A functorial isomorphism in derived category

Question

This question is a direct continuation of Question 1 in this post: Two basic questions on derived categories

Let $f\colon \mathcal{A}\to\mathcal{B}$ be a left exact functor between two abelian categories with sufficiently many injective objects (say push-forward on sheaves of vector spaces). Let $A^\bullet$ be a bounded from below complex in $\mathcal{A}$ such that for every $p$ $$R^qf(A^p)=0 \mbox{ unless } q\ne a,\,\,\,\,\,(\ast)$$ where $a$ is a fixed number. Let $Rf\colon D^+(\mathcal{A})\to D^+(\mathcal{B})$ be the derived functor.

Question. Does there exist a $\underline{functorial}$ isomorphism of objects in $D^+(\mathcal{B})$: of $Rf(A^\bullet)$ and of the complex $$[\dots\to R^af(A^i)\to R^af(A^{i+1})\to\dots],$$ where the differentials are induced by the differentials in $A^\bullet$? (The functoriality should be understood on the category of complexes with the above vanishing conditions $(\ast)$ for a fixed $a$.)

Remark. In an answer to the above mentioned post it was shown that the two objects are indeed isomoprhic in $D^+(\mathcal{B})$, but it seems to me that the isomorpism there is not functorial. Please correct me if I am wrong.

Answer 1

Please do not accept this as the answer as I think the answer was already given in the comments above. What I like about this question is that I do not know how to prove this without using double complexes. Does anybody? Anyway, I think the question becomes easier to understand if you formulate a stronger version:

Let $A^\bullet$ be a bounded below complex. Can one functorially compute $Rf(A^\bullet)$ as the total complex associated to a double complex, $B^{\bullet, \bullet}$ such that $B^{n, \bullet}$ is functorially isomorphic to $Rf(A^n)$ compatible with the maps $Rf(A^n) \to Rf(A^{n + 1})$ and the maps of complexes $B^{n, \bullet} \to B^{n + 1, \bullet}$? (OK, this formulation can still be made more precise, but I leave it up to you to do so.)

And this is basically exactly what Cartan-Eilenberg resolutions do for you (please read up on this). But instead of doing so, let's use a functorial injective resolution functor $j$ as suggested in the comments. This means that for every object $A$ there is a morphism $A[0] \to j(A)^\bullet$ which is functorial in $A$ as well as being an injective resolution. Such a functor $j$ exists as soon as you have functorial injective embeddings which happens for any Grothendieck abelian category, for example the category of sheaves of modules on a ringed space. Given $j$ we take $B^{n, m} = f(j(A^n)^m)$. To prove that this works (i.e., that this really does compute $Rf$; there is also the question here of how you define $Rf$ in the first place, which I am going to ignore), you apply a spectral sequence argument. Moreover, if you do it this way, then the whole $B^{\bullet, \bullet}$ thing is functorial on the category of bounded below complexes.

Having said all of the above, suppose that we have two bounded below complexes $A_i^\bullet$, $i = 1, 2$ and a map of complexes $\alpha : A_1^\bullet \to A_2^\bullet$ of complexes. Moreover, assume that we have $R^qf(A_i^n) = 0$ for $q \not = a$. Using the construction above we obtain two double complexes $B_i^{\bullet, \bullet}$ and we obtain a map of double complexes $\beta : B_1^{\bullet, \bullet} \to B_2^{\bullet, \bullet}$. Moreover, the associated map $Tot(\beta)$ on the associated total complexes, computes $Rf(\alpha)$ by construction. Since $\beta$ is a map of double complexes we see that it is compatible with truncations, etc, and it shows that Petersen's thing works fine.