The Stacks project

Lemma 21.25.3. In Situation 21.25.1 let $(K_ n)$ be an inverse system in $D_\mathcal {A}^+(\mathcal{O})$. Assume that for every $j$ the inverse system $(H^ j(K_ n))$ in $\mathcal{A}$ is eventually constant with value $\mathcal{H}^ j$. Then $H^ j(R\mathop{\mathrm{lim}}\nolimits K_ n) = \mathcal{H}^ j$ for all $j$.

Proof. Let $V \in \mathcal{B}$. Let $\{ V_ i \to V\} $ be in the set $\text{Cov}_ V$ of Situation 21.25.1. Because $K_ n$ is bounded below there is a spectral sequence

\[ E_2^{p, q} = H^ p(V_ i, H^ q(K_ n)) \]

converging to $H^{p + q}(V_ i, K_ n)$. See Derived Categories, Lemma 13.21.3. Observe that $E_2^{p, q} = 0$ for $p > d_ V$ by assumption. Pick $n_0$ such that

\[ \begin{matrix} \mathcal{H}^{j + 1} & = & H^{j + 1}(K_ n), \\ \mathcal{H}^ j & = & H^ j(K_ n), \\ \ldots , \\ \mathcal{H}^{j - d_ V - 2} & = & H^{j - d_ V - 2}(K_ n) \end{matrix} \]

for all $n \geq n_0$. Comparing the spectral sequences above for $K_ n$ and $K_{n_0}$, we see that for $n \geq n_0$ the cohomology groups $H^{j - 1}(V_ i, K_ n)$ and $H^ j(V_ i, K_ n)$ are independent of $n$. It follows that the map on sections $H^ j(R\mathop{\mathrm{lim}}\nolimits K_ n)(V) \to H^ j(K_ n)(V)$ is injective for $n$ large enough (depending on $V$), see Lemma 21.23.6. Since every object of $\mathcal{C}$ can be covered by elements of $\mathcal{B}$, we conclude that the map $H^ j(R\mathop{\mathrm{lim}}\nolimits K_ n) \to \mathcal{H}^ j$ is injective.

Surjectivity is shown in a similar manner. Namely, pick $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $\gamma \in \mathcal{H}^ j(U)$. We want to lift $\gamma $ to a section of $H^ j(R\mathop{\mathrm{lim}}\nolimits K_ n)$ after replacing $U$ by the members of a covering. Hence we may assume $U = V \in \mathcal{B}$ by property (1) of Situation 21.25.1. Pick $n_0$ such that

\[ \begin{matrix} \mathcal{H}^{j + 1} & = & H^{j + 1}(K_ n), \\ \mathcal{H}^ j & = & H^ j(K_ n), \\ \ldots , \\ \mathcal{H}^{j - d_ V - 2} & = & H^{j - d_ V - 2}(K_ n) \end{matrix} \]

for all $n \geq n_0$. Choose an element $\{ V_ i \to V\} $ of $\text{Cov}_ V$ such that $\gamma |_{V_ i} \in \mathcal{H}^ j(V_ i) = H^ j(K_{n_0})(V_ i)$ lifts to an element $\gamma _{n_0, i} \in H^ j(V_ i, K_{n_0})$. This is possible because $H^ j(K_{n_0})$ is the sheafification of $U \mapsto H^ j(U, K_{n_0})$ by Lemma 21.20.3. By the discussion in the first paragraph of the proof we have that $H^{j - 1}(V_ i, K_ n)$ and $H^ j(V_ i, K_ n)$ are independent of $n \geq n_0$. Hence $\gamma _{n_0, i}$ lifts to an element $\gamma _ i \in H^ j(V_ i, R\mathop{\mathrm{lim}}\nolimits K_ n)$ by Lemma 21.23.2. This finishes the proof. $\square$


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