Lemma 21.40.1. Assumptions and notation as in Situation 21.38.1. For $\mathcal{F}$ in $\textit{PAb}(\mathcal{C})$ and $n \geq 0$ consider the abelian sheaf $L_ n(\mathcal{F})$ on $\mathcal{D}$ which is the sheaf associated to the presheaf

$V \longmapsto H_ n(\mathcal{C}_ V, \mathcal{F}|_{\mathcal{C}_ V})$

with restriction maps as indicated in the proof. Then $L_ n(\mathcal{F}) = L_ n(\mathcal{F}^\# )$.

Proof. For a morphism $h : V' \to V$ of $\mathcal{D}$ there is a pullback functor $h^* : \mathcal{C}_ V \to \mathcal{C}_{V'}$ of fibre categories (Categories, Definition 4.33.6). Moreover for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ V)$ there is a strongly cartesian morphism $h^*U \to U$ covering $h$. Restriction along these strongly cartesian morphisms defines a transformation of functors

$\mathcal{F}|_{\mathcal{C}_ V} \longrightarrow \mathcal{F}|_{\mathcal{C}_{V'}} \circ h^*.$

By Example 21.39.3 we obtain the desired restriction map

$H_ n(\mathcal{C}_ V, \mathcal{F}|_{\mathcal{C}_ V}) \longrightarrow H_ n(\mathcal{C}_{V'}, \mathcal{F}|_{\mathcal{C}_{V'}})$

Let us denote $L_{n, p}(\mathcal{F})$ this presheaf, so that $L_ n(\mathcal{F}) = L_{n, p}(\mathcal{F})^\#$. The canonical map $\gamma : \mathcal{F} \to \mathcal{F}^+$ (Sites, Theorem 7.10.10) defines a canonical map $L_{n, p}(\mathcal{F}) \to L_{n, p}(\mathcal{F}^+)$. We have to prove this map becomes an isomorphism after sheafification.

Let us use the computation of homology given in Example 21.39.2. Denote $K_\bullet (\mathcal{F}|_{\mathcal{C}_ V})$ the complex associated to the restriction of $\mathcal{F}$ to the fibre category $\mathcal{C}_ V$. By the remarks above we obtain a presheaf $K_\bullet (\mathcal{F})$ of complexes

$V \longmapsto K_\bullet (\mathcal{F}|_{\mathcal{C}_ V})$

whose cohomology presheaves are the presheaves $L_{n, p}(\mathcal{F})$. Thus it suffices to show that

$K_\bullet (\mathcal{F}) \longrightarrow K_\bullet (\mathcal{F}^+)$

becomes an isomorphism on sheafification.

Injectivity. Let $V$ be an object of $\mathcal{D}$ and let $\xi \in K_ n(\mathcal{F})(V)$ be an element which maps to zero in $K_ n(\mathcal{F}^+)(V)$. We have to show there exists a covering $\{ V_ j \to V\}$ such that $\xi |_{V_ j}$ is zero in $K_ n(\mathcal{F})(V_ j)$. We write

$\xi = \sum (U_{i, n + 1} \to \ldots \to U_{i, 0}, \sigma _ i)$

with $\sigma _ i \in \mathcal{F}(U_{i, 0})$. We arrange it so that each sequence of morphisms $U_ n \to \ldots \to U_0$ of $\mathcal{C}_ V$ occurs are most once. Since the sums in the definition of the complex $K_\bullet$ are direct sums, the only way this can map to zero in $K_\bullet (\mathcal{F}^+)(V)$ is if all $\sigma _ i$ map to zero in $\mathcal{F}^+(U_{i, 0})$. By construction of $\mathcal{F}^+$ there exist coverings $\{ U_{i, 0, j} \to U_{i, 0}\}$ such that $\sigma _ i|_{U_{i, 0, j}}$ is zero. By our construction of the topology on $\mathcal{C}$ we can write $U_{i, 0, j} \to U_{i, 0}$ as the pullback (Categories, Definition 4.33.6) of some morphisms $V_{i, j} \to V$ and moreover each $\{ V_{i, j} \to V\}$ is a covering. Choose a covering $\{ V_ j \to V\}$ dominating each of the coverings $\{ V_{i, j} \to V\}$. Then it is clear that $\xi |_{V_ j} = 0$.

Surjectivity. Proof omitted. Hint: Argue as in the proof of injectivity. $\square$

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