The Stacks project

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

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

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

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

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