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Let $\mathcal{A}$ be a monoidal abelian category. Let $A$ and $B$ be simplicial objects in $\mathcal{A}$, and let $N_\ast(-)$ denote the normalized chain complex functor. Let $$AW_{A,B}\colon N_\ast(A\otimes B) \longrightarrow N_\ast(A)\otimes N_\ast(B)$$ and $$ EZ_{A,B}\colon N_\ast(A)\otimes N_\ast(B) \longrightarrow N_\ast(A\otimes B)$$ denote the Alexander-Whitney map and the Eilenberg-Zilber map respectively. Then on homology $AW_{A,B}$ and $EZ_{A,B}$ induce inverse isomorphisms. On the chain level they satisfy $$AW_{A,B}\circ EZ_{A,B} = Id_{N_\ast(A)\otimes N_\ast(B)}$$ and, at least in the case $\mathcal{A}=\mathrm{Ab}$, $$EZ_{A,B}\circ AW_{A,B}\sim Id_{N_\ast(A\otimes B)}.$$ Here $\sim$ denotes chain homotopy.

Is it true that $EZ_{A,B}\circ AW_{A,B}\sim Id_{N_\ast(A\otimes B)}$ for an arbitrary monoidal abelian category $\mathcal{A}?$ If so, does the result appear in the literature, or can a proof be quickly reconstructed from the literature?

Remark: What is most important for me is to know (1) how general $\mathcal{A}$ can be in order for the answer to be "yes", and to know (2) that there is a chain homotopy $EZ_{A,B}\circ AW_{A,B}\sim Id_{N_\ast(A\otimes B)}$, and not just that $AW_{A,B}$ and $EZ_{A,B}$ induce inverse quasi-isomorphisms.

Some more information: In the case $\mathcal{A}=\mathrm{Ab}$, the definition of $AW_{A,B}$ and $EZ_{A,B}$ can be found in Definitions 29.7 of Peter May's book Simplicial objects in algebraic topology and on the nLab page monoidal Dold-Kan correspondence. The definitions are the same for an arbitrary choice of $\mathcal{A}.$ Corollary 29.10 of May's book proves $AW_{A,B}\circ EZ_{A,B} = Id_{N_\ast(A)\otimes N_\ast(B)}$ and $EZ_{A,B}\circ AW_{A,B}\sim Id_{N_\ast(A\otimes B)}$, again in the case $\mathcal{A}=\mathrm{Ab}$. Section 8.5 of Weibel's book An introduction to homological algebra proves that the Alexander-Whitney and Eilenberg-Zilber maps are inverse quasi-isomorphisms, this time for an arbitrary abelian category $\mathcal{A}$, and in a more general bisimplicial setting. §1.2.3 in Lurie's Higher Algebra shows that $AW_{A,B}$ is a quasi-isomorphism in a somewhat more general setting. Are there Alexander-Whitney and shuffle maps for Dold-Kan for abelian categories? is a different mathoverflow question with a similar title to mine.

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  • $\begingroup$ This is surely true assuming the correct hypotheses on $\mathcal{A}$, which I suspect are: additive category with retracts equipped with biadditive monoidal structure. The key point to keep in mind is that the proofs that produce a chain homotopy $EZ\circ AW\sim Id$ actually show the existence of a natural chain homotopy, i.e., a collection of transformations $H_k\colon N(A\otimes B)_k\to N(A\otimes B)_{k+1}$ which are natural in $(A,B)$ ... $\endgroup$
    – Charles Rezk
    Commented Jun 28, 2018 at 21:41
  • $\begingroup$ When $\mathcal{A}$ is abelian groups, this condition means that $H_k$ must have a formula of the form: $a\otimes b\mapsto \sum_{f,g} m_{f,g}\; (af)\otimes (ag)$, where $f,g\colon [k+1]\to [k]$ are simplicial operators (which I think of as acting on simplices from the right), and $m_{f,g}$ is an integer. This is because you can classify natural transformations $A_k\otimes B_k\to A_{k+1}\otimes B_{k+1}$, and they all have this form .... $\endgroup$
    – Charles Rezk
    Commented Jun 28, 2018 at 21:44
  • $\begingroup$ (The terms of the normalized complex are natural summands of these objects, split off by idempotents which can be written as linear combinations of simplicial operators, so there is a similar statement for natural transformations $N(A\otimes B)_k\to N(A\otimes B)_{k+1}$.) So there is an explicit formula for abelian groups, but which clearly is meaingful in a general context, and very likely proves what you want ... $\endgroup$
    – Charles Rezk
    Commented Jun 28, 2018 at 21:46
  • $\begingroup$ It would be nice if somebody with time on their hands would produce an explicit formula for $H_k$, I doubt it is very difficult. $\endgroup$
    – Charles Rezk
    Commented Jun 28, 2018 at 21:47
  • $\begingroup$ @Charles Thanks very much for sketching this answer! $\endgroup$
    – Richard Hepworth
    Commented Jun 29, 2018 at 10:53

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