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 $\text{Tor}_A(k,k)$ is a bicommutative Hopf algebra. Is this a typo? Obviously, $\text{Tor}$ has a commutative algebra structure, but is it true that the coaddition is cocommutative?
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2$\begingroup$ What is the comultiplication? To get the multiplication we apply $\operatorname{Tor}_A(-,k)$ to $k \otimes k \rightarrow k$ and then precompose with the map $\operatorname{Tor}_A (k,k) \otimes \operatorname{Tor}_A (k,k) \rightarrow \operatorname{Tor}_A(k \otimes k,k)$. But to get the comultiplication it seems like we can't play the same game with $k \rightarrow k \otimes k$ since the canonical map $\operatorname{Tor}_A (k,k) \otimes \operatorname{Tor}_A (k,k) \rightarrow \operatorname{Tor}_A(k \otimes k,k)$ is in the wrong direction. $\endgroup$– Connor MalinCommented Apr 13, 2020 at 17:21
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2$\begingroup$ Proposition 7.10 gives the comultiplication: $[a_1,\dots, a_n] \rightarrow \Sigma [a_1, \dots, a_j] \otimes [a_{j+1},\dots,a_n]$. $\endgroup$– Connor MalinCommented Apr 13, 2020 at 17:56
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This is not true. Consider the algebra $A=T(V)/V^{\otimes 2}$, it is a commutative algebra whose augmentation ideal has zero multiplication. We have $\mathrm{Tor}_A(k,k)\cong T(V[1])$ 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 $V[1]$).
answered Apr 13, 2020 at 19:03
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$\begingroup$ 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? $\endgroup$ Commented Apr 13, 2020 at 19:19
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$\begingroup$ @ConnorMalin I can't guess your thinking process but one possibility is that you are confusing the concatenation product with the shuffle product? $\endgroup$ Commented Apr 13, 2020 at 19:30