The Stacks project

Proof. We are going to show that $H^ p(U, f^*\mathcal{F}) = H^ p_\tau (U, \mathcal{F}^ a)$ for any object $f : U \to S$ of the site $\mathcal{C}$. The result is true for $p = 0$ by the sheaf property.

Assume that $U$ is affine. Then we want to prove that $H^ p_\tau (U, \mathcal{F}^ a) = 0$ for all $p > 0$. We use induction on $p$.

1. Pick $\xi \in H^1_\tau (U, \mathcal{F}^ a)$. By Lemma 59.22.3, there exists an fpqc covering $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ such that $\xi |_{U_ i} = 0$ for all $i \in I$. Up to refining $\mathcal{U}$, we may assume that $\mathcal{U}$ is a standard $\tau $-covering. Applying the spectral sequence of Theorem 59.19.2, we see that $\xi $ comes from a cohomology class $\check\xi \in \check H^1(\mathcal{U}, \mathcal{F}^ a)$. Consider the covering $\mathcal{V} = \{ \coprod _{i\in I} U_ i \to U\} $. By Lemma 59.22.1, $\check H^\bullet (\mathcal{U}, \mathcal{F}^ a) = \check H^\bullet (\mathcal{V}, \mathcal{F}^ a)$. On the other hand, since $\mathcal{V}$ is a covering of the form $\{ \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)\} $ and $f^*\mathcal{F} = \widetilde{M}$ for some $A$-module $M$, we see the Čech complex $\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{F})$ is none other than the complex $(B/A)_\bullet \otimes _ A M$. Now by Lemma 59.16.4, $H^ p((B/A)_\bullet \otimes _ A M) = 0$ for $p > 0$, hence $\check\xi = 0$ and so $\xi = 0$.

2. Pick $\xi \in H^ p_\tau (U, \mathcal{F}^ a)$. By Lemma 59.22.3, there exists an fpqc covering $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ such that $\xi |_{U_ i} = 0$ for all $i \in I$. Up to refining $\mathcal{U}$, we may assume that $\mathcal{U}$ is a standard $\tau $-covering. We apply the spectral sequence of Theorem 59.19.2. Observe that the intersections $U_{i_0} \times _ U \ldots \times _ U U_{i_ p}$ are affine, so that by induction hypothesis the cohomology groups

   \[ E_2^{p, q} = \check H^ p(\mathcal{U}, \underline{H}^ q(\mathcal{F}^ a)) \]

   vanish for all $0 < q < p$. We see that $\xi $ must come from a $\check\xi \in \check H^ p(\mathcal{U}, \mathcal{F}^ a)$. Replacing $\mathcal{U}$ with the covering $\mathcal{V}$ containing only one morphism and using Lemma 59.16.4 again, we see that the Čech cohomology class $\check\xi $ must be zero, hence $\xi = 0$.

Next, assume that $U$ is separated. Choose an affine open covering $U = \bigcup _{i \in I} U_ i$ of $U$. The family $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ is then an fpqc covering, and all the intersections $U_{i_0} \times _ U \ldots \times _ U U_{i_ p}$ are affine since $U$ is separated. So all rows of the spectral sequence of Theorem 59.19.2 are zero, except the zeroth row. Therefore

where the last equality results from standard scheme theory, see Cohomology of Schemes, Lemma 30.2.6.

The general case is technical and (to extend the proof as given here) requires a discussion about maps of spectral sequences, so we won't treat it. It follows from Descent, Proposition 35.9.3 (whose proof takes a slightly different approach) combined with Cohomology on Sites, Lemma 21.7.1. $\square$