The Stacks project

Proof. Using assumption (1) and Lemma 60.18.2 we see that $R\Gamma (\text{Cris}(X/S), \mathcal{F})$ is isomorphic to $R\Gamma (\mathcal{C}, \mathcal{F})$. Note that the categories $\mathcal{C}$ used in Lemmas 60.18.2 and 60.18.3 agree. Let $f : T \to T'$ be a closed immersion as in (2). Surjectivity of $c_ f : f^*\mathcal{F}_{T'} \to \mathcal{F}_ T$ is equivalent to surjectivity of $\mathcal{F}_{T'} \to f_*\mathcal{F}_ T$. Hence, if $\mathcal{F}$ satisfies (1) and (2), then we obtain a short exact sequence

of quasi-coherent $\mathcal{O}_{T'}$-modules on $T'$, see Schemes, Section 26.24 and in particular Lemma 26.24.1. Thus, if $T'$ is affine, then we conclude that the restriction map $\mathcal{F}(U', T', \delta ') \to \mathcal{F}(U, T, \delta )$ is surjective by the vanishing of $H^1(T', \mathcal{K})$, see Cohomology of Schemes, Lemma 30.2.2. Hence the transition maps of the inverse systems in Lemma 60.18.3 are surjective. We conclude that $R^ pg_*(\mathcal{F}|_\mathcal {C}) = 0$ for all $p \geq 1$ where $g$ is as in Lemma 60.18.3. The object $D$ of the category $\mathcal{C}^\wedge$ satisfies the assumption of Lemma 60.18.4 by Lemma 60.5.7 with

in $\mathcal{C}$ because $D(n)$ is the $n + 1$-fold coproduct of $D$ in $\text{Cris}^\wedge (C/A)$, see Lemma 60.17.2. Thus we win. $\square$