**Proof.**
The result for $q = 0$ is clear from the definition of $\mathcal{F}^ a$. Let $\mathcal{C} = (\mathit{Sch}/S)_\tau $, or $\mathcal{C} = S_{\acute{e}tale}$, or $\mathcal{C} = S_{Zar}$.

We are going to apply Cohomology on Sites, Lemma 21.10.9 with $\mathcal{F} = \mathcal{F}^ a$, $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the set of affine schemes in $\mathcal{C}$, and $\text{Cov} \subset \text{Cov}_\mathcal {C}$ the set of standard affine $\tau $-coverings. Assumption (3) of the lemma is satisfied by Lemma 35.8.9. Hence we conclude that $H^ p(U, \mathcal{F}^ a) = 0$ for every affine object $U$ of $\mathcal{C}$.

Next, let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be any separated object. Denote $f : U \to S$ the structure morphism. Let $U = \bigcup U_ i$ be an affine open covering. We may also think of this as a $\tau $-covering $\mathcal{U} = \{ U_ i \to U\} $ of $U$ in $\mathcal{C}$. Note that $U_{i_0} \times _ U \ldots \times _ U U_{i_ p} = U_{i_0} \cap \ldots \cap U_{i_ p}$ is affine as we assumed $U$ separated. By Cohomology on Sites, Lemma 21.10.7 and the result above we see that

\[ H^ p(U, \mathcal{F}^ a) = \check{H}^ p(\mathcal{U}, \mathcal{F}^ a) = H^ p(U, f^*\mathcal{F}) \]

the last equality by Cohomology of Schemes, Lemma 30.2.6. In particular, if $S$ is separated we can take $U = S$ and $f = \text{id}_ S$ and the proposition is proved. We suggest the reader skip the rest of the proof (or rewrite it to give a clearer exposition).

Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $ on $S$. Choose an injective resolution $\mathcal{F}^ a \to \mathcal{J}^\bullet $ on $\mathcal{C}$. Denote $\mathcal{J}^ n|_ S$ the restriction of $\mathcal{J}^ n$ to opens of $S$; this is a sheaf on the topological space $S$ as open coverings are $\tau $-coverings. We get a complex

\[ 0 \to \mathcal{F} \to \mathcal{J}^0|_ S \to \mathcal{J}^1|_ S \to \ldots \]

which is exact since its sections over any affine open $U \subset S$ is exact (by the vanishing of $H^ p(U, \mathcal{F}^ a)$, $p > 0$ seen above). Hence by Derived Categories, Lemma 13.18.6 there exists map of complexes $\mathcal{J}^\bullet |_ S \to \mathcal{I}^\bullet $ which in particular induces a map

\[ R\Gamma (\mathcal{C}, \mathcal{F}^ a) = \Gamma (S, \mathcal{J}^\bullet ) \longrightarrow \Gamma (S, \mathcal{I}^\bullet ) = R\Gamma (S, \mathcal{F}). \]

Taking cohomology gives the map $H^ n(\mathcal{C}, \mathcal{F}^ a) \to H^ n(S, \mathcal{F})$ which we have to prove is an isomorphism. Let $\mathcal{U} : S = \bigcup U_ i$ be an affine open covering which we may think of as a $\tau $-covering also. By the above we get a map of double complexes

\[ \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{J}) = \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{J}|_ S) \longrightarrow \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}). \]

This map induces a map of spectral sequences

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

The first spectral sequence converges to $H^{p + q}(\mathcal{C}, \mathcal{F})$ and the second to $H^{p + q}(S, \mathcal{F})$. On the other hand, we have seen that the induced maps ${}^\tau \! E_2^{p, q} \to E_2^{p, q}$ are bijections (as all the intersections are separated being opens in affines). Whence also the maps $H^ n(\mathcal{C}, \mathcal{F}^ a) \to H^ n(S, \mathcal{F})$ are isomorphisms, and we win.
$\square$

## Comments (1)

Comment #1220 by David Corwin on