Acyclic directions of the Čech–Godement double complex

For each $q\ge 0$ define $A_x^0:={\mathcal{F}}_x$ and $A_x^q:=(Q^{q-1}(\mathcal{F}){)}_x$ for $q\ge 1$. By [F2], $G^q(V)={\prod }_{x\in V}{A}_x^q$. Consequently $D^{p,q}$ is the product over $x\in X$ of the coefficientwise Čech groups $C_x^p:={\prod }_{i_0<\cdots <i_p,x\in U_{i_0}\cap \cdots \cap U_{i_p}}{A}_x^q$: each intersection is precisely the set of points belonging to all the listed opens, and the restriction maps act by projection onto the relevant coordinates. The Čech differential therefore acts pointwise in $x$. Also $\mathrm{\Gamma }(X,G^q)={\prod }_{x\in X}{A}_x^q$ by the same Godement formula. Writing $S_x:=\{i\in I:x\in U_i\}$, which is nonempty because $\mathcal{U}$ covers $X$, the $x$-component of the augmented row $q$ has $A_x^q$ in degree $-1$, $C_x^p$ in degree $p$, augmentation the constant family $a\mapsto (a{)}_{i\in S_x}$, and differential the alternating sum $\delta$. Hence each row of $\tilde{D}$ is the product over $x\in X$ of these augmented coefficient complexes.

Here is the first use of [F7]: the family $(S_x{)}_{x\in X}$ consists of nonempty sets, so the Axiom of Choice provides a function $x\mapsto i_x$ with $i_x\in S_x$ for every $x\in X$. Fix such a choice for the rest of the proof.

Here is the bookkeeping lemma that turns exactness of the graded pieces of a filtration into acyclicity of the whole complex. Let $T^{\bullet }$ be a cochain complex with subcomplexes $F^s{T}^{\bullet }\subseteq T^{\bullet }$ for $s\in \mathbb{Z}$ such that $F^{s+1}\subseteq F^s$, that $F^s{T}^n=T^n$ for all $s\le 0$, and that for every fixed $n$ one has $F^s{T}^n=0$ for all $s$ large enough. If every quotient complex $F^s{T}^{\bullet }\mathrm{/}{F}^{s+1}{T}^{\bullet }$ is acyclic, then $T^{\bullet }$ is acyclic. Indeed, for each $s$ the short exact sequence $0\to F^{s+1}{T}^{\bullet }\to F^s{T}^{\bullet }\to F^s{T}^{\bullet }\mathrm{/}{F}^{s+1}{T}^{\bullet }\to 0$ of complexes gives by [F6] the exact sequence $\cdots \to H^n(F^{s+1})\to H^n(F^s)\to H^n(F^s\mathrm{/}{F}^{s+1})\to H^{n+1}(F^{s+1})\to \cdots$; the third term vanishes, so $H^n(F^{s+1})\to H^n(F^s)$ is an isomorphism for all $n$ and $s$. Fix $n$. Choose $M$ large enough that $F^M{T}^r=0$ for each of the three degrees $r=n-1,n,n+1$, which is possible by the pointwise eventual-vanishing hypothesis and a maximum of three bounds. Then $H^n(F^M{T}^{\bullet })=0$, and iterating the isomorphism over $s=0,1,\dots ,M-1$ gives $H^n(T)=H^n(F^0{T}^{\bullet })\cong H^n(F^M{T}^{\bullet })=0$.

Adjoin to $D$ the extra row $E^{p,-1}:=C^p(\mathcal{U},\mathcal{F})$ of [F1], with horizontal differential the Čech differential $\delta$ and vertical differential $w^p:C^p(\mathcal{U},\mathcal{F})\to D^{p,0}$ induced componentwise by $\epsilon :\mathcal{F}\to G^0$, and keep $D^{p,q}$ for $q\ge 0$ with its two differentials. Then $hv=vh$ on the new row too, because both the Čech differential and $\epsilon$ are natural with respect to the restriction maps, and $h^2=0$, $v^2=0$ hold there because ${\delta }^2=0$ on Čech cochains and $d^0\epsilon =0$ by exactness of the resolution [F3]; the total complex is $(\mathrm{Tot}E{)}^n=(\mathrm{Tot}D{)}^n\oplus E^{n+1,-1}=(\mathrm{Tot}D{)}^n\oplus C^{n+1}(\mathcal{U},\mathcal{F})$ in the convention of [F8]. The map $w:C^{\bullet }(\mathcal{U},\mathcal{F})\to \mathrm{Tot}D$ of the statement is the inclusion of this extra row: it is a cochain map, since for a Čech cochain $\alpha$ the vertical component of $d(w\alpha )$ is $d^0\epsilon (\alpha )=0$ and the horizontal component is $\delta (w\alpha )=w(\delta \alpha )$ by naturality of $\epsilon$.

