Lemma 15.86.6. Let $K = (K_ n^\bullet )$ be an object of $D(\textit{Ab}(\mathbf{N}))$. There exists a canonical distinguished triangle

\[ R\mathop{\mathrm{lim}}\nolimits K \to \prod \nolimits _ n K_ n^\bullet \to \prod \nolimits _ n K_ n^\bullet \to R\mathop{\mathrm{lim}}\nolimits K[1] \]

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 15.86.1 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 15.86.1 we see that the lemma holds in this case.

Next, assume $K = (K_ n^\bullet )$ is general. By Lemma 15.86.1 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 $K$.
$\square$

## Comments (0)