## 20.11 Čech cohomology and cohomology

Lemma 20.11.1. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be a covering. Let $\mathcal{I}$ be an injective $\mathcal{O}_ X$-module. 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. An injective $\mathcal{O}_ X$-module is also injective as an object in the category $\textit{PMod}(\mathcal{O}_ X)$ (for example since sheafification is an exact left adjoint to the inclusion functor, using Homology, Lemma 12.29.1). Hence we can apply Lemma 20.10.5 (or its proof) to see the result. $\square$

Lemma 20.11.2. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be a covering. There is a transformation

$\check{\mathcal{C}}^\bullet (\mathcal{U}, -) \longrightarrow R\Gamma (U, -)$

of functors $\textit{Mod}(\mathcal{O}_ X) \to D^{+}(\mathcal{O}_ X(U))$. In particular this provides canonical maps $\check{H}^ p(\mathcal{U}, \mathcal{F}) \to H^ p(U, \mathcal{F})$ for $\mathcal{F}$ ranging over $\textit{Mod}(\mathcal{O}_ X)$.

Proof. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet$. Consider the double complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet )$ with terms $\check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{I}^ q)$. There is a map of complexes

$\alpha : \Gamma (U, \mathcal{I}^\bullet ) \longrightarrow \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet ))$

coming from the maps $\mathcal{I}^ q(U) \to \check{H}^0(\mathcal{U}, \mathcal{I}^ q)$ and a map of complexes

$\beta : \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet ))$

coming from the map $\mathcal{F} \to \mathcal{I}^0$. We can apply Homology, Lemma 12.25.4 to see that $\alpha$ is a quasi-isomorphism. Namely, Lemma 20.11.1 implies that the $q$th row of the double complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet )$ is a resolution of $\Gamma (U, \mathcal{I}^ q)$. Hence $\alpha$ becomes invertible in $D^{+}(\mathcal{O}_ X(U))$ and the transformation of the lemma is the composition of $\beta$ followed by the inverse of $\alpha$. We omit the verification that this is functorial. $\square$

Lemma 20.11.3. Let $X$ be a topological space. Let $\mathcal{H}$ be an abelian sheaf on $X$. Let $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ be an open covering. The map

$\check{H}^1(\mathcal{U}, \mathcal{H}) \longrightarrow H^1(X, \mathcal{H})$

is injective and identifies $\check{H}^1(\mathcal{U}, \mathcal{H})$ via the bijection of Lemma 20.4.3 with the set of isomorphism classes of $\mathcal{H}$-torsors which restrict to trivial torsors over each $U_ i$.

Proof. To see this we construct an inverse map. Namely, let $\mathcal{F}$ be a $\mathcal{H}$-torsor whose restriction to $U_ i$ is trivial. By Lemma 20.4.2 this means there exists a section $s_ i \in \mathcal{F}(U_ i)$. On $U_{i_0} \cap U_{i_1}$ there is a unique section $s_{i_0i_1}$ of $\mathcal{H}$ such that $s_{i_0i_1} \cdot s_{i_0}|_{U_{i_0} \cap U_{i_1}} = s_{i_1}|_{U_{i_0} \cap U_{i_1}}$. A computation shows that $s_{i_0i_1}$ is a Čech cocycle and that its class is well defined (i.e., does not depend on the choice of the sections $s_ i$). The inverse maps the isomorphism class of $\mathcal{F}$ to the cohomology class of the cocycle $(s_{i_0i_1})$. We omit the verification that this map is indeed an inverse. $\square$

Lemma 20.11.4. Let $X$ be a ringed space. Consider the functor $i : \textit{Mod}(\mathcal{O}_ X) \to \textit{PMod}(\mathcal{O}_ X)$. 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 20.7.

Proof. It is clear that $i$ is left exact. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet$. 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 open $U$ 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 20.11.5. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be a covering. For any sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$ there is a spectral sequence $(E_ r, d_ r)_{r \geq 0}$ with

$E_2^{p, q} = \check{H}^ p(\mathcal{U}, \underline{H}^ q(\mathcal{F}))$

converging to $H^{p + q}(U, \mathcal{F})$. This spectral sequence is functorial in $\mathcal{F}$.

