Lemma 15.86.7. 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 (1)
Comment #9964 by Elías Guisado on
There are also: