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2 Typos

OK, I still doubt that if you divide out chain homotopies in (1) in the category of filtered complexcomplexes, you can get the category of complexes (and not the homotopy category) as a target in (2). I also doubt that you get an equivalence in this way unless you invert something in source and target. Sorry that I'm too lazy to look now at BBD.

Suppose that for any bounded complex $X$ in $\mathcal{P}$ you can construct a filtered complex of injectives $F$ in $\mathcal{A}$ such that each $F^{n}/F^{n+1}$ is an injective object in $\mathcal{P}$ and $gr^{\ast}(F)=F^{\ast}/F^{\ast+1}$ equipped with the differential $\delta$ obtained as in (2) is a bounded below complex quasi-isomorphic to $X$ in $C(\mathcal{P})$. You can probably check in BBD that all this is possible (at leas under reasonable assumptions).

Consider the following zig-zag of DG-algebra morphisms:

$\operatorname{End}_{C(\mathcal{P})}(gr^{\ast}(F),\delta)\leftarrow \operatorname{End}_{\mbox{filtered}}(F)\rightarrow \operatorname{End}_{C(\mathcal{A})}(F)$

By definition, the $\operatorname{Ext}$ algebra of $X$ is the cohomology of the DG-algebra on the left. According to (3), the $\operatorname{Ext}$ algebra of $real(X)$ is the cohomology of the DG-algebra on the right. Moreover, the $A$-infinity structures on these $\operatorname{Ext}$ algebras are obtained from these DG-algebras by the standard transfer procedure, choosing bases of cohomology vector spaces and representing cocycles for the elements in these bases. By definition, the transfer maps are $A$-infinity quasi-isomorphisms between the $\operatorname{Ext}$ $A$-infinity algebras and these DG-algebras.

Your claim in (2) should instead say, I think, that the functor you sketch induces an equivalence after inverting quasi-isomorphisms on the right, and something appropriate on the left (filtered quasi-isos?). Therefore $\leftarrow$ above is a quasi-isomorphism. Moreover, since you assume that $real$ is fully faithful, $\rightarrow$ above is also a quasi-isomorphism. Now you can get your desired $A$-infinity quasi-isomorphism between the $\operatorname{Ext}$ $A$-fininity algebras by inverting and composing.

1

OK, I still doubt that if you divide out chain homotopies in (1) in the category of filtered complex, you can get the category of complexes (and not the homotopy category) as a target in (2). I also doubt that you get an equivalence in this way unless you invert something in source and target. Sorry that I'm too lazy to look now at BBD.

Suppose that for any bounded complex $X$ in $\mathcal{P}$ you can construct a filtered complex of injectives $F$ in $\mathcal{A}$ such that each $F^{n}/F^{n+1}$ is an injective object in $\mathcal{P}$ and $gr^{\ast}(F)=F^{\ast}/F^{\ast+1}$ equipped with the differential $\delta$ obtained as in (2) is a bounded below complex quasi-isomorphic to $X$ in $C(\mathcal{P})$. You can probably check in BBD that all this is possible (at leas under reasonable assumptions).

Consider the following zig-zag of DG-algebra morphisms:

$\operatorname{End}_{C(\mathcal{P})}(gr^{\ast}(F),\delta)\leftarrow \operatorname{End}_{\mbox{filtered}}(F)\rightarrow \operatorname{End}_{C(\mathcal{A})}(F)$

By definition, the $\operatorname{Ext}$ algebra of $X$ is the cohomology of the DG-algebra on the left. According to (3), the $\operatorname{Ext}$ algebra of $real(X)$ is the cohomology of the DG-algebra on the right. Moreover, the $A$-infinity structures on these $\operatorname{Ext}$ algebras are obtained from these DG-algebras by the standard transfer procedure, choosing bases of cohomology vector spaces and representing cocycles for the elements in these bases. By definition, the transfer maps are $A$-infinity quasi-isomorphisms between the $\operatorname{Ext}$ $A$-infinity algebras and these DG-algebras.

Your claim in (2) should instead say, I think, that the functor you sketch induces an equivalence after inverting quasi-isomorphisms on the right, and something appropriate on the left (filtered quasi-isos?). Therefore $\leftarrow$ above is a quasi-isomorphism. Moreover, since you assume that $real$ is fully faithful, $\rightarrow$ above is also a quasi-isomorphism. Now you can get your desired $A$-infinity quasi-isomorphism between the $\operatorname{Ext}$ $A$-fininity algebras by inverting and composing.