
## 20.11 Čech cohomology as a functor on presheaves

Warning: In this section we work almost exclusively with presheaves and categories of presheaves and the results are completely wrong in the setting of sheaves and categories of sheaves!

Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}_ X$-modules. We have the Čech complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ of $\mathcal{F}$ just by thinking of $\mathcal{F}$ as a presheaf of abelian groups. However, each term $\check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{F})$ has a natural structure of a $\mathcal{O}_ X(U)$-module and the differential is given by $\mathcal{O}_ X(U)$-module maps. Moreover, it is clear that the construction

$\mathcal{F} \longmapsto \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$

is functorial in $\mathcal{F}$. In fact, it is a functor

20.11.0.1
$$\label{cohomology-equation-cech-functor} \check{\mathcal{C}}^\bullet (\mathcal{U}, -) : \textit{PMod}(\mathcal{O}_ X) \longrightarrow \text{Comp}^{+}(\text{Mod}_{\mathcal{O}_ X(U)})$$

see Derived Categories, Definition 13.8.1 for notation. Recall that the category of bounded below complexes in an abelian category is an abelian category, see Homology, Lemma 12.12.9.

Proof. For any open $W \subset U$ the functor $\mathcal{F} \mapsto \mathcal{F}(W)$ is an additive exact functor from $\textit{PMod}(\mathcal{O}_ X)$ to $\text{Mod}_{\mathcal{O}_ X(U)}$. The terms $\check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{F})$ of the complex are products of these exact functors and hence exact. Moreover a sequence of complexes is exact if and only if the sequence of terms in a given degree is exact. Hence the lemma follows. $\square$

Lemma 20.11.2. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. The functors $\mathcal{F} \mapsto \check{H}^ n(\mathcal{U}, \mathcal{F})$ form a $\delta$-functor from the abelian category of presheaves of $\mathcal{O}_ X$-modules to the category of $\mathcal{O}_ X(U)$-modules (see Homology, Definition 12.11.1).

Proof. By Lemma 20.11.1 a short exact sequence of presheaves of $\mathcal{O}_ X$-modules $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ is turned into a short exact sequence of complexes of $\mathcal{O}_ X(U)$-modules. Hence we can use Homology, Lemma 12.12.12 to get the boundary maps $\delta _{\mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3} : \check{H}^ n(\mathcal{U}, \mathcal{F}_3) \to \check{H}^{n + 1}(\mathcal{U}, \mathcal{F}_1)$ and a corresponding long exact sequence. We omit the verification that these maps are compatible with maps between short exact sequences of presheaves. $\square$

In the formulation of the following lemma we use the functor $j_{p!}$ of extension by $0$ for presheaves of modules relative to an open immersion $j : U \to X$. See Sheaves, Section 6.31. For any open $W \subset X$ and any presheaf $\mathcal{G}$ of $\mathcal{O}_ X|_ U$-modules we have

$(j_{p!}\mathcal{G})(W) = \left\{ \begin{matrix} \mathcal{G}(W) & \text{if } W \subset U \\ 0 & \text{else.} \end{matrix} \right.$

Moreover, the functor $j_{p!}$ is a left adjoint to the restriction functor see Sheaves, Lemma 6.31.8. In particular we have the following formula

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(j_{p!}\mathcal{O}_ U, \mathcal{F}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_ U, \mathcal{F}|_ U) = \mathcal{F}(U).$

Since the functor $\mathcal{F} \mapsto \mathcal{F}(U)$ is an exact functor on the category of presheaves we conclude that the presheaf $j_{p!}\mathcal{O}_ U$ is a projective object in the category $\textit{PMod}(\mathcal{O}_ X)$, see Homology, Lemma 12.25.2.

Note that if we are given open subsets $U \subset V \subset X$ with associated open immersions $j_ U, j_ V$, then we have a canonical map $(j_ U)_{p!}\mathcal{O}_ U \to (j_ V)_{p!}\mathcal{O}_ V$. It is the identity on sections over any open $W \subset U$ and $0$ else. In terms of the identification $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}((j_ U)_{p!}\mathcal{O}_ U, (j_ V)_{p!}\mathcal{O}_ V) = (j_ V)_{p!}\mathcal{O}_ V(U) = \mathcal{O}_ V(U)$ it corresponds to the element $1 \in \mathcal{O}_ V(U)$.

