Lemma 35.32.10. Let $S$ be a scheme. Let $\mathcal{U} = \{ U_ i \to S\} _{i \in I}$, and $\mathcal{V} = \{ V_ j \to S\} _{j \in J}$, be families of morphisms with target $S$. Let $\alpha : I \to J$, $\text{id} : S \to S$ and $g_ i : U_ i \to V_{\alpha (i)}$ be a morphism of families of maps with fixed target, see Sites, Definition 7.8.1. Assume that for each $j \in J$ the family $\{ g_ i : U_ i \to V_ j\} _{\alpha (i) = j}$ is a Zariski covering (see Topologies, Definition 34.3.1) of $V_ j$. Then the pullback functor

$\text{descent data relative to } \mathcal{V} \longrightarrow \text{descent data relative to } \mathcal{U}$

of Lemma 35.31.8 is an equivalence of categories. In particular, the category of schemes over $S$ is equivalent to the category of descent data relative to any Zariski covering of $S$.

Proof. The functor is faithful and fully faithful by Lemma 35.32.9. Let us indicate how to prove that it is essentially surjective. Let $(X_ i, \varphi _{ii'})$ be a descent datum relative to $\mathcal{U}$. Fix $j \in J$ and set $I_ j = \{ i \in I \mid \alpha (i) = j\}$. For $i, i' \in I_ j$ note that there is a canonical morphism

$c_{ii'} : U_ i \times _{g_ i, V_ j, g_{i'}} U_{i'} \to U_ i \times _ S U_{i'}.$

Hence we can pullback $\varphi _{ii'}$ by this morphism and set $\psi _{ii'} = c_{ii'}^*\varphi _{ii'}$ for $i, i' \in I_ j$. In this way we obtain a descent datum $(X_ i, \psi _{ii'})$ relative to the Zariski covering $\{ g_ i : U_ i \to V_ j\} _{i \in I_ j}$. Note that $\psi _{ii'}$ is an isomorphism from the open $X_{i, U_ i \times _{V_ j} U_{i'}}$ of $X_ i$ to the corresponding open of $X_{i'}$. It follows from Schemes, Section 26.14 that we may glue $(X_ i, \psi _{ii'})$ into a scheme $Y_ j$ over $V_ j$. Moreover, the morphisms $\varphi _{ii'}$ for $i \in I_ j$ and $i' \in I_{j'}$ glue to a morphism $\varphi _{jj'} : Y_ j \times _ S V_{j'} \to V_ j \times _ S Y_{j'}$ satisfying the cocycle condition (details omitted). Hence we obtain the desired descent datum $(Y_ j, \varphi _{jj'})$ relative to $\mathcal{V}$. $\square$

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