Proof. This is a Grothendieck spectral sequence (see Derived Categories, Lemma 13.22.2) for the functors

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

Namely, we have $\check{H}^0(\mathcal{U}, i(\mathcal{F})) = \mathcal{F}(U)$ by Lemma 20.9.2. We have that $i(\mathcal{I})$ is Čech acyclic by Lemma 20.11.1. And we have that $\check{H}^ p(\mathcal{U}, -) = R^ p\check{H}^0(\mathcal{U}, -)$ as functors on $\textit{PMod}(\mathcal{O}_ X)$ by Lemma 20.10.5. Putting everything together gives the lemma. $\square$

Lemma 20.11.6. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be a covering. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Assume that $H^ i(U_{i_0 \ldots i_ p}, \mathcal{F}) = 0$ for all $i > 0$, all $p \geq 0$ and all $i_0, \ldots , i_ p \in I$. Then $\check{H}^ p(\mathcal{U}, \mathcal{F}) = H^ p(U, \mathcal{F})$ as $\mathcal{O}_ X(U)$-modules.

Proof. We will use the spectral sequence of Lemma 20.11.5. The assumptions mean that $E_2^{p, q} = 0$ for all $(p, q)$ with $q \not= 0$. Hence the spectral sequence degenerates at $E_2$ and the result follows. $\square$

Lemma 20.11.7. Let $X$ be a ringed space. Let

$0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$

be a short exact sequence of $\mathcal{O}_ X$-modules. Let $U \subset X$ be an open subset. If there exists a cofinal system of open coverings $\mathcal{U}$ of $U$ such that $\check{H}^1(\mathcal{U}, \mathcal{F}) = 0$, then the map $\mathcal{G}(U) \to \mathcal{H}(U)$ is surjective.

Proof. Take an element $s \in \mathcal{H}(U)$. Choose an open covering $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ such that (a) $\check{H}^1(\mathcal{U}, \mathcal{F}) = 0$ and (b) $s|_{U_ i}$ is the image of a section $s_ i \in \mathcal{G}(U_ i)$. Since we can certainly find a covering such that (b) holds it follows from the assumptions of the lemma that we can find a covering such that (a) and (b) both hold. Consider the sections

$s_{i_0i_1} = s_{i_1}|_{U_{i_0i_1}} - s_{i_0}|_{U_{i_0i_1}}.$

Since $s_ i$ lifts $s$ we see that $s_{i_0i_1} \in \mathcal{F}(U_{i_0i_1})$. By the vanishing of $\check{H}^1(\mathcal{U}, \mathcal{F})$ we can find sections $t_ i \in \mathcal{F}(U_ i)$ such that

$s_{i_0i_1} = t_{i_1}|_{U_{i_0i_1}} - t_{i_0}|_{U_{i_0i_1}}.$

Then clearly the sections $s_ i - t_ i$ satisfy the sheaf condition and glue to a section of $\mathcal{G}$ over $U$ which maps to $s$. Hence we win. $\square$

Lemma 20.11.8. Let $X$ be a ringed space. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module such that

$\check{H}^ p(\mathcal{U}, \mathcal{F}) = 0$

for all $p > 0$ and any open covering $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ of an open of $X$. Then $H^ p(U, \mathcal{F}) = 0$ for all $p > 0$ and any open $U \subset X$.

Proof. Let $\mathcal{F}$ be a sheaf satisfying the assumption of the lemma. We will indicate this by saying “$\mathcal{F}$ has vanishing higher Čech cohomology for any open covering”. Choose an embedding $\mathcal{F} \to \mathcal{I}$ into an injective $\mathcal{O}_ X$-module. By Lemma 20.11.1 $\mathcal{I}$ has vanishing higher Čech cohomology for any open covering. Let $\mathcal{Q} = \mathcal{I}/\mathcal{F}$ so that we have a short exact sequence

$0 \to \mathcal{F} \to \mathcal{I} \to \mathcal{Q} \to 0.$

By Lemma 20.11.7 and our assumptions this sequence is actually exact as a sequence of presheaves! In particular we have a long exact sequence of Čech cohomology groups for any open covering $\mathcal{U}$, see Lemma 20.10.2 for example. This implies that $\mathcal{Q}$ is also an $\mathcal{O}_ X$-module with vanishing higher Čech cohomology for all open coverings.

