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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