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 (1)

Comment #8714 by Nico on

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


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