Cofinal Čech vanishing implies derived acyclicity

By [F3] the sheaf $\mathcal{F}$ embeds into an injective sheaf $\mathcal{I}$, the transfinite construction behind [F3] using the Axiom of Dependent Choice, which is available by [F11]; put $\mathcal{Q}:=\mathcal{I}\mathrm{/}\mathcal{F}$, so that $0\to \mathcal{F}\to \mathcal{I}\to \mathcal{Q}\to 0$ is exact, with $\mathcal{F}$ the kernel of $\mathcal{I}\to \mathcal{Q}$ and $\mathcal{Q}$ the cokernel by [F6]. The morphism $\mathcal{I}\to \mathcal{Q}$ is an epimorphism of sheaves, so by [F5] its stalk maps are surjective and every section of $\mathcal{Q}$ over an open set is locally in its image.

Let $U\in \mathcal{B}$ and $s\in \mathcal{Q}(U)$. Since $\mathcal{I}\to \mathcal{Q}$ is an epimorphism, for every $x\in U$ the germ $s_x$ lies in the image of ${\mathcal{I}}_x\to {\mathcal{Q}}_x$ by [F5], and because equality of germs of sections of a sheaf is a local condition there is an open neighbourhood $W_x\subseteq U$ of $x$ and $t_x\in \mathcal{I}(W_x)$ with image $s{\mathrm{\mid }}_{W_x}$; the family $(W_x{)}_{x\in U}$ is an open cover of $U$ and each $s{\mathrm{\mid }}_{W_x}$ is the image of a section of $\mathcal{I}$ over $W_x$. By cofinality of $\mathrm{C}\mathrm{o}\mathrm{v}(U)$ this cover has a refinement $\mathcal{U}=(U_i{)}_{i\in I}\in \mathrm{C}\mathrm{o}\mathrm{v}(U)$: for every $i$ the member $U_i$ is contained in some $W_x$, and restricting the corresponding $t_x$ yields $t_i\in \mathcal{I}(U_i)$ with image $s{\mathrm{\mid }}_{U_i}$. The cover $\mathcal{U}$ is finite and every finite intersection of its members lies in $\mathcal{B}$ by hypothesis, and ${\check{H}}^1(\mathcal{U},\mathcal{F})=0$.

With the data of [step 2.1], the sections $s_{ij}:=t_j{\mathrm{\mid }}_{U_i\cap U_j}-t_i{\mathrm{\mid }}_{U_i\cap U_j}$ for $i<j$ take values in $\mathcal{F}(U_i\cap U_j)$, because both $t_j$ and $t_i$ map to $s$ on the overlap and $\mathcal{F}$ is the kernel of $\mathcal{I}\to \mathcal{Q}$ by [F6]; the alternating family $s_{\bullet }$ indexed by increasing pairs is a Čech $1$-cocycle, since it is the Čech differential of the $0$-cochain $t_{\bullet }$ with values in $\mathcal{I}$, so its Čech differential vanishes by [F4]. Because ${\check{H}}^1(\mathcal{U},\mathcal{F})=\ker {\delta }^1\mathrm{/}\mathrm{im}{\delta }^0$ is zero by hypothesis and by [F1], it is a Čech coboundary: there are ${\tau }_i\in \mathcal{F}(U_i)$ with $s_{ij}={\tau }_j{\mathrm{\mid }}_{U_i\cap U_j}-{\tau }_i{\mathrm{\mid }}_{U_i\cap U_j}$ by [F4]. The sections $t_i-{\tau }_i\in \mathcal{I}(U_i)$ then satisfy $(t_j-{\tau }_j)\mathrm{\mid }=t_j\mathrm{\mid }-{\tau }_j\mathrm{\mid }=t_i\mathrm{\mid }-{\tau }_i\mathrm{\mid }=(t_i-{\tau }_i)\mathrm{\mid }$ on each overlap, so they glue to a section $t\in \mathcal{I}(U)$; since each ${\tau }_i$ lies in $\mathcal{F}$ and hence maps to $0$ in $\mathcal{Q}$, the image of $t$ is the family $(s{\mathrm{\mid }}_{U_i}{)}_i$, that is, $t$ is a lift of $s$ to $\mathcal{I}(U)$. Therefore $\mathcal{I}(U)\to \mathcal{Q}(U)$ is surjective for every $U\in \mathcal{B}$.

