The Stacks project

Lemma 35.32.9. 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 an fpqc covering 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 fully faithful.

Proof. Consider the morphism of schemes

\[ g : X = \coprod \nolimits _{i \in I} U_ i \longrightarrow Y = \coprod \nolimits _{j \in J} V_ j \]

over $S$ which on the $i$th component maps into the $\alpha (i)$th component via the morphism $g_{\alpha (i)}$. We claim that $\{ g : X \to Y\} $ is an fpqc covering of schemes. Namely, by Topologies, Lemma 34.9.3 for each $j$ the morphism $\{ \coprod _{\alpha (i) = j} U_ i \to V_ j\} $ is an fpqc covering. Thus for every affine open $V \subset V_ j$ (which we may think of as an affine open of $Y$) we can find finitely many affine opens $W_1, \ldots , W_ n \subset \coprod _{\alpha (i) = j} U_ i$ (which we may think of as affine opens of $X$) such that $V = \bigcup _{i = 1, \ldots , n} g(W_ i)$. This provides enough affine opens of $Y$ which can be covered by finitely many affine opens of $X$ so that Topologies, Lemma 34.9.2 part (3) applies, and the claim follows. Let us write $DD(X/S)$, resp. $DD(\mathcal{U})$ for the category of descent data with respect to $X/S$, resp. $\mathcal{U}$, and similarly for $Y/S$ and $\mathcal{V}$. Consider the diagram

\[ \xymatrix{ DD(Y/S) \ar[r] & DD(X/S) \\ DD(\mathcal{V}) \ar[u]^{\text{Lemma }023X} \ar[r] & DD(\mathcal{U}) \ar[u]_{\text{Lemma }023X} } \]

This diagram is commutative, see the proof of Lemma 35.31.8. The vertical arrows are equivalences. Hence the lemma follows from Lemma 35.32.4 which shows the top horizontal arrow of the diagram is fully faithful. $\square$


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