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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