Next, we look at the long exact cohomology sequence

$\xymatrix{ 0 \ar[r] & H^0(U, \mathcal{F}) \ar[r] & H^0(U, \mathcal{I}) \ar[r] & H^0(U, \mathcal{Q}) \ar[lld] \\ & H^1(U, \mathcal{F}) \ar[r] & H^1(U, \mathcal{I}) \ar[r] & H^1(U, \mathcal{Q}) \ar[lld] \\ & \ldots & \ldots & \ldots \\ }$

for any open $U \subset X$. Since $\mathcal{I}$ is injective we have $H^ n(U, \mathcal{I}) = 0$ for $n > 0$ (see Derived Categories, Lemma 13.20.4). By the above we see that $H^0(U, \mathcal{I}) \to H^0(U, \mathcal{Q})$ is surjective and hence $H^1(U, \mathcal{F}) = 0$. Since $\mathcal{F}$ was an arbitrary $\mathcal{O}_ X$-module with vanishing higher Čech cohomology we conclude that also $H^1(U, \mathcal{Q}) = 0$ since $\mathcal{Q}$ is another of these sheaves (see above). By the long exact sequence this in turn implies that $H^2(U, \mathcal{F}) = 0$. And so on and so forth. $\square$

Lemma 20.11.9. (Variant of Lemma 20.11.8.) Let $X$ be a ringed space. Let $\mathcal{B}$ be a basis for the topology on $X$. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Assume there exists a set of open coverings $\text{Cov}$ with the following properties:

1. For every $\mathcal{U} \in \text{Cov}$ with $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ we have $U, U_ i \in \mathcal{B}$ and every $U_{i_0 \ldots i_ p} \in \mathcal{B}$.

2. For every $U \in \mathcal{B}$ the open coverings of $U$ occurring in $\text{Cov}$ is a cofinal system of open coverings of $U$.

3. For every $\mathcal{U} \in \text{Cov}$ we have $\check{H}^ p(\mathcal{U}, \mathcal{F}) = 0$ for all $p > 0$.

Then $H^ p(U, \mathcal{F}) = 0$ for all $p > 0$ and any $U \in \mathcal{B}$.

Proof. Let $\mathcal{F}$ and $\text{Cov}$ be as in the lemma. We will indicate this by saying “$\mathcal{F}$ has vanishing higher Čech cohomology for any $\mathcal{U} \in \text{Cov}$”. Choose an embedding $\mathcal{F} \to \mathcal{I}$ into an injective $\mathcal{O}_ X$-module. By Lemma 20.11.1 $\mathcal{I}$ has vanishing higher Čech cohomology for any $\mathcal{U} \in \text{Cov}$. Let $\mathcal{Q} = \mathcal{I}/\mathcal{F}$ so that we have a short exact sequence

$0 \to \mathcal{F} \to \mathcal{I} \to \mathcal{Q} \to 0.$

By Lemma 20.11.7 and our assumption (2) this sequence gives rise to an exact sequence

$0 \to \mathcal{F}(U) \to \mathcal{I}(U) \to \mathcal{Q}(U) \to 0.$

for every $U \in \mathcal{B}$. Hence for any $\mathcal{U} \in \text{Cov}$ we get a short exact sequence of Čech complexes

$0 \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}) \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{Q}) \to 0$

since each term in the Čech complex is made up out of a product of values over elements of $\mathcal{B}$ by assumption (1). In particular we have a long exact sequence of Čech cohomology groups for any open covering $\mathcal{U} \in \text{Cov}$. This implies that $\mathcal{Q}$ is also an $\mathcal{O}_ X$-module with vanishing higher Čech cohomology for all $\mathcal{U} \in \text{Cov}$.

