Lemma 19.13.9. In the situation of Lemma 19.13.8 assume we have a second inverse system $\{ (E')^ i\} _{i \in \mathbf{Z}}$ and a compatible system of maps $(E')^ i \to E$. Then there exists a bi-filtered complex $K^\bullet $ of $\mathcal{A}$ such that $K^\bullet $ represents $E$, $F^ iK^\bullet $ represents $E^ i$, and $(F')^ iK^\bullet $ represents $(E')^ i$ compatibly with the given maps.

**Proof.**
Using the lemma we can first choose $K^\bullet $ and $F$. Then we can choose $(K')^\bullet $ and $F'$ which work for $\{ (E')^ i\} _{i \in \mathbf{Z}}$ and the maps $(E')^ i \to E$. Using Lemma 19.13.7 we can assume $K^\bullet $ is a K-injective complex. Then we can choose a map of complexes $(K')^\bullet \to K^\bullet $ corresponding to the given identifications $(K')^\bullet \cong E \cong K^\bullet $. We can additionally choose a termwise injective map $(K')^\bullet \to J^\bullet $ with $J^\bullet $ acyclic and K-injective. (To do this first map $(K')^\bullet $ to the cone on the identity and then apply Theorem 19.12.6.) Then $(K')^\bullet \to K^\bullet \times J^\bullet $ and $K^\bullet \to K^\bullet \times J^\bullet $ are both termwise injective and quasi-isomorphisms (as the product represents $E$ by Lemma 19.13.4). Then we can simply take the images of the filtrations on $K^\bullet $ and $(K')^\bullet $ under these maps to conclude.
$\square$

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