Lemma 20.10.5. 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 20.10.3 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 20.10.4 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 20.10.2) satisfies the assumptions of Homology, Lemma 12.12.4, 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 12.12.5 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 $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet )$ with terms $\check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{I}^ q)$. Consider the associated total complex $\text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet ))$, see Homology, Definition 12.18.3. There is a map of complexes

\[ \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet )) \]

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

\[ \check{H}^0(\mathcal{U}, \mathcal{I}^\bullet ) \longrightarrow \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet )) \]

coming from the maps $\check{H}^0(\mathcal{U}, \mathcal{I}^ q) \to \check{\mathcal{C}}^0(\mathcal{U}, \mathcal{I}^ q)$. Both of these maps are quasi-isomorphisms by an application of Homology, Lemma 12.25.4. Namely, the columns of the double complex are exact in positive degrees because the Čech complex as a functor is exact (Lemma 20.10.1) 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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