For each $x\in X$ write $\sigma :=i_x$ and define, for every $q\ge 0$, a homomorphism $k_x:C_x^{q+1}\to C_x^q$ by $(k_{x}c)(\tau ):=0$ if $\sigma \in \tau$ and $(k_{x}c)(\tau ):=(-1{)}^{r}c(\mathrm{sort}(\sigma \cup \tau ))$ if $\sigma \mathrm{\notin }\tau$, where $\tau$ runs over increasing $(q+1)$-tuples in $S_x$ and $r$ is the position of $\sigma$ in the increasing $(q+2)$-tuple $\mathrm{sort}(\sigma \cup \tau )$; this is the data fixed in [step 1.1] and [step 1.2]. Let $c\in C_x^{q+1}$ and let $\tau$ be an increasing $(q+2)$-tuple. If $\sigma \mathrm{\notin }\tau$ put $u:=\mathrm{sort}(\sigma \cup \tau )=(u_0<\cdots <u_{q+2})$ with $u_r=\sigma$; then $(\delta k_{x}c)(\tau )={\sum }_j(-1{)}^j(k_{x}c)(\tau \setminus j)$ and $(k_x\delta c)(\tau )=(-1{)}^r(\delta c)(u)=(-1{)}^r{\sum }_i(-1{)}^{i}c(u\setminus i)$. For an entry $u_i$ of $u$ that is an element ${\tau }_j$ of $\tau$ with ${\tau }_j<\sigma$ one has $i=j$ and the insertion position of $\sigma$ in $\mathrm{sort}(\sigma \cup (\tau \setminus j))$ is $r-1$, so the two occurrences of $c(u\setminus {\tau }_j)$ carry the signs $(-1{)}^{j+r-1}$ and $(-1{)}^{r+j}$ and cancel; for ${\tau }_j>\sigma$ one has $i=j+1$ and insertion position $r$, so the signs $(-1{)}^{j+r}$ and $(-1{)}^{r+j+1}$ cancel; the remaining term is $i=r$, contributing $(-1{)}^{2r}c(u\setminus \sigma )=c(\tau )$. If $\sigma \in \tau$, say ${\tau }_r=\sigma$, then $(k_x\delta c)(\tau )=0$ by the definition of $k_x$, while in the sum for $(\delta k_{x}c)(\tau )$ every $j\mathrm{\ne }r$ contributes $0$ because $\sigma \in \tau \setminus j$, and the term $j=r$ contributes $(-1{)}^r(k_{x}c)(\tau \setminus r)=(-1{)}^r(-1{)}^{r}c(\tau )=c(\tau )$, the insertion position of $\sigma$ in $\mathrm{sort}(\sigma \cup (\tau \setminus r))=\tau$ being $r$. In both cases $(\delta k_x+k_x\delta )(c)(\tau )=c(\tau )$, that is $\delta k_x+k_x\delta =\mathrm{id}$ on $C_x^{q+1}$ for every $q\ge 0$.

Each augmented complex ${\tilde{A}}_x^q$ is exact. If $c\in C_x^{q+1}$ is a cocycle, $\delta c=0$, then $c=\delta (k_{x}c)+k_x(\delta c)=\delta (k_{x}c)$ lies in the image of $\delta$, so every cohomology group $H^p({\tilde{A}}_x^q)$ vanishes for $p\ge 1$ by [step 2.1]. For $p=0$ suppose $\delta c=0$ with $c\in C_x^0$; for $i\in S_x$ the value at the two-element set $\{\sigma ,i\}$ gives $0=\pm (c(i)-c(\sigma ))$, so $c$ is the constant family with value $k(c):=c(\sigma )\in A_x^q$, that is $c$ is the image of the coefficient $k(c)$ under the augmentation; hence $H^0({\tilde{A}}_x^q)=0$. In degree $-1$ the augmentation $a\mapsto (a{)}_{i\in S_x}$ is injective because $S_x\mathrm{\ne }\mathrm{\varnothing }$: if the constant family is zero then $a=0$. Therefore $H^p({\tilde{A}}_x^q)=0$ for all $p\ge -1$, using [step 1.2] for the nonemptiness of $S_x$ and [step 2.1] for the positive degrees.

