For simplify, let's just consider the following category first:

  • object: "spectral sequences", that is a sequence of pages $(E_r)_{r\ge0}$, where each page $E_r$ is a differential object in an abelian category $\mathcal{A}$, equipped with a sequences of isomorphisms $H(E_r)\cong E_{r+1}$.
  • morphisms: a sequences of morphisms of differential objects $u_r\colon E_r\to E'_r$ such that the following diagrams commute. \begin{array} HH(E_r) & \stackrel{H(u_r)}{\longrightarrow} & H(E'_r) \\ \downarrow{\cong} & & \downarrow{\cong} \\ E_{r+1} & \stackrel{u_{r+1}}{\longrightarrow} & E'_{r+1} \end{array}

My question is: what can we say about this category? Is it an abelian category? If $\mathcal{A}$ satisfies more properties such as AB3,AB4,AB5, will the above category satisfy? If $\mathcal{A}$ is an abelian tensor category, or furthermore cocomplete tensor category, rigid tensor category etc., what about the above category?

If answer to aboves are positive, how about the following category?

  • object: convergent spectral sequences equipped with target object. Here the bigraded version of spectral sequences may be needed to state the conditional convergences.
  • morphisms: a morphism between spectral sequences together with a morphism between their target objects satisfying the compatible conditions.

Update I feel that in general these cannot be true. Unless certain taking limit/colimit functor is exact, it shouldn't expect the limit/colimit of spectral sequences exists or have good properties. For example, consider kernels and cokernels.

  • $\begingroup$ The category of spectral sequences is not abelian. This is a remark in Grothendieck's Tohoku paper. It is naturally $\mathsf{Ab}$-enriched. I believe it has biproducts and a zero object, since homology behaves well with biproducts. It is probably not possible to define kernels and cokernels pointwise, because they are not exact and behave not well with homology. If someone has an answer to this question after so many years, I would like to hear it. $\endgroup$
    – Nico
    May 19, 2021 at 9:15

1 Answer 1


I do not know if this answers the question exactly, but results about this category and its homotopy theory (for spectral sequences of $R$-modules for a commutative ring $R$) appear in Livernet and Whitehouse's paper "Homotopy theory of spectral sequences". I will summarize a few of these results here for anyone who may stumble across this question. Any errors are my own, not the authors'.

In section 3.2, it is proven that the category of spectral sequences of $R$-modules is not pre-abelian. In particular, example 3.2.5 provides a morphism of spectral sequences with no cokernel and example 3.2.6 provides a morphism of spectral sequences with no kernel.

Since this category is neither complete nor cocomplete, it cannot be given the structure of a model category, so the localization of this category with respect to "$E_r$-quasi-isomorphisms" (definition 5.1.1) must be studied using other techniques. Section 4 introduces the notion of "almost Brown categories," which serve as the replacement which will allow the authors to study the homotopy theory of spectral sequences.

In section 5, the authors prove (theorem 5.3.1) that the category of spectral sequences equipped with the class of $E_r$-quasi-isomorphisms and $r$-fibrations (definition 5.1.1) is an almost Brown category with functorial factorization. This allows one to do some amount of homotopy theory with spectral sequences! Some additional results on generation of (acyclic) $r$-fibrations are also shown in section 5.

Section 6 compares the category of spectral sequences (together with the almost Brown structures above) to corresponding structures on filtered complexes and multicomplexes. To give a rough summary, the spectral sequence functor from either filtered complexes or multicomplexes preserves weak equivalences and is a left exact functor of almost Brown categories, and preserves $r$-homotopy. However, at least in the situation of filtered complexes, it never induces an equivalence of homotopy categories.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.