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Remark 15.86.10. Let $(K_ n)$ be an inverse system of objects of $D(\textit{Ab})$. Let $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ be a derived limit of this system (see Derived Categories, Section 13.34). Such a derived limit exists because $D(\textit{Ab})$ has countable products (Derived Categories, Lemma 13.34.2). By Lemma 15.86.9 we can also lift $(K_ n)$ to an object $M$ of $D(\mathbf{N})$. Then $K \cong R\mathop{\mathrm{lim}}\nolimits M$ where $R\mathop{\mathrm{lim}}\nolimits $ is the functor (15.86.1.1) because $R\mathop{\mathrm{lim}}\nolimits M$ is also a derived limit of the system $(K_ n)$ by Lemma 15.86.7. Thus, although there may be many isomorphism classes of lifts $M$ of the system $(K_ n)$, the isomorphism type of $R\mathop{\mathrm{lim}}\nolimits M$ is independent of the choice because it is isomorphic to the derived limit $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ of the system. Thus we may apply results on $R\mathop{\mathrm{lim}}\nolimits $ proved in this section to derived limits. For example, for every $p \in \mathbf{Z}$ there is a canonical short exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(K_ n) \to H^ p(K) \to \mathop{\mathrm{lim}}\nolimits H^ p(K_ n) \to 0 \]

because we may apply Lemma 15.86.8 to $M$. This can also been seen directly, without invoking the existence of $M$, by applying the argument of the proof of Lemma 15.86.8 to the (defining) distinguished triangle $K \to \prod K_ n \to \prod K_ n \to K[1]$.


Comments (3)

Comment #8609 by nkym on

Maybe the last two references to 07KX should be to 07KY

Comment #9803 by ZL on

Typo: in the fourth line "to an object of ".


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