Let $(P_x,d_x{)}_{x\in X}$ be any family of cochain complexes of abelian groups with every $P_x$ exact. Then the product complex ${\prod }_x{P}_x$ with differential ${\prod }_x{d}_x$ is exact: $\ker ({\prod }_x{d}_x)={\prod }_x\ker d_x$ holds componentwise without any choice, while $\mathrm{im}({\prod }_x{d}_x)={\prod }_x\mathrm{im}{d}_x$ uses [F7], because a family $(b_x)$ with $b_x=d_x(a_x)$ for some $a_x$ is hit by the family of selected preimages; hence $H^p({\prod }_x{P}_x)={\prod }_x{H}^p(P_x)=0$. Applying this to the factor complexes ${\tilde{A}}_x^q$ of [step 3.1], and using the product description of the rows in [step 1.1], the row $q$ of $\tilde{D}$ is exact for every $q\ge 0$.

Apply [step 1.3] to $T^{\bullet }:=\mathrm{Tot}\tilde{D}$ with $F^s{T}^n:={⨁}_{p\ge -1,q\ge s,p+q=n}{\tilde{D}}^{p,q}$. These are subcomplexes: the horizontal differential $h$ raises $p$ and preserves $q$, and the vertical differential $(-1{)}^{p}v$ raises $q$ to $q+1\ge s$; the filtration is decreasing and $F^s{T}^n=T^n$ for $s\le 0$, while in total degree $n$ the second coordinate satisfies $q\le n+1$, so $F^s{T}^n=0$ for $s>n+1$. The quotient $F^s{T}^{\bullet }\mathrm{/}{F}^{s+1}{T}^{\bullet }$ is the row $s$ of $\tilde{D}$ with the induced differential $h$ (the vertical part lands in the next quotient), which is exact by [step 4.1]. Hence $\mathrm{Tot}\tilde{D}$ is acyclic.

Fix $p\ge 0$ and an increasing $(p+1)$-tuple $\alpha =(i_0<\cdots <i_p)$, and put $W_{\alpha }:=U_{i_0}\cap \cdots \cap U_{i_p}$. By the definition of $D^{p,q}$, the column $D^{p,\bullet }$ is the product over all such tuples $\alpha$ of the complexes $\mathrm{\Gamma }(W_{\alpha },G^{\bullet }{\mathrm{\mid }}_{W_{\alpha }})$, with differentials induced by the Godement differentials. Each restriction $G^q{\mathrm{\mid }}_{W_{\alpha }}$ is flasque on the open subspace $W_{\alpha }$, since every restriction map between opens of $W_{\alpha }$ is a restriction map of $G^q$, and the resolution restricts to an exact complex $0\to \mathcal{F}{\mathrm{\mid }}_{W_{\alpha }}\to G^{\bullet }{\mathrm{\mid }}_{W_{\alpha }}$. By [F4] each term is $\mathrm{\Gamma }$-acyclic on $W_{\alpha }$. The supplied datum of Enough injective abelian sheaves covers all of $\mathrm{A}\mathrm{b}(W_{\alpha })$, including $\mathcal{F}{\mathrm{\mid }}_{W_{\alpha }}$ and every cycle of the restricted Godement resolution, so the acyclic resolution theorem [F5] applies (its hypothesis DC is available by [F11]) and identifies $H^q(\mathrm{\Gamma }(W_{\alpha },G^{\bullet }{\mathrm{\mid }}_{W_{\alpha }}))$ with $H^q(W_{\alpha },\mathcal{F}{\mathrm{\mid }}_{W_{\alpha }})$. Taking the product over $\alpha$ and using [step 4.1] gives $H^q(D^{p,\bullet })={\prod }_{\alpha }{H}^q(W_{\alpha },\mathcal{F}{\mathrm{\mid }}_{W_{\alpha }})$. In degree zero the identification is explicit: exactness at $G^0$ and left exactness of $\mathrm{\Gamma }$ [F12] give $\ker (G^0(W_{\alpha })\to G^1(W_{\alpha }))=\epsilon (\mathcal{F}(W_{\alpha }))$, and $\epsilon$ is injective on sections; hence the map $w$ of the statement induces an isomorphism $C^p(\mathcal{U},\mathcal{F})\to H^0(D^{p,\bullet })={\prod }_{\alpha }\ker (G^0(W_{\alpha })\to G^1(W_{\alpha }))$. If $\mathcal{U}$ is $\mathcal{F}$-acyclic then by [F14] each factor $H^q(W_{\alpha },\mathcal{F}{\mathrm{\mid }}_{W_{\alpha }})$ vanishes for $q>0$, so $H^q(D^{p,\bullet })=0$ for $q>0$. This is assertion 2.