Fix $U\in \mathcal{B}$ and $\mathcal{U}\in \mathrm{C}\mathrm{o}\mathrm{v}(U)$. Every finite intersection $W$ of members of $\mathcal{U}$ lies in $\mathcal{B}$ by hypothesis, so $\mathcal{I}(W)\to \mathcal{Q}(W)$ is surjective by [step 3.1]; taking products over the finitely many tuples of the finite cover, the induced map of Čech complexes $C^{\bullet }(\mathcal{U},\mathcal{I})\to C^{\bullet }(\mathcal{U},\mathcal{Q})$ is surjective in each degree, and by functoriality [F2] the sequence $0\to C^{\bullet }(\mathcal{U},\mathcal{F})\to C^{\bullet }(\mathcal{U},\mathcal{I})\to C^{\bullet }(\mathcal{U},\mathcal{Q})\to 0$ of cochain complexes is exact. Next, ${\check{H}}^p(\mathcal{U},\mathcal{I})=0$ for $p>0$: an injective sheaf is flasque by [F7] and hence $\mathrm{\Gamma }$-acyclic on every open subspace by [F7], so every nonempty finite intersection of members of $\mathcal{U}$ is $\mathcal{I}$-acyclic and the Leray theorem [F8] identifies ${\check{H}}^p(\mathcal{U},\mathcal{I})$ with $H^p(U,\mathcal{I}{\mathrm{\mid }}_U)=0$. From the long exact sequence of the short exact sequence of complexes [F4, F12] we get ${\check{H}}^1(\mathcal{U},\mathcal{Q})\hookrightarrow {\check{H}}^2(\mathcal{U},\mathcal{F})=0$ and isomorphisms ${\check{H}}^p(\mathcal{U},\mathcal{Q})\cong {\check{H}}^{p+1}(\mathcal{U},\mathcal{F})$ for $p\ge 1$, so ${\check{H}}^p(\mathcal{U},\mathcal{Q})=0$ for every $p>0$. Thus $\mathcal{Q}$ satisfies the same Čech vanishing hypothesis as $\mathcal{F}$, for the same basis and the same cofinal families.

Let $U\in \mathcal{B}$. The short exact sequence $0\to \mathcal{F}\to \mathcal{I}\to \mathcal{Q}\to 0$ restricted to $U$ gives by [F9] the exact sequence $H^0(U,\mathcal{I}{\mathrm{\mid }}_U)\to H^0(U,\mathcal{Q}{\mathrm{\mid }}_U)\to H^1(U,\mathcal{F}{\mathrm{\mid }}_U)\to H^1(U,\mathcal{I}{\mathrm{\mid }}_U)$; here $H^0(U,\mathcal{I}{\mathrm{\mid }}_U)=\mathcal{I}(U)$ and $H^0(U,\mathcal{Q}{\mathrm{\mid }}_U)=\mathcal{Q}(U)$ by [F10], so the first map is the surjection of [step 3.1], and $H^1(U,\mathcal{I}{\mathrm{\mid }}_U)=0$ because $\mathcal{I}$ is injective, hence flasque, hence acyclic on the open subspace $U$ by [F7]. Exactness therefore forces $H^1(U,\mathcal{F}{\mathrm{\mid }}_U)=0$.

We prove by induction on $n\ge 1$ the statement $P(n)$: for every abelian sheaf $\mathcal{G}$ satisfying the Čech vanishing hypothesis of the lemma, $H^q(U,\mathcal{G}{\mathrm{\mid }}_U)=0$ for all $U\in \mathcal{B}$ and all $1\le q\le n$. The case $n=1$ is [step 4.2], whose proof used only that hypothesis, the basis and the cofinal families. For the induction step let $\mathcal{G}$ satisfy the hypothesis; the argument of [step 1.1], [step 2.1] and [step 4.1] applied to $\mathcal{G}$ in place of $\mathcal{F}$ produces an injective ${\mathcal{I}}_{\mathcal{G}}$ with quotient ${\mathcal{G}}^{\mathrm{\prime }}={\mathcal{I}}_{\mathcal{G}}\mathrm{/}\mathcal{G}$ that again satisfies the Čech vanishing hypothesis, and the long exact sequence gives $H^{n+1}(U,\mathcal{G}{\mathrm{\mid }}_U)\cong H^n(U,{\mathcal{G}}^{\mathrm{\prime }}{\mathrm{\mid }}_U)$ for every $U\in \mathcal{B}$, because the neighbouring term $H^{n+1}(U,{\mathcal{I}}_{\mathcal{G}}{\mathrm{\mid }}_U)$ vanishes by acyclicity of injectives [F7]. By $P(n)$ applied to ${\mathcal{G}}^{\mathrm{\prime }}$ the group on the right is $0$, so $H^{n+1}(U,\mathcal{G}{\mathrm{\mid }}_U)=0$ and $P(n+1)$ holds. Induction gives $H^q(U,\mathcal{F}{\mathrm{\mid }}_U)=0$ for every $q\ge 1$ and every $U\in \mathcal{B}$. ∎