Lemma 15.85.6. Let $K = (K_ n^\bullet )$ be an object of $D(\textit{Ab}(\mathbf{N}))$. There exists a canonical distinguished triangle

$R\mathop{\mathrm{lim}}\nolimits K \to \prod \nolimits _ n K_ n^\bullet \to \prod \nolimits _ n K_ n^\bullet \to R\mathop{\mathrm{lim}}\nolimits K$

in $D(\textit{Ab})$. In other words, $R\mathop{\mathrm{lim}}\nolimits K$ is a derived limit of the inverse system $(K_ n^\bullet )$ of $D(\textit{Ab})$, see Derived Categories, Definition 13.34.1.

Proof. Suppose that for each $p$ the inverse system $(K_ n^ p)$ is right acyclic for $\mathop{\mathrm{lim}}\nolimits$. By Lemma 15.85.1 this gives a short exact sequence

$0 \to \mathop{\mathrm{lim}}\nolimits _ n K^ p_ n \to \prod \nolimits _ n K^ p_ n \to \prod \nolimits _ n K^ p_ n \to 0$

for each $p$. Since the complex consisting of $\mathop{\mathrm{lim}}\nolimits _ n K^ p_ n$ computes $R\mathop{\mathrm{lim}}\nolimits K$ by Lemma 15.85.1 we see that the lemma holds in this case.

Next, assume $K = (K_ n^\bullet )$ is general. By Lemma 15.85.1 there is a quasi-isomorphism $K \to L$ in $D(\textit{Ab}(\mathbf{N}))$ such that $(L_ n^ p)$ is acyclic for each $p$. Then $\prod K_ n^\bullet$ is quasi-isomorphic to $\prod L_ n^\bullet$ as products are exact in $\textit{Ab}$, whence the result for $L$ (proved above) implies the result for $K$. $\square$

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