
## 21.13 Cohomology of modules

Everything that was said for cohomology of abelian sheaves goes for cohomology of modules, since the two agree.

Lemma 21.13.1. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. An injective sheaf of modules is also injective as an object in the category $\textit{PMod}(\mathcal{O})$.

Proof. Apply Homology, Lemma 12.26.1 to the categories $\mathcal{A} = \textit{Mod}(\mathcal{O})$, $\mathcal{B} = \textit{PMod}(\mathcal{O})$, the inclusion functor and sheafification. (See Modules on Sites, Section 18.11 to see that all assumptions of the lemma are satisfied.) $\square$

Lemma 21.13.2. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Consider the functor $i : \textit{Mod}(\mathcal{C}) \to \textit{PMod}(\mathcal{C})$. It is a left exact functor with right derived functors given by

$R^ pi(\mathcal{F}) = \underline{H}^ p(\mathcal{F}) : U \longmapsto H^ p(U, \mathcal{F})$

see discussion in Section 21.8.

Proof. It is clear that $i$ is left exact. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet$ in $\textit{Mod}(\mathcal{O})$. By definition $R^ pi$ is the $p$th cohomology presheaf of the complex $\mathcal{I}^\bullet$. In other words, the sections of $R^ pi(\mathcal{F})$ over an object $U$ of $\mathcal{C}$ are given by

$\frac{\mathop{\mathrm{Ker}}(\mathcal{I}^ n(U) \to \mathcal{I}^{n + 1}(U))}{\mathop{\mathrm{Im}}(\mathcal{I}^{n - 1}(U) \to \mathcal{I}^ n(U))}.$

which is the definition of $H^ p(U, \mathcal{F})$. $\square$

Lemma 21.13.3. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ be a covering of $\mathcal{C}$. Let $\mathcal{I}$ be an injective $\mathcal{O}$-module, i.e., an injective object of $\textit{Mod}(\mathcal{O})$. Then

$\check{H}^ p(\mathcal{U}, \mathcal{I}) = \left\{ \begin{matrix} \mathcal{I}(U) & \text{if} & p = 0 \\ 0 & \text{if} & p > 0 \end{matrix} \right.$

Proof. Lemma 21.10.3 gives the first equality in the following sequence of equalities

\begin{align*} \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}) & = \mathop{Mor}\nolimits _{\textit{PAb}(\mathcal{C})}( \mathbf{Z}_{\mathcal{U}, \bullet }, \mathcal{I}) \\ & = \mathop{Mor}\nolimits _{\textit{PMod}(\mathbf{Z})}( \mathbf{Z}_{\mathcal{U}, \bullet }, \mathcal{I}) \\ & = \mathop{Mor}\nolimits _{\textit{PMod}(\mathcal{O})}( \mathbf{Z}_{\mathcal{U}, \bullet } \otimes _{p, \mathbf{Z}} \mathcal{O}, \mathcal{I}) \end{align*}

The third equality by Modules on Sites, Lemma 18.9.2. By Lemma 21.13.1 we see that $\mathcal{I}$ is an injective object in $\textit{PMod}(\mathcal{O})$. Hence $\mathop{\mathrm{Hom}}\nolimits _{\textit{PMod}(\mathcal{O})}(-, \mathcal{I})$ is an exact functor. By Lemma 21.10.5 we see the vanishing of higher Čech cohomology groups. For the zeroth see Lemma 21.9.2. $\square$

Lemma 21.13.4. Let $\mathcal{C}$ be a site. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}$. Let $\mathcal{F}$ be an $\mathcal{O}$-module, and denote $\mathcal{F}_{ab}$ the underlying sheaf of abelian groups. Then we have

$H^ i(\mathcal{C}, \mathcal{F}_{ab}) = H^ i(\mathcal{C}, \mathcal{F})$

and for any object $U$ of $\mathcal{C}$ we also have

$H^ i(U, \mathcal{F}_{ab}) = H^ i(U, \mathcal{F}).$

Here the left hand side is cohomology computed in $\textit{Ab}(\mathcal{C})$ and the right hand side is cohomology computed in $\textit{Mod}(\mathcal{O})$.

Proof. By Derived Categories, Lemma 13.20.4 the $\delta$-functor $(\mathcal{F} \mapsto H^ p(U, \mathcal{F}))_{p \geq 0}$ is universal. The functor $\textit{Mod}(\mathcal{O}) \to \textit{Ab}(\mathcal{C})$, $\mathcal{F} \mapsto \mathcal{F}_{ab}$ is exact. Hence $(\mathcal{F} \mapsto H^ p(U, \mathcal{F}_{ab}))_{p \geq 0}$ is a $\delta$-functor also. Suppose we show that $(\mathcal{F} \mapsto H^ p(U, \mathcal{F}_{ab}))_{p \geq 0}$ is also universal. This will imply the second statement of the lemma by uniqueness of universal $\delta$-functors, see Homology, Lemma 12.11.5. Since $\textit{Mod}(\mathcal{O})$ has enough injectives, it suffices to show that $H^ i(U, \mathcal{I}_{ab}) = 0$ for any injective object $\mathcal{I}$ in $\textit{Mod}(\mathcal{O})$, see Homology, Lemma 12.11.4.

