Lemma 8.12.2. Let $u : \mathcal{C} \to \mathcal{D}$ be a continuous functor of sites. Let $p : \mathcal{S} \to \mathcal{D}$ be a stack over $\mathcal{D}$. Then $u^ p\mathcal{S}$ is a stack over $\mathcal{C}$.

**Proof.**
We have seen in Lemma 8.12.1 that $u^ p\mathcal{S}$ is a fibred category over $\mathcal{C}$. Moreover, in the proof of that lemma we have seen that a morphism $(a, \beta )$ of $u^ p\mathcal{S}$ is strongly cartesian if and only $\beta $ is strongly cartesian in $\mathcal{S}$. Hence, given a morphism $a : U \to U'$ of $\mathcal{C}$, not only do we have the equalities $(u^ p\mathcal{S})_ U = \mathcal{S}_ U$ and $(u^ p\mathcal{S})_{U'} = \mathcal{S}_{U'}$, but via these equalities the pullback functors agree; in a formula $a^*(U', y') = (U, u(a)^*y')$.

Having said this, let $\mathcal{U} = \{ U_ i \to U\} $ be a covering of $\mathcal{C}$. As $u$ is continuous we see that $\mathcal{V} = \{ u(U_ i) \to u(U)\} $ is a covering of $\mathcal{D}$, and that $u(U_ i \times _ U U_ j) = u(U_ i) \times _{u(U)} u(U_ j)$ and similarly for the triple fibre products $U_ i \times _ U U_ j \times _ U U_ k$. As we have the identifications of fibre categories and pullbacks we see that descend data relative to $\mathcal{U}$ are identical to descend data relative to $\mathcal{V}$. Since by assumption we have effective descent in $\mathcal{S}$ we conclude the same holds for $u^ p\mathcal{S}$. $\square$

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## Comments (2)

Comment #1100 by S.Carnahan on

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