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Jun 29, 2018 at 10:53 comment added Richard Hepworth @Charles Thanks very much for sketching this answer!
Jun 28, 2018 at 21:47 comment added Charles Rezk 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.
Jun 28, 2018 at 21:46 comment added Charles Rezk (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 ...
Jun 28, 2018 at 21:44 comment added Charles Rezk 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 ....
Jun 28, 2018 at 21:41 comment added Charles Rezk 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 $(A,B)$ ...
Jun 28, 2018 at 14:51 review First posts
Jun 28, 2018 at 15:15
Jun 28, 2018 at 14:50 history asked Richard Hepworth CC BY-SA 4.0