Lemma 27.2.1. Let $S$ be a scheme. Let $\mathcal{B}$ be a basis for the topology of $S$. Suppose given the following data:

1. For every $U \in \mathcal{B}$ a scheme $f_ U : X_ U \to U$ over $U$.

2. For $U, V \in \mathcal{B}$ with $V \subset U$ a morphism $\rho ^ U_ V : X_ V \to X_ U$ over $U$.

Assume that

1. each $\rho ^ U_ V$ induces an isomorphism $X_ V \to f_ U^{-1}(V)$ of schemes over $V$,

2. whenever $W, V, U \in \mathcal{B}$, with $W \subset V \subset U$ we have $\rho ^ U_ W = \rho ^ U_ V \circ \rho ^ V_ W$.

Then there exists a morphism $f : X \to S$ of schemes and isomorphisms $i_ U : f^{-1}(U) \to X_ U$ over $U \in \mathcal{B}$ such that for $V, U \in \mathcal{B}$ with $V \subset U$ the composition

$\xymatrix{ X_ V \ar[r]^{i_ V^{-1}} & f^{-1}(V) \ar[rr]^{inclusion} & & f^{-1}(U) \ar[r]^{i_ U} & X_ U }$

is the morphism $\rho ^ U_ V$. Moreover $X$ is unique up to unique isomorphism over $S$.

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

$F(T) = \left\{ \begin{matrix} (g, \{ h_ U\} _{U \in \mathcal{B}}), \ g : T \to S, \ h_ U : g^{-1}(U) \to X_ U, \\ f_ U \circ h_ U = g|_{g^{-1}(U)}, \ h_ U|_{g^{-1}(V)} = \rho ^ U_ V \circ h_ V \ \forall \ V, U \in \mathcal{B}, V \subset U \end{matrix} \right\} .$

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

$F_ W \longrightarrow \mathop{\mathrm{Mor}}\nolimits (-, X_ W), \ (g, \{ h_ U\} ) \longmapsto h_ W$

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. $\square$

Comment #567 by Will Chen on

In the proof, "Say $\xi_i = (g_i,\{f_{i,U}\}$", the $f_{i,U}$ should be $h_{i,U}$.

Comment #4017 by Jonas Ehrhard on

Do the morphisms $\rho^U_V$ have to be morphisms over $U$? Otherwise I don't see why $\rho^U_V: X_V \to X_U$ ends up in $f_U^{-1}(V)$.

Comment #5896 by Taro konno on

In the second paragraph of the proof , two typos ;

1. $h_ U : U \to X_ U$ should be $h_ U :g^{-1}(U) \to X_ U$ in "Hence the morphisms $h_{i, U} : g_ i^{-1}(U) \to X_ U$ glue to a unique morphism $h_ U : U \to X_ U$. "

2. $(g, \lbrace f_ U \rbrace )$ should be $(g, \lbrace h_ U \rbrace)$ in "It is easy to verify that the system $(g, \lbrace f_ U\rbrace )$ is an element of $F(T)$. "

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