The Stacks project

Proof. Apply Homology, Lemma 12.29.1 to the categories $\mathcal{A} = \textit{Ab}(\mathcal{C})$, $\mathcal{B} = \textit{PAb}(\mathcal{C})$, the inclusion functor and sheafification. (See Modules on Sites, Section 18.3 to see that all assumptions of the lemma are satisfied.) $\square$

Proof. By Lemma 21.10.1 we see that $\mathcal{I}$ is an injective object in $\textit{PAb}(\mathcal{C})$. Hence we can apply Lemma 21.9.6 (or its proof) to see the vanishing of higher Čech cohomology group. For the zeroth see Lemma 21.8.2. $\square$

Proof. Let $\mathcal{F}$ be an abelian sheaf. 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 $\check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{I}^ q)$. Next, 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

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 21.10.2 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 $D^{+}(\mathbf{Z})$ 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 $\mathcal{G}|_ U$-torsor on $\mathcal{C}/U$ whose restriction to $\mathcal{C}/U_ i$ is trivial. By Lemma 21.4.2 this means there exists a section $s_ i \in \mathcal{F}(U_ i)$. On $U_{i_0} \times _ U U_{i_1}$ there is a unique section $s_{i_0i_1}$ of $\mathcal{G}$ such that $s_{i_0i_1} \cdot s_{i_0}|_{U_{i_0} \times _ U U_{i_1}} = s_{i_1}|_{U_{i_0} \times _ U U_{i_1}}$. An easy 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 object $U$ of $\mathcal{C}$ 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 21.8.2. We have that $i(\mathcal{I})$ is Čech acyclic by Lemma 21.10.2. And we have that $\check{H}^ p(\mathcal{U}, -) = R^ p\check{H}^0(\mathcal{U}, -)$ as functors on $\textit{PAb}(\mathcal{C})$ by Lemma 21.9.6. Putting everything together gives the lemma. $\square$

Proof. We will use the spectral sequence of Lemma 21.10.6. 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 a covering $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ 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 $s_{i_0i_1} \in \mathcal{F}(U_{i_0} \times _ U U_{i_1})$. 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 $\mathcal{F}$ 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 abelian sheaf. By Lemma 21.10.2 $\mathcal{I}$ 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 21.10.8 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 covering $\mathcal{U} \in \text{Cov}$. This implies that $\mathcal{Q}$ is also an abelian sheaf 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 abelian sheaf 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. $\square$