Lemma 85.12.2. In Situation 85.3.3.

1. A sheaf $\mathcal{F}$ of sets or abelian groups is cartesian if and only if the maps $(f_{\delta ^ n_ j})^{-1}\mathcal{F}_{n - 1} \to \mathcal{F}_ n$ are isomorphisms.

2. An object $K$ of $D(\mathcal{C}_{total})$ is cartesian if and only if the maps $(f_{\delta ^ n_ j})^{-1}K_{n - 1} \to K_ n$ are isomorphisms.

3. If $\mathcal{O}$ is a sheaf of rings on $\mathcal{C}_{total}$ a sheaf $\mathcal{F}$ of $\mathcal{O}$-modules is cartesian if and only if the maps $(f_{\delta ^ n_ j})^*\mathcal{F}_{n - 1} \to \mathcal{F}_ n$ are isomorphisms.

4. If $\mathcal{O}$ is a sheaf of rings on $\mathcal{C}_{total}$ an object $K$ of $D(\mathcal{O})$ is cartesian if and only if the maps $L(f_{\delta ^ n_ j})^*K_{n - 1} \to K_ n$ are isomorphisms.

Proof. In each case the key is that the pullback functors compose to pullback functor; for part (4) see Cohomology on Sites, Lemma 21.18.3. We show how the argument works in case (1) and omit the proof in the other cases. The category $\Delta$ is generated by the morphisms the morphisms $\delta ^ n_ j$ and $\sigma ^ n_ j$, see Simplicial, Lemma 14.2.2. Hence we only need to check the maps $(f_{\delta ^ n_ j})^{-1}\mathcal{F}_{n - 1} \to \mathcal{F}_ n$ and $(f_{\sigma ^ n_ j})^{-1}\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ are isomorphisms, see Simplicial, Lemma 14.3.2 for notation. Since $\sigma ^ n_ j \circ \delta _ j^{n + 1} = \text{id}_{[n]}$ the composition

$\mathcal{F}_ n = (f_{\sigma ^ n_ j})^{-1} (f_{\delta _ j^{n + 1}})^{-1} \mathcal{F}_ n \to (f_{\sigma ^ n_ j})^{-1} \mathcal{F}_{n + 1} \to \mathcal{F}_ n$

is the identity. Thus the result for $\delta ^{n + 1}_ j$ implies the result for $\sigma ^ n_ j$. $\square$

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