I believe that there isn't a commutative model for the DGA of cochains on a space, because of cohomology operations. This question has some nice explanations of why this is so. For example, there is a way of constructing Steenrod operations on the cohomology of any $E_{\infty}$ DGA over $\mathbb{F}_p$, and if this DGA is strictly graded commutative then most of the Steenrod operations, including $Sq^0$, vanish.

However, I recently became aware of a result of Guggenheim and May (Differential Torsion Products, Theorem 4.1) which states that for any commutative ring $R$, there is a quasi-isomorphism of DGAs $g:C^* (BT,R) \to H^*(BT,R)$, where $T$ is a torus.

Question 1 Why does this result not contradict that there isn't a commutative model for cochains on a space?

I can guess what the answer will be: by "model" I should mean an equivalent $E_{\infty}$ DGA, and Guggenheim and May state that this equivalence $g$ kills all $\smile_1$ products, showing that this $g$ is not a map of $E_{\infty}$ DGAs, but I would appreciate more detail, especially in regards to why this fact about $\smile_1$ products shows that the equivalence is not $E_{\infty}$.

Question 2 This is a softer question: what, if anything, do I lose if I replace $C^*(BT,\mathbb{F}_p)$ with $H^*(BT,\mathbb{F}_p)$?

Of course this depends on what I am trying to do; I am asking this question in the context of trying to understand the singular Cartan model for torus-equivariant cohomology, but I would appreciate any wisdom.

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    $\begingroup$ From a highly categorical point of view, being commutative is not a property but a structure. Cochains form an $E_\infty$-algebra, and also cohomology in a rather trivial way, but the Guggenheim-May quasi-isomorphism is only an $A_\infty$-map, not compatible with the $E_\infty$ structures. Both $E_\infty$-algebras are not quasi-isomorphic as such, the Steenrod operations argument works. Concerning your second question, what you loose is all that higher structure. $\endgroup$ – Fernando Muro Feb 15 '18 at 9:16
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    $\begingroup$ The $\cup_i$ products encode (a lot of) the $E_{\infty}$ structure. In particular, $sq^i(x):=x\cup_{\vert x\vert -i} x$. So if a map doesn't respect $\cup_i$ products then it can not respect Steenrod operations. Also, $sq^0$ is nonzero on degree $0$ elements of a genuinely commutative DGA, right? $\endgroup$ – Sean Tilson Feb 15 '18 at 11:12
  • $\begingroup$ That helps, thanks. @SeanTilson yes, it is nonzero on degree zero elements, I guess it sends them to the identity. $\endgroup$ – James Cameron Feb 15 '18 at 14:09
  • $\begingroup$ I think it is the frobenous in degree 0,because it is the cup square. Do you know about Dyer-Lashof operations? Those might be a better perspective... $\endgroup$ – Sean Tilson Feb 16 '18 at 15:06
  • $\begingroup$ @SeanTilson Yes, I was confused. Thanks for the pointer to Bob Bruner's adams spectral sequence book! $\endgroup$ – James Cameron Feb 16 '18 at 22:24

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