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.

  • $\begingroup$ If you contruct Rf(A) by constructing functorially a Cartan-Eilenberg resolution and using it to compute, one of the hypercohomology spectral sequences gives you those isomorphisms naturally, no? $\endgroup$ Oct 18, 2014 at 10:06
  • $\begingroup$ I do not see how the spectral sequence argument could imply that. $\endgroup$
    – asv
    Oct 18, 2014 at 10:09
  • $\begingroup$ Why do you think my answer (to the other question) is not functorial? The truncations $\tau_{\geq n}$ are certainly functorial, and a zig-zag of quasi-isomorphisms in $\mathrm{Ch}(\mathcal A)$ defines functorially a morphism in $D(\mathcal A)$. And as Mariano remarks, injective resolutions can be made functorial under mild hypotheses on $\mathcal A$ (certainly satisfied for sheaves on spaces). $\endgroup$ Oct 18, 2014 at 13:14
  • $\begingroup$ @DanPetersen: May be I am missing something. But morphisms between injective resolutions by bi-complexes are defined only up to homotopy of bi-complexes. I think that the vertical truncation is not compatible with such homotopies. Am I wrong? $\endgroup$
    – asv
    Oct 18, 2014 at 14:25

1 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.

  • $\begingroup$ I have one remark and one question. The remark is that the functor $j$ takes values in the category $K^+(\mathcal{B})$ of bounded below complexes up to homotopy. More precisely the object $j(A^\bullet)$ is defined up to an isomoprhism, unique up to homotopy. That implies that the composition $$j(A^m)\to j(A^{m+1})\to j(A^{m+2})$$ is not zero but only homotopic to zero. Hence $B^{\bullet,\bullet}$ is not a bi-complex. However this difficulty can be avoided if one uses the usual Cartan-Eilenberg resolution. But moprhisms of such resolutions are defined up to homotopy of bi-complexes only. $\endgroup$
    – asv
    Oct 19, 2014 at 8:18
  • $\begingroup$ The question is as follows. The morphism of bi-complexes $\beta\colon B_1^{\bullet,\bullet}\to B_2^{\bullet,\bullet}$ is defined only up to homotopy of bi-complexes. In the answer to the previous post it was suggested to use a vertical truncation of bi-complexes. This solves that question, but apparently the vertical truncation is not compatible with homotopies: namely a homotopy between two morphism of bi-complexes does not induce homotopy of vertically truncated bi-complexes (am I wrong?). On the other hand I do not see how to truncate $Tot(B^{\bullet,\bullet})$ to get the required complex. $\endgroup$
    – asv
    Oct 19, 2014 at 8:32
  • 1
    $\begingroup$ I don't think $j$ necessary takes values in $K^+(\mathcal B)$. Look at stacks.math.columbia.edu/tag/0140 $\endgroup$ Oct 19, 2014 at 9:13
  • $\begingroup$ @DanPetersen: Thanks for the reference! This indeed makes my remark unnecessary. However I still do not see if it resolves my question. $\endgroup$
    – asv
    Oct 19, 2014 at 9:30
  • $\begingroup$ I think I understood the argument of Vlad the Impaler combined with the comment of Dan Petersen. This is very nice and does answer my question. The minor point is whether this functorial isomorphism depends on the choice of resolution functor j. $\endgroup$
    – asv
    Oct 19, 2014 at 16:23

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