In the paper ["Poincaré duality and commutative differential graded algebras"](https://eudml.org/doc/272242), Lambrechts and Stanley constructed PD model for cdga with simply connected cohomology. My question is: if $A$ and $B$ are two quasi-isomoprhic CDGA, and $\hat{A}$ and $\hat{B}$ be their PD models respectively, are $\hat{A}$ and $\hat{B}$ quasi-isomorphic as PD algebras? 



Naturality of PD model of a CDGA
I guess it depends what you mean by "quasi-isomorphic as Poincaré duality algebras". I suppose you mean: does there exist a zigzag $\hat{A} \gets \cdot \to \cdot \gets \cdot \to \hat{B}$ such that all the intermediary algebras are Poincaré duality algebras, and each map is a quasi-isomorphism that commutes with the fixed orientations?


In the paper of Lambrechts & Stanley, they show (Theorem 7.1) that this is true for Poincaré duality algebras of dimension $\ge 7$ which are 2-connected and homologically 3-connected (i.e. $A^0 = \Bbbk$, $A^1 = A^2 = 0$, $H^3(A) = 0$). They explicitly build a third Poincaré duality algebra $\hat{C}$ and the zigzag $\hat{A} \to \hat{C} \gets \hat{B}$.


To my knowledge, the question remains open in general. Note that this is a difficult question. Indeed a quasi-isomorphism of (connected for simplicity) Poincaré duality algebras $f : A \to B$ is necessarily injective, for example. If $A^n = \Bbbk\langle\omega\_A\rangle$ and $B^n = \Bbbk\langle\omega\_B\rangle$ are the respective top classes, then $f(\omega\_A) = \omega\_B$. Now if $x \in A^k$ is any nonzero element, there exists $y \in A^{n-k}$ such that $xy = \omega\_A$, so $f(x) f(y) = \omega\_B \neq 0$ so $f(x) \neq 0$. This severely restrict your possibilities to build the zigzag, as it's hard to take quotients etc.




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Another plausible notion of "quasi-isomorphic as Poincaré duality algebras" would be a zigzag $\hat{A} \xleftarrow{f} R \xrightarrow{g} \hat{B}$ of quasi-isomorphisms of cdgas such that $\varepsilon\_A \circ f = \varepsilon\_B \circ g$ (where $\varepsilon\_A : \hat{A}^n \to \Bbbk$ (resp $\varepsilon\_B$) is the fixed orientation). In other words you don't require a Poincaré duality algebra in the middle but you still have some compatibility.


I showed in [my thesis](https://ori-nuxeo.univ-lille1.fr/nuxeo/site/esupversions/0c76f6d6-6909-40b4-936c-14056253fadc) in Proposition 2.4.34 that in characteristic zero, for simply connected Poincaré duality algebras, you can always find such a zigzag. Despite what's written there, the argument is purely algebraic and $\hat{A}$, $\hat{B}$ don't have to be models of some closed manifold. (The statement is also going to appear in [the corresponding paper](https://arxiv.org/abs/1608.08054) but I'm waiting on a referee report to update the arXiv version.)


