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Let A be a commutative (or graded commutative) algebra over a field k. In some sources, such as Mcleary's book on spectral sequences, Corollary 7.12, pg. 248, it is claimed that TorA(k,k) is a bicommutative Hopf algebra. Is this a typo? Obviously, Tor has a commutative algebra structure, but is it true that the coaddition is cocommutative?

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    What is the comultiplication? To get the multiplication we apply TorA(,k) to kkk and then precompose with the map TorA(k,k)TorA(k,k)TorA(kk,k). But to get the comultiplication it seems like we can't play the same game with kkk since the canonical map TorA(k,k)TorA(k,k)TorA(kk,k) is in the wrong direction. Commented Apr 13, 2020 at 17:21
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    Proposition 7.10 gives the comultiplication: [a1,,an]Σ[a1,,aj][aj+1,,an]. Commented Apr 13, 2020 at 17:56

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This is not true. Consider the algebra A=T(V)/V2, it is a commutative algebra whose augmentation ideal has zero multiplication. We have TorA(k,k)T(V[1]) with the shuffle product and deconcatenation coproduct, so the coproduct is very much noncommutative (it is the coproduct of the cofree conilpotent coalgebra co-generated by V[1]).

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  • To me, this coproduct looks like the dual of something like the cup product which makes me think it should be cocommutative. Is this completely wrongheaded? Commented Apr 13, 2020 at 19:19
  • @ConnorMalin I can't guess your thinking process but one possibility is that you are confusing the concatenation product with the shuffle product? Commented Apr 13, 2020 at 19:30

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