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.
$\square$

## Comments (1)

Comment #1108 by Evan Warner on