Lemma 20.11.3. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be a covering. Denote $j_{i_0\ldots i_ p} : U_{i_0 \ldots i_ p} \to X$ the open immersion. Consider the chain complex $K(\mathcal{U})_\bullet$ of presheaves of $\mathcal{O}_ X$-modules

$\ldots \to \bigoplus _{i_0i_1i_2} (j_{i_0i_1i_2})_{p!}\mathcal{O}_{U_{i_0i_1i_2}} \to \bigoplus _{i_0i_1} (j_{i_0i_1})_{p!}\mathcal{O}_{U_{i_0i_1}} \to \bigoplus _{i_0} (j_{i_0})_{p!}\mathcal{O}_{U_{i_0}} \to 0 \to \ldots$

where the last nonzero term is placed in degree $0$ and where the map

$(j_{i_0\ldots i_{p + 1}})_{p!}\mathcal{O}_{U_{i_0\ldots i_{p + 1}}} \longrightarrow (j_{i_0\ldots \hat i_ j \ldots i_{p + 1}})_{p!} \mathcal{O}_{U_{i_0\ldots \hat i_ j \ldots i_{p + 1}}}$

is given by $(-1)^ j$ times the canonical map. Then there is an isomorphism

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(K(\mathcal{U})_\bullet , \mathcal{F}) = \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$

functorial in $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\textit{PMod}(\mathcal{O}_ X))$.

Proof. We saw in the discussion just above the lemma that

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}( (j_{i_0\ldots i_ p})_{p!}\mathcal{O}_{U_{i_0\ldots i_ p}}, \mathcal{F}) = \mathcal{F}(U_{i_0\ldots i_ p}).$

Hence we see that it is indeed the case that the direct sum

$\bigoplus \nolimits _{i_0 \ldots i_ p} (j_{i_0 \ldots i_ p})_{p!}\mathcal{O}_{U_{i_0 \ldots i_ p}}$

represents the functor

$\mathcal{F} \longmapsto \prod \nolimits _{i_0\ldots i_ p} \mathcal{F}(U_{i_0\ldots i_ p}).$

Hence by Categories, Yoneda Lemma 4.3.5 we see that there is a complex $K(\mathcal{U})_\bullet$ with terms as given. It is a simple matter to see that the maps are as given in the lemma. $\square$

Lemma 20.11.4. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be a covering. Let $\mathcal{O}_\mathcal {U} \subset \mathcal{O}_ X$ be the image presheaf of the map $\bigoplus j_{p!}\mathcal{O}_{U_ i} \to \mathcal{O}_ X$. The chain complex $K(\mathcal{U})_\bullet$ of presheaves of Lemma 20.11.3 above has homology presheaves

$H_ i(K(\mathcal{U})_\bullet ) = \left\{ \begin{matrix} 0 & \text{if} & i \not= 0 \\ \mathcal{O}_\mathcal {U} & \text{if} & i = 0 \end{matrix} \right.$

Proof. Consider the extended complex $K^{ext}_\bullet$ one gets by putting $\mathcal{O}_\mathcal {U}$ in degree $-1$ with the obvious map $K(\mathcal{U})_0 = \bigoplus _{i_0} (j_{i_0})_{p!}\mathcal{O}_{U_{i_0}} \to \mathcal{O}_\mathcal {U}$. It suffices to show that taking sections of this extended complex over any open $W \subset X$ leads to an acyclic complex. In fact, we claim that for every $W \subset X$ the complex $K^{ext}_\bullet (W)$ is homotopy equivalent to the zero complex. Write $I = I_1 \amalg I_2$ where $W \subset U_ i$ if and only if $i \in I_1$.

If $I_1 = \emptyset$, then the complex $K^{ext}_\bullet (W) = 0$ so there is nothing to prove.

If $I_1 \not= \emptyset$, then $\mathcal{O}_\mathcal {U}(W) = \mathcal{O}_ X(W)$ and

$K^{ext}_ p(W) = \bigoplus \nolimits _{i_0 \ldots i_ p \in I_1} \mathcal{O}_ X(W).$

This is true because of the simple description of the presheaves $(j_{i_0 \ldots i_ p})_{p!}\mathcal{O}_{U_{i_0 \ldots i_ p}}$. Moreover, the differential of the complex $K^{ext}_\bullet (W)$ is given by

$d(s)_{i_0 \ldots i_ p} = \sum \nolimits _{j = 0, \ldots , p + 1} \sum \nolimits _{i \in I_1} (-1)^ j s_{i_0 \ldots i_{j - 1} i i_ j \ldots i_ p}.$

The sum is finite as the element $s$ has finite support. Fix an element $i_{\text{fix}} \in I_1$. Define a map

$h : K^{ext}_ p(W) \longrightarrow K^{ext}_{p + 1}(W)$

by the rule

$h(s)_{i_0 \ldots i_{p + 1}} = \left\{ \begin{matrix} 0 & \text{if} & i_0 \not= i \\ s_{i_1 \ldots i_{p + 1}} & \text{if} & i_0 = i_{\text{fix}} \end{matrix} \right.$

