Lemma 21.23.12. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $(K_ n)$ be an inverse system of objects of $D(\mathcal{O})$. Let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be a subset. Assume

every object of $\mathcal{C}$ has a covering whose members are elements of $\mathcal{B}$,

for all $U \in \mathcal{B}$ and all $q \in \mathbf{Z}$ we have

$H^ p(U, H^ q(K_ n)) = 0$ for $p > 0$,

the inverse system $H^0(U, H^ q(K_ n))$ has vanishing $R^1\mathop{\mathrm{lim}}\nolimits $.

Then $H^ q(R\mathop{\mathrm{lim}}\nolimits K_ n) = \mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$ for $q \in \mathbf{Z}$.

**Proof.**
Set $K = R\mathop{\mathrm{lim}}\nolimits K_ n$. We will use notation as in Remark 21.23.4. Let $U \in \mathcal{B}$. By Lemma 21.23.11 and (2)(a) we have $H^ q(U, K_ n) = H^0(U, H^ q(K_ n))$. Using that the functor $R\Gamma (U, -)$ commutes with derived limits we have

\[ H^ q(U, K) = H^ q(R\mathop{\mathrm{lim}}\nolimits R\Gamma (U, K_ n)) = \mathop{\mathrm{lim}}\nolimits H^0(U, H^ q(K_ n)) \]

where the final equality follows from More on Algebra, Remark 15.77.9 and assumption (2)(b). Thus $H^ q(U, K)$ is the inverse limit the sections of the sheaves $H^ q(K_ n)$ over $U$. Since $\mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$ is a sheaf we find using assumption (1) that $H^ q(K)$, which is the sheafification of the presheaf $U \mapsto H^ q(U, K)$, is equal to $\mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$. This proves the lemma.
$\square$

## Comments (0)