$(A^{-2}_ n \to A^{-1}_ n \to A^0_ n \to A^1_ n)$

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 $(H_ n^{-1})$, $(H_ n^0)$ 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

1. $(A^{-2}_ n)$ and $(A^{-1}_ n)$ have vanishing $R^1\mathop{\mathrm{lim}}\nolimits$,

2. $(H^{-1}_ n)$ has vanishing $R^1\mathop{\mathrm{lim}}\nolimits$,

then $H^0 = \mathop{\mathrm{lim}}\nolimits H_ n^0$.

Proof. Let $K \in D(\textit{Ab}(\mathbf{N}))$ be the object represented by the system of complexes whose $n$th constituent is the complex $A^{-2}_ n \to A^{-1}_ n \to A^0_ n \to A^1_ n$. We will compute $H^0(R\mathop{\mathrm{lim}}\nolimits K)$ using both spectral sequences1 of Derived Categories, Lemma 13.21.3. The first has $E_1$-page

$\begin{matrix} 0 & 0 & R^1\mathop{\mathrm{lim}}\nolimits A^0_ n & R^1\mathop{\mathrm{lim}}\nolimits A^1_ n \\ A^{-2} & A^{-1} & A^0 & A^1 \end{matrix}$

with horizontal differentials and all higher differentials are zero. The second has $E_2$ page

$\begin{matrix} R^1\mathop{\mathrm{lim}}\nolimits H^{-2}_ n & 0 & R^1\mathop{\mathrm{lim}}\nolimits H^0_ n & R^1 \mathop{\mathrm{lim}}\nolimits H^1_ n \\ \mathop{\mathrm{lim}}\nolimits H^{-2}_ n & \mathop{\mathrm{lim}}\nolimits H^{-1}_ n & \mathop{\mathrm{lim}}\nolimits H^0_ n & \mathop{\mathrm{lim}}\nolimits H^1_ n \end{matrix}$

and degenerates at this point. The result follows. $\square$

 To use these spectral sequences we have to show that $\textit{Ab}(\mathbf{N})$ has enough injectives. A inverse system $(I_ n)$ of abelian groups is injective if and only if each $I_ n$ is an injective abelian group and the transition maps are split surjections. Every system embeds in one of these. Details omitted.

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