Lemma 15.82.1. In the situation above. The functor $\mathop{\mathrm{lim}}\nolimits : \textit{Mod}(\mathbf{N}, (A_ n)) \to \text{Mod}_ A$ has a right derived functor

$R\mathop{\mathrm{lim}}\nolimits : D(\textit{Mod}(\mathbf{N}, (A_ n))) \longrightarrow D(A)$

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

2. the object $R\mathop{\mathrm{lim}}\nolimits M_ n$ of $D(\text{Mod}_ A)$ is represented by the complex

$\prod M_ n \to \prod M_ n,\quad (x_ n) \mapsto (x_ n - f_{n + 1}(x_{n + 1}))$

sitting in degrees $0$ and $1$,

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

4. every $K^\bullet \in D(\textit{Mod}(\mathbf{N}, (A_ n)))$ 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. The proof of this is word for word the same as the proof of Lemma 15.81.1. $\square$

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