**Proof.**
By Theorem 19.12.6 we obtain quasi-isomorphisms $i : K^\bullet \to I^\bullet $ and $i^ p : F^ pK^\bullet \to I^{p, \bullet }$ as well as commutative diagrams

\[ \vcenter { \xymatrix{ K^\bullet \ar[d]_ i & F^ pK^\bullet \ar[l] \ar[d]_{i^ p} \\ I^\bullet & I^{p, \bullet } \ar[l]_{\alpha ^ p} } } \quad \text{and}\quad \vcenter { \xymatrix{ F^{p'}K^\bullet \ar[d]_{i^{p'}} & F^ pK^\bullet \ar[l] \ar[d]_{i^ p} \\ I^{p', \bullet } & I^{p, \bullet } \ar[l]_{\alpha ^{p p'}} } } \quad \text{for }p' \leq p \]

such that $\alpha ^ p \circ \alpha ^{p' p} = \alpha ^{p'}$ and $\alpha ^{p'p''} \circ \alpha ^{pp'} = \alpha ^{pp''}$. The problem is that the maps $\alpha ^ p : I^{p, \bullet } \to I^\bullet $ need not be injective. For each $p$ we choose an injection $t^ p : I^{p, \bullet } \to J^{p, \bullet }$ into an acyclic K-injective complex $J^{p, \bullet }$ whose terms are injective objects of $\mathcal{A}$ (first map to the cone on the identity and then use the theorem). Choose a map of complexes $s^ p : I^\bullet \to J^{p, \bullet }$ such that the following diagram commutes

\[ \xymatrix{ K^\bullet \ar[d]_ i & F^ pK^\bullet \ar[l] \ar[d]_{i^ p} \\ I^\bullet \ar[rd]_{s^ p} & I^{p, \bullet } \ar[d]^{t^ p} \\ & J^{p, \bullet } } \]

This is possible: the composition $F^ pK^\bullet \to J^{p, \bullet }$ is homotopic to zero because $J^{p, \bullet }$ is acyclic and K-injective (Derived Categories, Lemma 13.31.2). Since the objects $J^{p, n - 1}$ are injective and since $F^ pK^ n \to K^ n \to I^ n$ are injective morphisms, we can lift the maps $F^ pK^ n \to J^{p, n - 1}$ giving the homotopy to a map $h^ n : I^ n \to J^{p, n - 1}$. Then we set $s^ p$ equal to $h \circ \text{d} + \text{d} \circ h$. (Warning: It will not be the case that $t^ p = s^ p \circ \alpha ^ p$, so we have to be careful not to use this below.)

Consider

\[ J^\bullet = I^\bullet \times \prod \nolimits _ p J^{p, \bullet } \]

Because products in $D(\mathcal{A})$ are given by taking products of K-injective complexes (Lemma 19.13.4) and since $J^{p, \bullet }$ is isomorphic to $0$ in $D(\mathcal{A})$ we see that $J^\bullet \to I^\bullet $ is an isomorphism in $D(\mathcal{A})$. Consider the map

\[ j = i \times (s^ p \circ i)_{p \in \mathbf{Z}} : K^\bullet \longrightarrow I^\bullet \times \prod \nolimits _ p J^{p, \bullet } = J^\bullet \]

By our remarks above this is a quasi-isomorphism. It is also injective. For $p \in \mathbf{Z}$ we let $F^ pJ^\bullet \subset J^\bullet $ be

\[ \mathop{\mathrm{Im}}\left( \alpha ^ p \times (t^{p'} \circ \alpha ^{pp'})_{p' \leq p} : I^{p, \bullet } \to I^\bullet \times \prod \nolimits _{p' \leq p} J^{p', \bullet } \right) \times \prod \nolimits _{p' > p} J^{p', \bullet } \]

This complex is isomorphic to the complex $I^{p, \bullet } \times \prod _{p' > p} J^{p, \bullet }$ as $\alpha ^{pp} = \text{id}$ and $t^ p$ is injective. Hence $F^ pJ^\bullet $ is quasi-isomorphic to $I^{p, \bullet }$ (argue as above). We have $j(F^ pK^\bullet ) \subset F^ pJ^\bullet $ because of the commutativity of the diagram above. The corresponding map of complexes $F^ pK^\bullet \to F^ pJ^\bullet $ is a quasi-isomorphism by what we just said. Finally, to see that $F^{p + 1}J^\bullet \subset F^ pJ^\bullet $ use that $\alpha ^{p + 1p} \circ \alpha ^{pp'} = \alpha ^{p + 1p'}$ and the commutativity of the first displayed diagram in the first paragraph of the proof.

We claim that $j : K^\bullet \to J^\bullet $ is a solution to the problem posed by the lemma. Namely, $F^ pJ^ n$ is an injective object of $\mathcal{A}$ because it is isomorphic to $I^{p, n} \times \prod _{p' > p} J^{p', n}$ and products of injectives are injective. Then the injective map $F^ pJ^ n \to J^ n$ splits and hence the quotient $J^ n/F^ pJ^ n$ is injective as well as a direct summand of the injective object $J^ n$. Similarly for $F^ pJ^ n/F^{p'}J^ n$. This in particular means that $0 \to F^ pJ^\bullet \to J^\bullet \to J^\bullet /F^ pJ^\bullet \to 0$ is a termwise split short exact sequence of complexes, hence defines a distinguished triangle in $K(\mathcal{A})$ by fiat. Since $J^\bullet $ and $F^ pJ^\bullet $ are K-injective complexes we see that the same is true for $J^\bullet /F^ pJ^\bullet $ by Derived Categories, Lemma 13.31.3. A similar argument shows that $F^ pJ^\bullet /F^{p'}J^\bullet $ is K-injective. By construction $j : K^\bullet \to J^\bullet $ and the induced maps $F^ pK^\bullet \to F^ pJ^\bullet $ are quasi-isomorphisms. Using the long exact cohomology sequences of the complexes in play we find that the same holds for $K^\bullet /F^ pK^\bullet \to J^\bullet /F^ pJ^\bullet $ and $F^ pK^\bullet /F^{p'}K^\bullet \to F^ pJ^\bullet /F^{p'}J^\bullet $.
$\square$

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