The Stacks project

Proof. This follows because

and the fact that $\mathcal{F}(\coprod _ a V_ a) = \prod _ a \mathcal{F}(V_ a)$ since disjoint unions are $\tau $-coverings. $\square$

Proof. By Lemma 35.9.1 we see that $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}^ a)$ is isomorphic to $\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{F}^ a)$ where $\mathcal{V} = \{ V \to U\} $ with $V = \coprod _{i = 1, \ldots n} U_ i$ affine also. Set $A = \Gamma (V, \mathcal{O}_ V)$. Since $\{ V \to U\} $ is a $\tau $-covering we see that $R \to A$ is faithfully flat. On the other hand, by definition of $\mathcal{F}^ a$ we have that the degree $p$ term $\check{\mathcal{C}}^ p(\mathcal{V}, \mathcal{F}^ a)$ is

We omit the verification that the maps of the Čech complex agree with the maps in the complex $s((A/R)_\bullet \otimes _ R M)$. The vanishing of cohomology is Lemma 35.3.6. $\square$

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 35.9.2. 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

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

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

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

This map induces a map of spectral sequences

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. $\square$

Proof. Part (1) follows from the discussion in Remark 35.8.6. Part (2) is just a warning, and can be explained in the following way: First the statement cannot be made precise since $f_*$ does not transform quasi-coherent sheaves into quasi-coherent sheaves in general. Even if this is the case for $f$ (and any base change of $f$), then the compatibility over the big sites would mean that formation of $f_*\mathcal{F}$ commutes with any base change, which does not hold in general. An explicit example is the quasi-compact open immersion $j : X = \mathbf{A}^2_ k \setminus \{ 0\} \to \mathbf{A}^2_ k = Y$ where $k$ is a field. We have $j_*\mathcal{O}_ X = \mathcal{O}_ Y$ but after base change to $\mathop{\mathrm{Spec}}(k)$ by the $0$ map we see that the pushforward is zero.

Let us prove (3) in case $\tau = {\acute{e}tale}$. Note that $f$, and any base change of $f$, transforms quasi-coherent sheaves into quasi-coherent sheaves, see Schemes, Lemma 26.24.1. The equality $f_{small, *}(\mathcal{F}^ a) = (f_*\mathcal{F})^ a$ means that for any étale morphism $g : U \to S$ we have $\Gamma (U, g^*f_*\mathcal{F}) = \Gamma (U \times _ S T, (g')^*\mathcal{F})$ where $g' : U \times _ S T \to T$ is the projection. This is true by Cohomology of Schemes, Lemma 30.5.2. $\square$

Proof. We prove this for the étale topology; we omit the proof in the case of the Zariski topology. By Cohomology of Schemes, Lemma 30.4.5 the sheaves $R^ if_*\mathcal{F}$ are quasi-coherent so that the assertion makes sense. The sheaf $R^ if_{small, *}\mathcal{F}^ a$ is the sheaf associated to the presheaf

where $g : U \to S$ is an object of $S_{\acute{e}tale}$, see Cohomology on Sites, Lemma 21.7.4. By our conventions the right hand side is the étale cohomology of the restriction of $\mathcal{F}^ a$ to the localization $T_{\acute{e}tale}/U \times _ S T$ which equals $(U \times _ S T)_{\acute{e}tale}$. By Proposition 35.9.3 this is presheaf the same as the presheaf

where $g' : U \times _ S T \to T$ is the projection. If $U$ is affine then this is the same as $H^0(U, R^ if'_*(g')^*\mathcal{F})$, see Cohomology of Schemes, Lemma 30.4.6. By Cohomology of Schemes, Lemma 30.5.2 this is equal to $H^0(U, g^*R^ if_*\mathcal{F})$ which is the value of $(R^ if_*\mathcal{F})^ a$ on $U$. Thus the values of the sheaves of modules $R^ if_{small, *}(\mathcal{F}^ a)$ and $(R^ if_*\mathcal{F})^ a$ on every affine object of $S_{\acute{e}tale}$ are canonically isomorphic which implies they are canonically isomorphic. $\square$