Lemma 12.31.7. Let
be an inverse system of complexes of abelian groups and denote $A^{-2} \to A^{-1} \to A^0 \to A^1$ its limit. Denote $(H_ i^{-1})$, $(H_ i^0)$ the inverse systems of cohomologies, and denote $H^{-1}$, $H^0$ the cohomologies of $A^{-2} \to A^{-1} \to A^0 \to A^1$. If $(A^{-2}_ i)$ and $(A^{-1}_ i)$ are ML and $(H^{-1}_ i)$ is essentially constant, then $H^0 = \mathop{\mathrm{lim}}\nolimits H_ i^0$.
Proof. Let $Z^ j_ i = \mathop{\mathrm{Ker}}(A^ j_ i \to A^{j + 1}_ i)$ and $I^ j_ i = \mathop{\mathrm{Im}}(A^{j - 1}_ i \to A^ j_ i)$. Note that $\mathop{\mathrm{lim}}\nolimits Z^0_ i = \mathop{\mathrm{Ker}}(\mathop{\mathrm{lim}}\nolimits A^0_ i \to \mathop{\mathrm{lim}}\nolimits A^1_ i)$ as taking kernels commutes with limits. The systems $(I^{-1}_ i)$ and $(I^0_ i)$ have ML as quotients of the systems $(A^{-2}_ i)$ and $(A^{-1}_ i)$, see Lemma 12.31.3. Thus an exact sequence
of inverse systems where $(I^{-1}_ i)$ has ML and where $(H^{-1}_ i)$ is essentially constant by assumption. Hence $(Z^{-1}_ i)$ has ML by Lemma 12.31.6. The exact sequence
and an application of Lemma 12.31.3 shows that $\mathop{\mathrm{lim}}\nolimits A^{-1}_ i \to \mathop{\mathrm{lim}}\nolimits I^0_ i$ is surjective. Finally, the exact sequence
and Lemma 12.31.3 show that $\mathop{\mathrm{lim}}\nolimits I^0_ i \to \mathop{\mathrm{lim}}\nolimits Z^0_ i \to \mathop{\mathrm{lim}}\nolimits H^0_ i \to 0$ is exact. Putting everything together we win. $\square$
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