Lemma 15.87.7. Let $(A_ n)$ be an inverse system of rings. Every $K \in D(\textit{Mod}(\mathbf{N}, (A_ n)))$ can be represented by a system of complexes $(M_ n^\bullet )$ such that all the transition maps $M_{n + 1}^\bullet \to M_ n^\bullet$ are surjective.

Proof. Let $K$ be represented by the system $(K_ n^\bullet )$. Set $M_1^\bullet = K_1^\bullet$. Suppose we have constructed surjective maps of complexes $M_ n^\bullet \to M_{n - 1}^\bullet \to \ldots \to M_1^\bullet$ and homotopy equivalences $\psi _ e : K_ e^\bullet \to M_ e^\bullet$ such that the diagrams

$\xymatrix{ K_{e + 1}^\bullet \ar[d] \ar[r] & K_ e^\bullet \ar[d] \\ M_{e + 1}^\bullet \ar[r] & M_ e^\bullet }$

commute for all $e < n$. Then we consider the diagram

$\xymatrix{ K_{n + 1}^\bullet \ar[r] & K_ n^\bullet \ar[d] \\ & M_ n^\bullet }$

By Derived Categories, Lemma 13.9.8 we can factor the composition $K_{n + 1}^\bullet \to M_ n^\bullet$ as $K_{n + 1}^\bullet \to M_{n + 1}^\bullet \to M_ n^\bullet$ such that the first arrow is a homotopy equivalence and the second a termwise split surjection. The lemma follows from this and induction. $\square$

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).