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Skip to the bottom for my questions, first some discussion:

It is a celebrated theorem of Kadeišvili that $A_{\infty}$-algebra structures can be transferred along homotopy equivalences so that the homotopy equivalences are $A_{\infty}$ maps. There are some results in this direction where 'homotopy equivalence' is replaced by 'quasi-isomorphism', for example Burke shows in Transfer of A-infinity structures to projective resolutions that under some mild hypotheses if $R$ is a $Q$-algebra then a projective $Q$-resolution $P_{\bullet}$ of $R$ has an $A_{\infty}$-algebra structure so that the map $P_{\bullet} \to R$ is $A_{\infty}$.

I am interested in these types of results for $E_{\infty}$-algebras in chain complexes, and after looking at the literature I am a little confused. First, I believe that there can't be such a general result for $E_{\infty}$-algebras because if $X$ is any space then the singular cochain complex $S^{*}(X,k)$ over a field $k$ is an $E_{\infty}$-algebra that is homotopy equivalent to its cohomology algebra $H^*(X,k)$ (which as a graded commutative ring is also an $E_{\infty}$-algebra). The homotopy equivalence $S^{*}(X,k) \to H^*(X,k)$ can't be an $E_{\infty}$ map because of the action of the Steenrod squares-- the homology of any $E_{\infty}$-algebra has Steenrod operations, but all operations except for $Sq^j$ on classes of degree $j$ are zero for a strictly commutative algebra.

However, I have seen more general theorems that appear to give as a special case that $E_{\infty}$-algebra structures can always be transferred along quasi-isomorphisms. For example, in Axiomatic Homotopy Theory for Operads, Theorem 3.5, I understand Berger and Moerdijk to state that :

If $f:X \to Y$ is a weak equivalence between bifibrant objects in a monoidal model category $\mathcal{M}$ where the operads carry a transferred model structure and $\mathcal{P}$ is a cofibrant operad, then any $\mathcal{P}$- algebra structure on $X$ (resp. $Y$) induces a $\mathcal{P}$-algebra structure on $Y$ (resp. $X$) in such a way that $f$ preserves the $\mathcal{P}$-algebra structures up to homotopy.

(In the topological setting this theorem is the 'Homotopy Invariance Property' of Boardmann and Vogt)

This theorem seems to be in contradiction with the map $S^{\bullet}(X,k) \to H^*(X,k)$ not being $E_{\infty}$--I believe that chain complexes of $k$-modules with the projective model structure for $\mathcal{M}$ and an $E_{\infty}$ operad for $\mathcal{P}$ satisfy the necessary hypotheses.

So, my questions are:

Question One: Working with chain complexes of $Q$-modules (where $Q$ is a commutative ring), given a quasi-isomorphism $X \to Y$ under what conditions does an $E_{\infty}$-algebra structure on $Y$ lift to a compatible one on $X$?

Question Two: If you cannot always compatibly lift an $E_{\infty}$-algebra structure, what is lacking with chain complexes and $E_{\infty}$ operads in order to apply the homotopy invariance property?

Question Three: If you can always lift an $E_{\infty}$-algebra structure, how does this work for the homotopy equivalence between the singular chain complex of a space and its cohomology?

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    $\begingroup$ So $H^*(X;k)$ can get two $E_\infty$ structures. One is transferred from $C^*(X;k)$. The other is the trivial one, coming from the fact that $H^*(X;k)$ is a graded-commutative ring. The issue (which one also finds in the $A_\infty$-vs-associative case) is that these two aren't equivalent $E_\infty$ structures even though they have the same binary product. The Steenrod squares can be defined in terms of operations (extra data, beyond the product) that vanish for the second structure but not the first. $\endgroup$ Commented Jan 21, 2022 at 20:13
  • $\begingroup$ @TylerLawson That's really interesting, thank you! I didn't know a strictly commutative object could have an interesting $E_{\infty}$ structure, but I suppose I shouldn't be surprised since the same thing happens in the $A_{\infty}$ case as you say. Do you know if anyone has done computations of the interesting $E_{\infty}$ structure for $H^*(X)$ for particular spaces? I assume that this is more information than just the Steenrod algebra structure? Also, do you know what you get if you transfer the trivial $E_{\infty}$ structure on $H^*(X)$ to $C^*(X,k)$? $\endgroup$
    – J Cameron
    Commented Jan 21, 2022 at 21:45

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