Lemma 115.8.6. 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. Let d \in \mathbf{N}. Assume
K_ n is an object of D^+(\mathcal{O}) for all n,
for q \in \mathbf{Z} there exists n(q) such that H^ q(K_{n + 1}) \to H^ q(K_ n) is an isomorphism for n \geq n(q),
every object of \mathcal{C} has a covering whose members are elements of \mathcal{B},
for every U \in \mathcal{B} we have H^ p(U, H^ q(K_ n)) = 0 for p > d and all q.
Then we have H^ m(R\mathop{\mathrm{lim}}\nolimits K_ n) = \mathop{\mathrm{lim}}\nolimits H^ m(K_ n) for all m \in \mathbf{Z}.
Proof.
Set K = R\mathop{\mathrm{lim}}\nolimits K_ n. Let U \in \mathcal{B}. For each n there is a spectral sequence
H^ p(U, H^ q(K_ n)) \Rightarrow H^{p + q}(U, K_ n)
which converges as K_ n is bounded below, see Derived Categories, Lemma 13.21.3. If we fix m \in \mathbf{Z}, then we see from our assumption (4) that only H^ p(U, H^ q(K_ n)) contribute to H^ m(U, K_ n) for 0 \leq p \leq d and m - d \leq q \leq m. By assumption (2) this implies that H^ m(U, K_{n + 1}) \to H^ m(U, K_ n) is an isomorphism as soon as n \geq \max {n(m), \ldots , n(m - d)}. The functor R\Gamma (U, -) commutes with derived limits by Injectives, Lemma 19.13.6. Thus we have
H^ m(U, K) = H^ m(R\mathop{\mathrm{lim}}\nolimits R\Gamma (U, K_ n))
On the other hand we have just seen that the complexes R\Gamma (U, K_ n) have eventually constant cohomology groups. Thus by More on Algebra, Remark 15.86.10 we find that H^ m(U, K) is equal to H^ m(U, K_ n) for all n \gg 0 for some bound independent of U \in \mathcal{B}. Pick such an n. Finally, recall that H^ m(K) is the sheafification of the presheaf U \mapsto H^ m(U, K) and H^ m(K_ n) is the sheafification of the presheaf U \mapsto H^ m(U, K_ n). On the elements of \mathcal{B} these presheaves have the same values. Therefore assumption (3) guarantees that the sheafifications are the same too. The lemma follows.
\square
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