Lemma 96.19.11. Let $f : \mathcal{U} \to \mathcal{X}$ and $g : \mathcal{X} \to \mathcal{Y}$ be composable $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Let $\tau \in \{ Zar, {\acute{e}tale}, smooth, syntomic, \linebreak[0] fppf\} $. Assume
$\mathcal{F}$ is an abelian sheaf on $\mathcal{X}_\tau $,
for every object $x$ of $\mathcal{X}$ there exists a covering $\{ x_ i \to x\} $ in $\mathcal{X}_\tau $ such that each $x_ i$ is isomorphic to $f(u_ i)$ for some object $u_ i$ of $\mathcal{U}$,
the category $\mathcal{U}$ has equalizers, and
the functor $f$ is faithful.
Then there is a first quadrant spectral sequence of abelian sheaves on $\mathcal{Y}_\tau $
\[ E_1^{p, q} = R^ q(g \circ f_ p)_*f_ p^{-1}\mathcal{F} \Rightarrow R^{p + q}g_*\mathcal{F} \]
where all higher direct images are computed in the $\tau $-topology.
Proof.
Note that the assumptions on $f : \mathcal{U} \to \mathcal{X}$ and $\mathcal{F}$ are identical to those in Lemma 96.19.8. Hence the preliminary remarks made in the proof of that lemma hold here also. These remarks imply in particular that
\[ 0 \to g_*\mathcal{I} \to (g \circ f_0)_*f_0^{-1}\mathcal{I} \to (g \circ f_1)_*f_1^{-1}\mathcal{I} \to \ldots \]
is exact if $\mathcal{I}$ is an injective object of $\textit{Ab}(\mathcal{X}_\tau )$. Having said this, consider the two spectral sequences of Homology, Section 12.25 associated to the double complex $\mathcal{C}^{\bullet , \bullet }$ with terms
\[ \mathcal{C}^{p, q} = (g \circ f_ p)_*\mathcal{I}^ q \]
where $\mathcal{F} \to \mathcal{I}^\bullet $ is an injective resolution in $\textit{Ab}(\mathcal{X}_\tau )$. The first spectral sequence implies, via Homology, Lemma 12.25.4, that $g_*\mathcal{I}^\bullet $ is quasi-isomorphic to the total complex associated to $\mathcal{C}^{\bullet , \bullet }$. Since $f_ p^{-1}\mathcal{I}^\bullet $ is an injective resolution of $f_ p^{-1}\mathcal{F}$ (see Lemma 96.17.5) the second spectral sequence has terms $E_1^{p, q} = R^ q(g \circ f_ p)_*f_ p^{-1}\mathcal{F}$ as in the statement of the lemma.
$\square$
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