Let $\mathcal{I}$ be an injective object of $\textit{Mod}(\mathcal{O})$. Apply Lemma 21.11.9 with $\mathcal{F} = \mathcal{I}$, $\mathcal{B} = \mathcal{C}$ and $\text{Cov} = \text{Cov}_\mathcal {C}$. Assumption (3) of that lemma holds by Lemma 21.13.3. Hence we see that $H^ i(U, \mathcal{I}_{ab}) = 0$ for every object $U$ of $\mathcal{C}$.

If $\mathcal{C}$ has a final object then this also implies the first equality. If not, then according to Sites, Lemma 7.29.5 we see that the ringed topos $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ is equivalent to a ringed topos where the underlying site does have a final object. Hence the lemma follows. $\square$

Lemma 21.13.5. Let $\mathcal{C}$ be a site. Let $I$ be a set. For $i \in I$ let $\mathcal{F}_ i$ be an abelian sheaf on $\mathcal{C}$. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. The canonical map

$H^ p(U, \prod \nolimits _{i \in I} \mathcal{F}_ i) \longrightarrow \prod \nolimits _{i \in I} H^ p(U, \mathcal{F}_ i)$

is an isomorphism for $p = 0$ and injective for $p = 1$.

Proof. The statement for $p = 0$ is true because the product of sheaves is equal to the product of the underlying presheaves, see Sites, Lemma 7.10.1. Proof for $p = 1$. Set $\mathcal{F} = \prod \mathcal{F}_ i$. Let $\xi \in H^1(U, \mathcal{F})$ map to zero in $\prod H^1(U, \mathcal{F}_ i)$. By locality of cohomology, see Lemma 21.8.3, there exists a covering $\mathcal{U} = \{ U_ j \to U\}$ such that $\xi |_{U_ j} = 0$ for all $j$. By Lemma 21.11.4 this means $\xi$ comes from an element $\check\xi \in \check H^1(\mathcal{U}, \mathcal{F})$. Since the maps $\check H^1(\mathcal{U}, \mathcal{F}_ i) \to H^1(U, \mathcal{F}_ i)$ are injective for all $i$ (by Lemma 21.11.4), and since the image of $\xi$ is zero in $\prod H^1(U, \mathcal{F}_ i)$ we see that the image $\check\xi _ i = 0$ in $\check H^1(\mathcal{U}, \mathcal{F}_ i)$. However, since $\mathcal{F} = \prod \mathcal{F}_ i$ we see that $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ is the product of the complexes $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}_ i)$, hence by Homology, Lemma 12.29.1 we conclude that $\check\xi = 0$ as desired. $\square$

Lemma 21.13.6. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $a : U' \to U$ be a monomorphism in $\mathcal{C}$. Then for any injective $\mathcal{O}$-module $\mathcal{I}$ the restriction mapping $\mathcal{I}(U) \to \mathcal{I}(U')$ is surjective.

Proof. Let $j : \mathcal{C}/U \to \mathcal{C}$ and $j' : \mathcal{C}/U' \to \mathcal{C}$ be the localization morphisms (Modules on Sites, Section 18.19). Since $j_!$ is a left adjoint to restriction we see that for any sheaf $\mathcal{F}$ of $\mathcal{O}$-modules

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

Similarly, the sheaf $j'_!\mathcal{O}_{U'}$ represents the functor $\mathcal{F} \mapsto \mathcal{F}(U')$. Moreover below we describe a canonical map of $\mathcal{O}$-modules

$j'_!\mathcal{O}_{U'} \longrightarrow j_!\mathcal{O}_ U$

which corresponds to the restriction mapping $\mathcal{F}(U) \to \mathcal{F}(U')$ via Yoneda's lemma (Categories, Lemma 4.3.5). It suffices to prove the displayed map of modules is injective, see Homology, Lemma 12.24.2.

To construct our map it suffices to construct a map between the presheaves which assign to an object $V$ of $\mathcal{C}$ the $\mathcal{O}(V)$-module

$\bigoplus \nolimits _{\varphi ' \in \mathop{Mor}\nolimits _\mathcal {C}(V, U')} \mathcal{O}(V) \quad \text{and}\quad \bigoplus \nolimits _{\varphi \in \mathop{Mor}\nolimits _\mathcal {C}(V, U)} \mathcal{O}(V)$

see Modules on Sites, Lemma 18.19.2. We take the map which maps the summand corresponding to $\varphi '$ to the summand corresponding to $\varphi = a \circ \varphi '$ by the identity map on $\mathcal{O}(V)$. As $a$ is a monomorphism, this map is injective. As sheafification is exact, the result follows. $\square$

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