Lemma 13.34.5. Let $\mathcal{A}$ be an abelian category with countable products and enough injectives. Let $K^\bullet$ be a complex. Let $I_ n^\bullet$ be the inverse system of bounded below complexes of injectives produced by Lemma 13.29.3. Then $I^\bullet = \mathop{\mathrm{lim}}\nolimits I_ n^\bullet$ exists, is K-injective, represents $R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet$ in $D(\mathcal{A})$, and the following are equivalent

1. the map $K^\bullet \to I^\bullet$ (see proof) is a quasi-isomorphism,

2. the map $K^\bullet \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet$ of Remark 13.34.4 is an isomorphism in $D(\mathcal{A})$.

Proof. The statement of the lemma makes sense as $R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet$ exists by Lemma 13.34.3. Each complex $I_ n^\bullet$ is K-injective by Lemma 13.31.4. Choose direct sum decompositions $I_{n + 1}^ p = C_{n + 1}^ p \oplus I_ n^ p$ for all $n \geq 1$. Set $C_1^ p = I_1^ p$. The complex $I^\bullet = \mathop{\mathrm{lim}}\nolimits I_ n^\bullet$ exists because we can take $I^ p = \prod _{n \geq 1} C_ n^ p$. Fix $p \in \mathbf{Z}$. We claim there is a split short exact sequence

$0 \to I^ p \to \prod I_ n^ p \to \prod I_ n^ p \to 0$

of objects of $\mathcal{A}$. Here the first map is given by the projection maps $I^ p \to I_ n^ p$ and the second map by $(x_ n) \mapsto (x_ n - f^ p_{n + 1}(x_{n + 1}))$ where $f^ p_ n : I_ n^ p \to I_{n - 1}^ p$ are the transition maps. The splitting comes from the map $\prod I_ n^ p \to \prod C_ n^ p = I^ p$. We obtain a termwise split short exact sequence of complexes

$0 \to I^\bullet \to \prod I_ n^\bullet \to \prod I_ n^\bullet \to 0$

Hence a corresponding distinguished triangle in $K(\mathcal{A})$ and $D(\mathcal{A})$. By Lemma 13.31.5 the products are K-injective and represent the corresponding products in $D(\mathcal{A})$. It follows that $I^\bullet$ represents $R\mathop{\mathrm{lim}}\nolimits I_ n^\bullet$ (Definition 13.34.1). Since $R\mathop{\mathrm{lim}}\nolimits I_ n^\bullet \cong R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet$ as derived limits are defined on the level of the derived category, we see that $I^\bullet$ represents $R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet$. Moreover, the complex $I^\bullet$ is K-injective by Lemma 13.31.3. By the commutative diagram of Lemma 13.29.3 and since $K^ i = (\tau _{\geq -n}K^\bullet )^ i$ for $n \gg 0$ we see that we get a unique map $\gamma : K^\bullet \to I^\bullet$ such that the diagrams

$\xymatrix{ K^\bullet \ar[r] \ar[d]_\gamma & \tau _{\geq -n} K^\bullet \ar[d] \\ I^\bullet \ar[r] & I_ n^\bullet }$

commute. It follows that $\gamma$ is a map of complexes which represents the map $c : K^\bullet \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet$ of Remark 13.34.4 in $D(\mathcal{A})$. In other words, the diagram

$\xymatrix{ K^\bullet \ar[r]_-c \ar[d]_\gamma & R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} K^\bullet \ar[d]^{\cong } \\ I^\bullet \ar[r]^-{\cong } & R\mathop{\mathrm{lim}}\nolimits I_ n^\bullet }$

is commutative in $D(\mathcal{A})$. The lemma follows. $\square$

Comment #2463 by anonymous on

In item (2) of the lemma it should be $\tau_{\geq -n}$ and not $\leq$.

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