The Stacks project

Lemma 15.86.1. The functor $\mathop{\mathrm{lim}}\nolimits : \textit{Ab}(\mathbf{N}) \to \textit{Ab}$ has a right derived functor

15.86.1.1
\begin{equation} \label{more-algebra-equation-Rlim} R\mathop{\mathrm{lim}}\nolimits : D(\textit{Ab}(\mathbf{N})) \longrightarrow D(\textit{Ab}) \end{equation}

As usual we set $R^ p\mathop{\mathrm{lim}}\nolimits (K) = H^ p(R\mathop{\mathrm{lim}}\nolimits (K))$. Moreover, we have

  1. for any $(A_ n)$ in $\textit{Ab}(\mathbf{N})$ we have $R^ p\mathop{\mathrm{lim}}\nolimits A_ n = 0$ for $p > 1$,

  2. the object $R\mathop{\mathrm{lim}}\nolimits A_ n$ of $D(\textit{Ab})$ is represented by the complex

    \[ \prod A_ n \to \prod A_ n,\quad (x_ n) \mapsto (x_ n - f_{n + 1}(x_{n + 1})) \]

    sitting in degrees $0$ and $1$,

  3. if $(A_ n)$ is ML, then $R^1\mathop{\mathrm{lim}}\nolimits A_ n = 0$, i.e., $(A_ n)$ is right acyclic for $\mathop{\mathrm{lim}}\nolimits $,

  4. every $K^\bullet \in D(\textit{Ab}(\mathbf{N}))$ is quasi-isomorphic to a complex whose terms are right acyclic for $\mathop{\mathrm{lim}}\nolimits $, and

  5. if each $K^ p = (K^ p_ n)$ is right acyclic for $\mathop{\mathrm{lim}}\nolimits $, i.e., of $R^1\mathop{\mathrm{lim}}\nolimits _ n K^ p_ n = 0$, then $R\mathop{\mathrm{lim}}\nolimits K$ is represented by the complex whose term in degree $p$ is $\mathop{\mathrm{lim}}\nolimits _ n K_ n^ p$.

Proof. Let $(A_ n)$ be an arbitrary inverse system. Let $(B_ n)$ be the inverse system with

\[ B_ n = A_ n \oplus A_{n - 1} \oplus \ldots \oplus A_1 \]

and transition maps given by projections. Let $A_ n \to B_ n$ be given by $(1, f_ n, f_{n - 1} \circ f_ n, \ldots , f_2 \circ \ldots \circ f_ n)$ where $f_ i : A_ i \to A_{i - 1}$ are the transition maps. In this way we see that every inverse system is a subobject of a ML system (Homology, Section 12.31). It follows from Derived Categories, Lemma 13.15.6 using Homology, Lemma 12.31.3 that every ML system is right acyclic for $\mathop{\mathrm{lim}}\nolimits $, i.e., (3) holds. This already implies that $RF$ is defined on $D^+(\textit{Ab}(\mathbf{N}))$, see Derived Categories, Proposition 13.16.8. Set $C_ n = A_{n - 1} \oplus \ldots \oplus A_1$ for $n > 1$ and $C_1 = 0$ with transition maps given by projections as well. Then there is a short exact sequence of inverse systems $0 \to (A_ n) \to (B_ n) \to (C_ n) \to 0$ where $B_ n \to C_ n$ is given by $(x_ i) \mapsto (x_ i - f_{i + 1}(x_{i + 1}))$. Since $(C_ n)$ is ML as well, we conclude that (2) holds (by proposition reference above) which also implies (1). Finally, this implies by Derived Categories, Lemma 13.32.2 that $R\mathop{\mathrm{lim}}\nolimits $ is in fact defined on all of $D(\textit{Ab}(\mathbf{N}))$. In fact, the proof of Derived Categories, Lemma 13.32.2 proceeds by proving assertions (4) and (5). $\square$


Comments (2)

Comment #8714 by Nico on

In the fourth line of the proof, the is missing a parenthesis.


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07KW. Beware of the difference between the letter 'O' and the digit '0'.