We will use the shorthand $h(s)_{i_0 \ldots i_{p + 1}} = (i_0 = i_{\text{fix}}) s_{i_1 \ldots i_ p}$ for this. Then we compute

\begin{eqnarray*} & & (dh + hd)(s)_{i_0 \ldots i_ p} \\ & = & \sum _ j \sum _{i \in I_1} (-1)^ j h(s)_{i_0 \ldots i_{j - 1} i i_ j \ldots i_ p} + (i = i_0) d(s)_{i_1 \ldots i_ p} \\ & = & s_{i_0 \ldots i_ p} + \sum _{j \geq 1}\sum _{i \in I_1} (-1)^ j (i_0 = i_{\text{fix}}) s_{i_1 \ldots i_{j - 1} i i_ j \ldots i_ p} + (i_0 = i_{\text{fix}}) d(s)_{i_1 \ldots i_ p} \end{eqnarray*}

which is equal to $s_{i_0 \ldots i_ p}$ as desired. $\square$

Lemma 20.11.5. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering of $U \subset X$. The Čech cohomology functors $\check{H}^ p(\mathcal{U}, -)$ are canonically isomorphic as a $\delta$-functor to the right derived functors of the functor

$\check{H}^0(\mathcal{U}, -) : \textit{PMod}(\mathcal{O}_ X) \longrightarrow \text{Mod}_{\mathcal{O}_ X(U)}.$

Moreover, there is a functorial quasi-isomorphism

$\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow R\check{H}^0(\mathcal{U}, \mathcal{F})$

where the right hand side indicates the right derived functor

$R\check{H}^0(\mathcal{U}, -) : D^{+}(\textit{PMod}(\mathcal{O}_ X)) \longrightarrow D^{+}(\mathcal{O}_ X(U))$

of the left exact functor $\check{H}^0(\mathcal{U}, -)$.

Proof. Note that the category of presheaves of $\mathcal{O}_ X$-modules has enough injectives, see Injectives, Proposition 19.8.5. Note that $\check{H}^0(\mathcal{U}, -)$ is a left exact functor from the category of presheaves of $\mathcal{O}_ X$-modules to the category of $\mathcal{O}_ X(U)$-modules. Hence the derived functor and the right derived functor exist, see Derived Categories, Section 13.20.

Let $\mathcal{I}$ be a injective presheaf of $\mathcal{O}_ X$-modules. In this case the functor $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(-, \mathcal{I})$ is exact on $\textit{PMod}(\mathcal{O}_ X)$. By Lemma 20.11.3 we have

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(K(\mathcal{U})_\bullet , \mathcal{I}) = \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}).$

By Lemma 20.11.4 we have that $K(\mathcal{U})_\bullet$ is quasi-isomorphic to $\mathcal{O}_\mathcal {U}[0]$. Hence by the exactness of Hom into $\mathcal{I}$ mentioned above we see that $\check{H}^ i(\mathcal{U}, \mathcal{I}) = 0$ for all $i > 0$. Thus the $\delta$-functor $(\check{H}^ n, \delta )$ (see Lemma 20.11.2) satisfies the assumptions of Homology, Lemma 12.11.4, and hence is a universal $\delta$-functor.

By Derived Categories, Lemma 13.20.4 also the sequence $R^ i\check{H}^0(\mathcal{U}, -)$ forms a universal $\delta$-functor. By the uniqueness of universal $\delta$-functors, see Homology, Lemma 12.11.5 we conclude that $R^ i\check{H}^0(\mathcal{U}, -) = \check{H}^ i(\mathcal{U}, -)$. This is enough for most applications and the reader is suggested to skip the rest of the proof.

Let $\mathcal{F}$ be any presheaf of $\mathcal{O}_ X$-modules. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet$ in the category $\textit{PMod}(\mathcal{O}_ X)$. Consider the double complex $A^{\bullet , \bullet }$ with terms

$A^{p, q} = \check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{I}^ q).$

Consider the simple complex $sA^\bullet$ associated to this double complex. There is a map of complexes

$\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow sA^\bullet$

coming from the maps $\check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{F}) \to A^{p, 0} = \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^0)$ and there is a map of complexes

$\check{H}^0(\mathcal{U}, \mathcal{I}^\bullet ) \longrightarrow sA^\bullet$

coming from the maps $\check{H}^0(\mathcal{U}, \mathcal{I}^ q) \to A^{0, q} = \check{\mathcal{C}}^0(\mathcal{U}, \mathcal{I}^ q)$. Both of these maps are quasi-isomorphisms by an application of Homology, Lemma 12.22.7. Namely, the columns of the double complex are exact in positive degrees because the Čech complex as a functor is exact (Lemma 20.11.1) and the rows of the double complex are exact in positive degrees since as we just saw the higher Čech cohomology groups of the injective presheaves $\mathcal{I}^ q$ are zero. Since quasi-isomorphisms become invertible in $D^{+}(\mathcal{O}_ X(U))$ this gives the last displayed morphism of the lemma. We omit the verification that this morphism is functorial. $\square$

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