The Stacks project

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. $\square$

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