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By a theorem of Adams, if $A = C^*(X;\mathbb{Q})$ is the CDGA of rational cochains on $X$ then the cohomology of the bar complex of $A$ is isomorphic to $H^*(\Omega X; \mathbb{Q})$ as a coalgebra (see e.g. Félix–Oprea–Tanré, Algebraic models in geometry, Theorem 5.52 for this formulation).

Maybe this is a naive question, but if I take a rational CDGA model (in the sense of Sullivan) $A$ of a space $X$, then $BA$ becomes a Hopf algebra. Is it then true that 1/ $BA$ is a rational CDGA model for $\Omega X$, and (if so) 2/, that the coproduct of $BA$ represents the product of $\Omega X$ upon realization?

(I am aware of this previous question, but if I understand it correctly it doesn't really answer my question... I've also been reading Majewski's book Rational homotopical models and uniqueness but I wasn't able to directly apply his results to my question.)

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Najib Idrissi
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  • We have a problem at the very beginning. Rational cochains do not form a CDGA. With Sullivan models everything works as expected, assuming usual finiteness and connectivity conditions. – Fernando Muro Feb 05 '16 at 09:37
  • It's possible I misunderstood the book and C∗(X;Q)C∗(X;Q) isn't the rational (singular?) cochains, sorry. So if I understand correctly, under finiteness and connectivity conditions, 1/ and 2/ are true? (Sorry for the two pings, I mistakenly deleted the other comment...) – Najib Idrissi Feb 05 '16 at 09:49
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    Why doesn't your linked question provide the answers you're looking for? – Theo Johnson-Freyd Feb 05 '16 at 15:01
  • @Theo If I understand correctly, for example in Manuel Rivera's answer, Chen's map is said to induce a Hopf algebra map on cohomology (and it's for real forms). But it's not necessarily the case that it's compatible with the coproduct before taking homology, right? Only up to homotopy. To be fair it couldn't possibly be since, as Fernando points out, $C^(\Omega X)$ isn't a cdga, but what about cdga models...? Mark Grant's answer is also after* passing to cohomology... (I feel it's one of these situations where the answer is right before my eyes but I can't see it...) – Najib Idrissi Feb 05 '16 at 15:13
  • @NajibIdrissi The iterated integral map works for rational differential forms. If $A$ is the CDGA of rational differential forms on $X$ then we have a map (see Richard Hain's work) of DG coalgebras $BA \to C^(\Omega X; \mathbb{Q})$ assuming $\Omega X$ is the Moore based loop space of $X$. On homology this is a map of algebras (shuffle on $BA$, cup on $C^(\Omega X; \mathbb{Q})$. At the level of cochains this is probably a map of $A_{\infty}-algebras$, just like the De Rham map from forms to cochains. – Manuel Rivera Sep 06 '16 at 20:20
  • @NajibIdrissi $C^(\Omega X; \mathbb{Q})$ is not a CDGA but one can symmetrize to obtain a commutative product, but then you loose strict associativity, but I think you can define a $C_{\infty}$-structure on $C^(\Omega X; \mathbb{Q})$ extending this symmetrized cup product. – Manuel Rivera Sep 06 '16 at 20:22

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