Next, we look at the long exact cohomology sequence

$\xymatrix{ 0 \ar[r] & H^0(U, \mathcal{F}) \ar[r] & H^0(U, \mathcal{I}) \ar[r] & H^0(U, \mathcal{Q}) \ar[lld] \\ & H^1(U, \mathcal{F}) \ar[r] & H^1(U, \mathcal{I}) \ar[r] & H^1(U, \mathcal{Q}) \ar[lld] \\ & \ldots & \ldots & \ldots \\ }$

for any $U \in \mathcal{B}$. Since $\mathcal{I}$ is injective we have $H^ n(U, \mathcal{I}) = 0$ for $n > 0$ (see Derived Categories, Lemma 13.20.4). By the above we see that $H^0(U, \mathcal{I}) \to H^0(U, \mathcal{Q})$ is surjective and hence $H^1(U, \mathcal{F}) = 0$. Since $\mathcal{F}$ was an arbitrary $\mathcal{O}_ X$-module with vanishing higher Čech cohomology for all $\mathcal{U} \in \text{Cov}$ we conclude that also $H^1(U, \mathcal{Q}) = 0$ since $\mathcal{Q}$ is another of these sheaves (see above). By the long exact sequence this in turn implies that $H^2(U, \mathcal{F}) = 0$. And so on and so forth. $\square$

Lemma 20.11.10. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{I}$ be an injective $\mathcal{O}_ X$-module. Then

1. $\check{H}^ p(\mathcal{V}, f_*\mathcal{I}) = 0$ for all $p > 0$ and any open covering $\mathcal{V} : V = \bigcup _{j \in J} V_ j$ of $Y$.

2. $H^ p(V, f_*\mathcal{I}) = 0$ for all $p > 0$ and every open $V \subset Y$.

In other words, $f_*\mathcal{I}$ is right acyclic for $\Gamma (V, -)$ (see Derived Categories, Definition 13.15.3) for any $V \subset Y$ open.

Proof. Set $\mathcal{U} : f^{-1}(V) = \bigcup _{j \in J} f^{-1}(V_ j)$. It is an open covering of $X$ and

$\check{\mathcal{C}}^\bullet (\mathcal{V}, f_*\mathcal{I}) = \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}).$

This is true because

$f_*\mathcal{I}(V_{j_0 \ldots j_ p}) = \mathcal{I}(f^{-1}(V_{j_0 \ldots j_ p})) = \mathcal{I}(f^{-1}(V_{j_0}) \cap \ldots \cap f^{-1}(V_{j_ p})) = \mathcal{I}(U_{j_0 \ldots j_ p}).$

Thus the first statement of the lemma follows from Lemma 20.11.1. The second statement follows from the first and Lemma 20.11.8. $\square$

The following lemma implies in particular that $f_* : \textit{Ab}(X) \to \textit{Ab}(Y)$ transforms injective abelian sheaves into injective abelian sheaves.

Lemma 20.11.11. Let $f : X \to Y$ be a morphism of ringed spaces. Assume $f$ is flat. Then $f_*\mathcal{I}$ is an injective $\mathcal{O}_ Y$-module for any injective $\mathcal{O}_ X$-module $\mathcal{I}$.

Proof. In this case the functor $f^*$ transforms injections into injections (Modules, Lemma 17.19.2). Hence the result follows from Homology, Lemma 12.29.1. $\square$

Lemma 20.11.12. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $I$ be a set. For $i \in I$ let $\mathcal{F}_ i$ be an $\mathcal{O}_ X$-module. Let $U \subset X$ be open. 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 Sheaves, Section 6.29. 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 20.7.2, there exists an open covering $\mathcal{U} : U = \bigcup U_ j$ such that $\xi |_{U_ j} = 0$ for all $j$. By Lemma 20.11.3 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 20.11.3), 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 12.32.1 we conclude that $\check\xi = 0$ as desired. $\square$

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