The Stacks project

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. $\square$

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

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

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. $\square$

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. $\square$

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

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

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

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. $\square$

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. $\square$

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

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

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. $\square$

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

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

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. $\square$

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

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

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

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

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. $\square$

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

This is true because

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. $\square$

Proof. In this case the functor $f^*$ transforms injections into injections (Modules, Lemma 17.20.2). Hence the result follows from Homology, Lemma 12.29.1. $\square$

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. $\square$