ h_ U\} )$ is an element of $F(T)$. Hence $F$ satisfies the sheaf property for the Zariski topology.

Next we verify that each $F_ W$, $W \in \mathcal{B}$ is representable. Namely, we claim that the transformation of functors

is an isomorphism. To see this suppose that $T$ is a scheme and $\alpha : T \to X_ W$ is a morphism. Set $g = f_ W \circ \alpha $. For any $U \in \mathcal{B}$ such that $U \subset W$ we can define $h_ U : g^{-1}(U) \to X_ U$ be the composition $(\rho ^ W_ U)^{-1} \circ \alpha |_{g^{-1}(U)}$. This works because the image $\alpha (g^{-1}(U))$ is contained in $f_ W^{-1}(U)$ and condition (a) of the lemma. It is clear that $f_ U \circ h_ U = g|_{g^{-1}(U)}$ for such a $U$. Moreover, if also $V \in \mathcal{B}$ and $V \subset U \subset W$, then $\rho ^ U_ V \circ h_ V = h_ U|_{g^{-1}(V)}$ by property (b) of the lemma. We still have to define $h_ U$ for an arbitrary element $U \in \mathcal{B}$. Since $\mathcal{B}$ is a basis for the topology on $S$ we can find an open covering $U \cap W = \bigcup U_ i$ with $U_ i \in \mathcal{B}$. Since $g$ maps into $W$ we have $g^{-1}(U) = g^{-1}(U \cap W) = \bigcup g^{-1}(U_ i)$. Consider the morphisms $h_ i = \rho ^ U_{U_ i} \circ h_{U_ i} : g^{-1}(U_ i) \to X_ U$. It is a simple matter to use condition (b) of the lemma to prove that $h_ i|_{g^{-1}(U_ i) \cap g^{-1}(U_ j)} = h_ j|_{g^{-1}(U_ i) \cap g^{-1}(U_ j)}$. Hence these morphisms glue to give the desired morphism $h_ U : g^{-1}(U) \to X_ U$. We omit the (easy) verification that the system $(g, \{ h_ U\} )$ is an element of $F_ W(T)$ which maps to $\alpha $ under the displayed arrow above.

Next, we verify each $F_ W \subset F$ is representable by open immersions. This is clear from the definitions.

Finally we have to verify the collection $(F_ W)_{W \in \mathcal{B}}$ covers $F$. This is clear by construction and the fact that $\mathcal{B}$ is a basis for the topology of $S$.

Let $X$ be a scheme representing the functor $F$. Let $(f, \{ i_ U\} ) \in F(X)$ be a “universal family”. Since each $F_ W$ is representable by $X_ W$ (via the morphism of functors displayed above) we see that $i_ W : f^{-1}(W) \to X_ W$ is an isomorphism as desired. The lemma is proved. $\square$