The map $u:\mathrm{\Gamma }(X,G^{\bullet })\to \mathrm{Tot}D$ is a cochain map: its component in $D^{0,q}$ is the restriction map of [F1], the Čech differential of the family $(s{\mathrm{\mid }}_{U_i}{)}_{i\in I}$ vanishes because the family is the restriction of the single section $s$ to a cover, and the vertical differential of the family is $(d^{q}s){\mathrm{\mid }}_{U_i}$ by naturality of the morphism $d^q:G^q\to G^{q+1}$ with respect to restrictions, which is $u$ of the Godement differential. With the total-complex convention of [F8] the differential of $\mathrm{Tot}\tilde{D}$ sends $(x,c)\in (\mathrm{Tot}D{)}^n\oplus \mathrm{\Gamma }(X,G^{n+1})$ to $(dx+u(c),-d_{G}c)$: the horizontal coefficient in degree $p=-1$ is $1$, and the vertical coefficient is $(-1{)}^{-1}=-1$. This is exactly the cone differential $(d^D(y)+f_{n-1}(x),-d_{n-1}^C(x))$ of [F9] for $f=u$, after the standard reindexing of cochain complexes as chain complexes and the identification $\mathrm{\Gamma }(X,G^{n+1})$ with $C[1{]}^n$; hence $\mathrm{Tot}\tilde{D}=\mathrm{Cone}(u)$ is acyclic by [step 5.1], and the cone criterion [F10] shows that $u$ is a quasi-isomorphism in the sense of [F13]. This is the first half of assertion 3.

Assume now that $\mathcal{U}$ is $\mathcal{F}$-acyclic, and filter $\mathrm{Tot}E$ by $F^s{T}^n:={⨁}_{p\ge s,p+q=n}{E}^{p,q}$. These are subcomplexes because $h$ raises $p$ and the vertical differential keeps $p$ fixed, the filtration is decreasing and finite in each total degree, and the quotient $F^s{T}^{\bullet }\mathrm{/}{F}^{s+1}{T}^{\bullet }$ is the column $s$ of $E$ with differential $(-1{)}^{s}v$ (the horizontal part leaves the quotient). By [step 5.2] that quotient has vanishing cohomology: its degree $-1$ cohomology is the kernel of the injective map $C^s(\mathcal{U},\mathcal{F})\to D^{s,0}$ and hence zero, while its degree $0$ cohomology is $\ker (D^{s,0}\to D^{s,1})\mathrm{/}\mathrm{im}(C^s(\mathcal{U},\mathcal{F})\to D^{s,0})=0$ by the explicit $H^0$ identification of [step 5.2], and its $q>0$ cohomology is $H^q(D^{s,\bullet })=0$. Hence $\mathrm{Tot}E$ is acyclic by [step 1.3].

With the total differential of [F8], an element $(x,\alpha )\in (\mathrm{Tot}D{)}^n\oplus C^{n+1}(\mathcal{U},\mathcal{F})$ of $\mathrm{Tot}E$ has differential $(dx+(-1{)}^{n+1}w(\alpha ),\delta \alpha )$: the extra row sits in bidegree $(p,q)=(n+1,-1)$, so its horizontal differential enters with coefficient $1$ and its vertical differential with the coefficient $(-1{)}^p=(-1{)}^{n+1}$. Conjugating the second summand by the sign $(-1{)}^{n+1}$ turns this into the cone differential $(dx+w(\alpha ),-\delta \alpha )$ of [F9] for $f=w$, so $\mathrm{Tot}E$ is isomorphic to the mapping cone of $w$; since $\mathrm{Tot}E$ is acyclic by [step 6.2], the cone criterion [F10] makes $w$ a quasi-isomorphism. This is the second half of assertion 3, and together with [step 6.1] and [step 5.2] it proves assertions 1, 2 and 3. ∎