Lemma 8.3.7. Let \mathcal{C} be a category. Let \mathcal{V} = \{ V_ j \to U\} _{j \in J} \to \mathcal{U} = \{ U_ i \to U\} _{i \in I} be a morphism of families of maps with fixed target of \mathcal{C} given by \text{id} : U \to U, \alpha : J \to I and f_ j : V_ j \to U_{\alpha (j)}. Let p : \mathcal{S} \to \mathcal{C} be a fibred category. If
for 0 \leq p \leq 3 and 0 \leq q \leq 3 with p + q \geq 2 and i_1, \ldots , i_ p \in I and j_1, \ldots , j_ q \in J the fibre products U_{i_1} \times _ U \ldots \times _ U U_{i_ p} \times _ U V_{j_1} \times _ U \ldots \times _ U V_{j_ q} exist,
the functor \mathcal{S}_ U \to DD(\mathcal{V}) is an equivalence,
for every i \in I the functor \mathcal{S}_{U_ i} \to DD(\mathcal{V}_ i) is fully faithful, and
for every i, i' \in I the functor \mathcal{S}_{U_ i \times _ U U_{i'}} \to DD(\mathcal{V}_{ii'}) is faithful.
Here \mathcal{V}_ i = \{ U_ i \times _ U V_ j \to U_ i\} _{j \in J} and \mathcal{V}_{ii'} = \{ U_ i \times _ U U_{i'} \times _ U V_ j \to U_ i \times _ U U_{i'}\} _{j \in J}. Then \mathcal{S}_ U \to DD(\mathcal{U}) is an equivalence.
Proof.
Condition (1) guarantees we have enough fibre products so that the statement makes sense. We will show that the functor \mathcal{S}_ U \to DD(\mathcal{U}) is essentially surjective. Suppose given a descent datum (X_ i, \varphi _{ii'}) relative to \mathcal{U}. By Lemma 8.3.3 we can pull this back to a descent datum (X_ j, \varphi _{jj'}) for \mathcal{V}. By assumption (2) this descent datum is effective, hence we get an object X of \mathcal{S}_ U such that the pullback of the trivial descent datum (X, \text{id}_ X) by the morphism \mathcal{V} \to \{ U \to U\} is isomorphic to (X_ j, \varphi _{jj'}). Next, observe that we have a diagram
\xymatrix{ \mathcal{V}_ i \ar[r] \ar[d] & \mathcal{V} \ar[r] & \mathcal{U} \ar[d] \\ \{ U_ i \to U_ i\} \ar[rr] \ar[rru] & & \{ U \to U\} }
of morphisms of families of maps with fixed target of \mathcal{C}. This diagram does not commute, but by Lemma 8.3.3 the pullback functors on descent data one gets are canonically isomorphic. Hence (X, \text{id}_ X) and (X_ i, \text{id}_{X_ i}) pull back to isomorphic objects in DD(\mathcal{V}_ i). Hence by assumption (3) we obtain an isomorphism (U_ i \to U)^*X \to X_ i in the category \mathcal{S}_{U_ i}. We omit the verification that these arrows are compatible with the morphisms \varphi _{ii'}; hint: use the faithfulness of the functors in condition (4). We also omit the verification that the functor \mathcal{S}_ U \to DD(\mathcal{U}) is fully faithful.
\square
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