**Proof.**
By Theorem 19.12.6 we can choose a K-injective complex $I^\bullet $ representing $E$ all of whose terms $I^ n$ are injective objects of $\mathcal{A}$. Choose a complex $G^{0, \bullet }$ representing $E^0$. Choose a map of complexes $\varphi ^0 : G^{0, \bullet } \to I^\bullet $ representing $E^0 \to E$. For $i > 0$ we inductively represent $E^ i \to E^{i - 1}$ by a map of complexes $\delta : G^{i, \bullet } \to G^{i - 1, \bullet }$ and we set $\varphi ^ i = \delta \circ \varphi ^{i - 1}$. For $i < 0$ we inductively represent $E^{i + 1} \to E^ i$ by a termwise injective map of complexes $\delta : G^{i + 1, \bullet } \to G^{i, \bullet }$ (for example you can use Derived Categories, Lemma 13.9.6). Claim: we can find a map of complexes $\varphi ^ i : G^{i, \bullet } \to I^\bullet $ representing the map $E^ i \to E$ and fitting into the commutative diagram

\[ \xymatrix{ G^{i + 1, \bullet } \ar[r]_\delta \ar[d]_{\varphi ^{i + 1}} & G^{i, \bullet } \ar[ld]^{\varphi ^ i} \\ I^\bullet } \]

Namely, we first choose any map of complexes $\varphi : G^{i, \bullet } \to I^\bullet $ representing the map $E^ i \to E$. Then we see that $\varphi \circ \delta $ and $\varphi ^{i + 1}$ are homotopic by some homotopy $h^ p : G^{i + 1, p} \to I^{p - 1}$. Since the terms of $I^\bullet $ are injective and since $\delta $ is termwise injective, we can lift $h^ p$ to $(h')^ p : G^{i, p} \to I^{p - 1}$. Then we set $\varphi ^ i = \varphi + h' \circ d + d \circ h'$ and we get what we claimed.

Next, we choose for every $i$ a termwise injective map of complexes $a^ i : G^{i, \bullet } \to J^{i, \bullet }$ with $J^{i, \bullet }$ acyclic, K-injective, with $J^{i, p}$ injective objects of $\mathcal{A}$. To do this first map $G^{i, \bullet }$ to the cone on the identity and then apply the theorem cited above. Arguing as above we can find maps of complexes $\delta ' : J^{i, \bullet } \to J^{i - 1, \bullet }$ such that the diagrams

\[ \xymatrix{ G^{i, \bullet } \ar[r]_\delta \ar[d]_{a^ i} & G^{i - 1, \bullet } \ar[d]^{a^{i - 1}} \\ J^{i, \bullet } \ar[r]^{\delta '} & J^{i - 1, \bullet } } \]

commute. (You could also use the functoriality of cones plus the functoriality in the theorem to get this.) Then we consider the maps

\[ \xymatrix{ G^{i + 1, \bullet } \times \prod \nolimits _{p > i + 1} J^{p, \bullet } \ar[r] \ar[rd] & G^{i, \bullet } \times \prod \nolimits _{p > i} J^{p, \bullet } \ar[r] \ar[d] & G^{i - 1, \bullet } \times \prod \nolimits _{p > i - 1} J^{p, \bullet } \ar[ld] \\ & I^\bullet \times \prod \nolimits _ p J^{p, \bullet } } \]

Here the arrows on $J^{p, \bullet }$ are the obvious ones (identity or zero). On the factor $G^{i, \bullet }$ we use $\delta : G^{i, \bullet } \to G^{i - 1, \bullet }$, the map $\varphi ^ i : G^{i, \bullet } \to I^\bullet $, the zero map $0 : G^{i, \bullet } \to J^{p, \bullet }$ for $p > i$, the map $a^ i : G^{i, \bullet } \to J^{p, \bullet }$ for $p = i$, and $(\delta ')^{i - p} \circ a^ i = a^ p \circ \delta ^{i - p} : G^{i, \bullet } \to J^{p, \bullet }$ for $p < i$. We omit the verification that all the arrows in the diagram are termwise injective. Thus we obtain a filtered complex. Because products in $D(\mathcal{A})$ are given by taking products of K-injective complexes (Lemma 19.13.4) and because $J^{p, \bullet }$ is zero in $D(\mathcal{A})$ we conclude this diagram represents the given diagram in the derived category. This finishes the proof.
$\square$

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