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Space-level homotopy induces (co)chain homotopy, but I've never heard of the converse. I am not sure if it is simply not true?

However, for good enough spaces (finite type nilpotent), Mandell proved that weak homotopy types are determined by cochain complexes as an $E_\infty$-algebras up to quasi-isomorphism. Does this suggest that there are (co)chain homotopies in the sense of $E_\infty$, such that all of them backward induce homotopies between maps between spaces?

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  • $\begingroup$ I read this question as: "What does the functor $C^*(-;\mathbb{Z})$ from finite,type nilpotent spaces to E_infty algebras do on morphism sets?. Is it injective and or surjective?" and I think it is a good question. $\endgroup$
    – HenrikRüping
    Commented May 14, 2020 at 6:40
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    $\begingroup$ In fact Mandell answers this question in his paper. This functor is split injective on morphisms. Moreover, for any map of $E_\infty$-algebras $f:C^*(Y)\to C^*(X)$, one can construct a map $g:X\to Y$ such that $H^*(g)=H^*(f)$ (but $C^*(g)$ is different from $f$ in general). $\endgroup$
    – Geoffroy Horel
    Commented May 14, 2020 at 9:24
  • $\begingroup$ @GeoffroyHorel I don't have the paper by my hand. Does that include chain homotopies $f$, with a degree shift? Also.. though $H^*(g) = H^*(f)$, I think it's still weaker than the statement that f and g are homotopic.. $\endgroup$
    – Student
    Commented May 14, 2020 at 11:13
  • $\begingroup$ chain maps that are not of degree 0 can't be compatible with $E_\infty$-structure. Yes I agree that $H^*(g)=H^*(f)$ does not imply that the two maps are homotopic. In fact there are in general many maps of $E_\infty$-algebras $C^*(Y)\to C^*(X)$ that are not homotopic in the $E_\infty$-sense to maps of the form $C^*(g)$ for $g:X\to Y$ a map of spaces. $\endgroup$
    – Geoffroy Horel
    Commented May 15, 2020 at 16:16

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