The Stacks project

Lemma 21.12.4. Let $\mathcal{C}$ be a site. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}$. Let $\mathcal{F}$ be an $\mathcal{O}$-module, and denote $\mathcal{F}_{ab}$ the underlying sheaf of abelian groups. Then we have

\[ H^ i(\mathcal{C}, \mathcal{F}_{ab}) = H^ i(\mathcal{C}, \mathcal{F}) \]

and for any object $U$ of $\mathcal{C}$ we also have

\[ H^ i(U, \mathcal{F}_{ab}) = H^ i(U, \mathcal{F}). \]

Here the left hand side is cohomology computed in $\textit{Ab}(\mathcal{C})$ and the right hand side is cohomology computed in $\textit{Mod}(\mathcal{O})$.

Proof. By Derived Categories, Lemma 13.20.4 the $\delta $-functor $(\mathcal{F} \mapsto H^ p(U, \mathcal{F}))_{p \geq 0}$ is universal. The functor $\textit{Mod}(\mathcal{O}) \to \textit{Ab}(\mathcal{C})$, $\mathcal{F} \mapsto \mathcal{F}_{ab}$ is exact. Hence $(\mathcal{F} \mapsto H^ p(U, \mathcal{F}_{ab}))_{p \geq 0}$ is a $\delta $-functor also. Suppose we show that $(\mathcal{F} \mapsto H^ p(U, \mathcal{F}_{ab}))_{p \geq 0}$ is also universal. This will imply the second statement of the lemma by uniqueness of universal $\delta $-functors, see Homology, Lemma 12.12.5. Since $\textit{Mod}(\mathcal{O})$ has enough injectives, it suffices to show that $H^ i(U, \mathcal{I}_{ab}) = 0$ for any injective object $\mathcal{I}$ in $\textit{Mod}(\mathcal{O})$, see Homology, Lemma 12.12.4.

Let $\mathcal{I}$ be an injective object of $\textit{Mod}(\mathcal{O})$. Apply Lemma 21.10.9 with $\mathcal{F} = \mathcal{I}$, $\mathcal{B} = \mathcal{C}$ and $\text{Cov} = \text{Cov}_\mathcal {C}$. Assumption (3) of that lemma holds by Lemma 21.12.3. Hence we see that $H^ i(U, \mathcal{I}_{ab}) = 0$ for every object $U$ of $\mathcal{C}$.

If $\mathcal{C}$ has a final object then this also implies the first equality. If not, then according to Sites, Lemma 7.29.5 we see that the ringed topos $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ is equivalent to a ringed topos where the underlying site does have a final object. Hence the lemma follows. $\square$

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