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3 why was that cofibrant replacement there? why am I not in bed?

Your theorem needs at least one further assumption. Otherwise, we can let F be the functor on spaces sending $X$ to the commutative DGA $H^*(X;R)$, with zero differential.

I'm not sure what statement to add. The thing I initially wanted to write is that your functor should take homotopy pushouts to homotopy pullbacks. However, commutative DGAs aren't closed under homotopy pullback.

EDIT: Let's suppose you make this assumption. Take the diagram $* \leftarrow \mathbb{CP}^\infty \rightarrow *$, form the homotopy pushout $\Sigma \mathbb{CP}^\infty$, and apply your functor $F$ to the associated square pushout diagram. You get a commutative square of commutative differential graded algebras, which is in particular a commutative square of $E_\infty$-algebras. The category of $E_\infty$-algebras has homotopy pullbacks, and so you can construct by this method a zigzag $weak equivalence$F(\Sigma \mathbb{CP}^\infty) \leftarrow C \rightarrow P $$of weak equivalences P of E_\infty-algebras, where C is a cofibrant replacement and P is the homotopy pullback. However, in any characteristic the Steenrod operations are stable operations and an invariant of weak equivalence of E_\infty-algebras. If pR = 0, the element x \in H^2(F(\mathbb{CP}^\infty)) supports nonzero power operations (namely, powers) at any prime, and so the associated element$$\sigma x \in H^3(P) \cong H^3(F(\Sigma \mathbb{CP}^\infty))$$would also support nonzero power operations. But for commutative DGAs those operations are automatically zero. Ideally one could make this for more general R where p is not invertible using a secondary operation and B\mathbb{Z}/p instead, but the kids were up early and I'm too tired to figure out how right now. 2 added 1371 characters in body EDIT: Let's suppose you make this assumption. Take the diagram * \leftarrow \mathbb{CP}^\infty \rightarrow *, form the homotopy pushout \Sigma \mathbb{CP}^\infty, and apply your functor F to the associated square pushout diagram. You get a commutative square of commutative differential graded algebras, which is in particular a commutative square of E_\infty-algebras. The category of E_\infty-algebras has homotopy pullbacks, and so you can construct by this method a zigzag$$F(\Sigma \mathbb{CP}^\infty) \leftarrow C \rightarrow P$$of weak equivalences of E_\infty-algebras, where C is a cofibrant replacement and P is the homotopy pullback. However, in any characteristic the Steenrod operations are stable operations and an invariant of weak equivalence of E_\infty-algebras. If pR = 0, the element x \in H^2(F(\mathbb{CP}^\infty)) supports nonzero power operations (namely, powers) at any prime, and so the associated element$$\sigma x \in H^3(P) \cong H^3(F(\Sigma \mathbb{CP}^\infty))would also support nonzero power operations. But for commutative DGAs those operations are automatically zero.

Ideally one could make this for more general $R$ where $p$ is not invertible using a secondary operation and $B\mathbb{Z}/p$ instead, but the kids were up early and I'm too tired to figure out how right now.

1

Your theorem needs at least one further assumption. Otherwise, we can let F be the functor on spaces sending $X$ to the commutative DGA $H^*(X;R)$, with zero differential.

I'm not sure what statement to add. The thing I initially wanted to write is that your functor should take homotopy pushouts to homotopy pullbacks. However, commutative DGAs aren't closed under homotopy pullback.