Lemma 84.23.3. Let $(\mathcal{C}, \mathcal{O}_\mathcal {C})$ be a ringed site. Assume given weak Serre subcategories $\mathcal{A}_ U \subset \textit{Mod}(\mathcal{O}_ U)$ satisfying conditions (1), (2), and (3) above. Assume $\mathcal{C}$ has equalizers and fibre products and let $K$ be a hypercovering. Let $((\mathcal{C}/K)_{total}, \mathcal{O})$ be as in Remark 84.16.5. Let $\mathcal{A}_{total} \subset \textit{Mod}(\mathcal{O})$ denote the weak Serre subcategory of cartesian $\mathcal{O}$-modules $\mathcal{F}$ whose restriction $\mathcal{F}_ n$ is in $\mathcal{A}_{K_ n}$ for all $n$ (as defined above). Then the functor $La^*$ defines an equivalence

\[ D_\mathcal {A}(\mathcal{O}_\mathcal {C}) \longrightarrow D_{\mathcal{A}_{total}}(\mathcal{O}) \]

with quasi-inverse $Ra_*$.

**Proof.**
The cartesian $\mathcal{O}$-modules form a weak Serre subcategory by Lemma 84.12.6 (the required hypotheses hold by the discussion in Remark 84.16.5). Since the restriction functor $g_ n^* : \textit{Mod}(\mathcal{O}) \to \textit{Mod}(\mathcal{O}_ n)$ are exact, it follows that $\mathcal{A}_{total}$ is a weak Serre subcategory.

Let us show that $a^* : \mathcal{A} \to \mathcal{A}_{total}$ is an equivalence of categories with inverse given by $La_*$. We already know that $La_*a^*\mathcal{F} = \mathcal{F}$ by the bounded version (Lemma 84.18.4). It is clear that $a^*\mathcal{F}$ is in $\mathcal{A}_{total}$ for $\mathcal{F}$ in $\mathcal{A}$. Conversely, assume that $\mathcal{G} \in \mathcal{A}_{total}$. Because $\mathcal{G}$ is cartesian we see that $\mathcal{G} = a^*\mathcal{F}$ for some $\mathcal{O}_\mathcal {C}$-module $\mathcal{F}$ by Lemma 84.18.1. We want to show that $\mathcal{F}$ is in $\mathcal{A}$. Take $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. We have to show that the restriction of $\mathcal{F}$ to $\mathcal{C}/U$ is in $\mathcal{A}_ U$. As usual, write $K_0 = \{ U_{0, i}\} _{i \in I_0}$. Since $K$ is a hypercovering, the map $\coprod _{i \in I_0} h_{U_{0, i}} \to *$ becomes surjective after sheafification. This implies there is a covering $\{ U_ j \to U\} _{j \in J}$ and a map $\tau : J \to I_0$ and for each $j \in J$ a morphism $\varphi _ j : U_ j \to U_{0, \tau (j)}$. Since $\mathcal{G}_0 = a_0^*\mathcal{F}$ we find that the restriction of $\mathcal{F}$ to $\mathcal{C}/U_ j$ is equal to the restriction of the $\tau (j)$th component of $\mathcal{G}_0$ to $\mathcal{C}/U_ j$ via the morphism $\varphi _ j : U_ j \to U_{0, \tau (i)}$. Hence by (1) we find that $\mathcal{F}|_{\mathcal{C}/U_ j}$ is in $\mathcal{A}_{U_ j}$ and in turn by (2) we find that $\mathcal{F}|_{\mathcal{C}/U}$ is in $\mathcal{A}_ U$.

In particular the statement of the lemma makes sense. The lemma now follows from Cohomology on Sites, Lemma 21.28.6. Assumption (1) is clear (see Remark 84.16.5). Assumptions (2) and (3) we proved in the preceding paragraph. Assumption (4) is immediate from (3). For assumption (5) let $\mathcal{B}_{total}$ be the set of objects $U/U_{n, i}$ of the site $(\mathcal{C}/K)_{total}$ such that $U \in \mathcal{B}$ where $\mathcal{B}$ is as in (3). Here we use the description of the site $(\mathcal{C}/K)_{total}$ given in Section 84.16. Moreover, we set $\text{Cov}_{U/U_{n, i}}$ equal to $\text{Cov}_ U$ and $d_{U/U_{n, i}}$ equal $d_ U$ where $\text{Cov}_ U$ and $d_ U$ are given to us by (3). Then we claim that condition (5) holds with these choices. This follows immediately from Lemma 84.16.3 and the fact that $\mathcal{F} \in \mathcal{A}_{total}$ implies $\mathcal{F}_ n \in \mathcal{A}_{K_ n}$ and hence $\mathcal{F}_{n, i} \in \mathcal{A}_{U_{n, i}}$. (The reader who worries about the difference between cohomology of abelian sheaves versus cohomology of sheaves of modules may consult Cohomology on Sites, Lemma 21.12.4.)
$\square$

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