\begin{document}

    \author{A. Milivojevi\'c}
	\address{Max-Planck-Institut f\"ur Mathematik,
		Vivatsgasse 7,
		53111 Bonn}
	\email{milivojevic@mpim-bonn.mpg.de}
	\author{J. Stelzig}
	\address{ Mathematisches Institut der Ludwig-Maximilians-Universit\"at M\"unchen,
		Theresienstraße 39, 80993 M\"unchen}
	\email{jonas.stelzig@math.lmu.de}
	\author{L. Zoller}
	\address{ Mathematisches Institut der Ludwig-Maximilians-Universit\"at M\"unchen,
		Theresienstraße 39, 80993 M\"unchen}
	\email{zoller@math.lmu.de}
	\mathfrak{su}bjclass[2020]{55P62, 55S30, 57N65}
	\keywords{Poincar\'e duality algebras, Massey products, non-zero degree maps, formality}
	
	\begin{abstract}
	    A result of Taylor shows that triple Massey products pull back non-trivially under non-zero degree maps of rational Poincar\'e duality spaces. Generally one can wonder whether formality, or the lack of certain obstructions to it, is preserved under non-zero degree maps, in accordance with the general heuristic that the domain of a dominant map should be more complicated than its target. We show that Taylor's result does not directly generalize to the case of quadruple Massey products; namely, we construct a non-zero degree map of rational Poincar\'e duality spaces where a non-trivial quadruple Massey product on the target becomes trivial when pulled back to the domain. For the construction we introduce a method to turn a cdga into one that satisfies Poincar\'e duality, which is natural in certain situations. 
	\end{abstract}

\title{Poincar\'e dualization and formal domination}

\setcounter{tocdepth}{1}
\tableofcontents

\section{Introduction: Formality and non-zero degree maps}

    The existence of non-zero degree maps between closed manifolds, giving a relation going by the name of \emph{domination} or \emph{Gromov's partial order}, has been of substantial interest going back at least to work of Gromov, Milnor, and Thurston in the 1970's, see \cite[p.173]{CT89}. The general empirical observation is that the domain of a non-zero degree map should be ``more complicated'' than its target, see e.g. loc. cit. In view of this heuristic, one might expect that rational homotopy theoretic formality is preserved under such maps.

Indeed, formal spaces are those having the ``simplest'' rational homotopy type with a given cohomology algebra, where simplicity refers to the vanishing of certain higher structures on the cohomology, the most prominent incarnation of which are Massey products.
	
This embeds into the more general question of under which circumstances a map $Y\xrightarrow{f} X$ of topological spaces with formal domain $Y$ lets us deduce something about the formality of $X$, or at least about the vanishing of certain Massey products on $X$. A naive first guess could be that positive conclusions can be drawn as soon as the map on cohomology $H(X)\rightarrow H(Y)$ is injective, since then the vanishing Massey products on $H(Y)$ together with naturality might imply the vanishing of Massey products on $H(X)$. However, while this certainly works for the binary product, i.e. the cup product, it is not hard to see that the argument fails in general for triple and higher products, due to the presence of indeterminacy (see Remark \ref{noPD}).

If we restrict ourselves to the more geometric setting of closed orientable manifolds, or more generally rational Poincar\'e duality spaces, and non-zero degree maps between them, then, as predicted by the aforementioned heuristic of domination, there are indeed a number of positive results in the above direction:
	
	\begin{itemize}    
	\item 	Taylor \cite{Ta10} proved that a non-vanishing Massey triple product on $X$ pulls back to a non-vanishing Massey triple product on $Y$. That is, if $0 \not \in \langle x, y, z \rangle$ on $X$, then $0 \not \in \langle f^*x, f^*y, f^*z \rangle$ on $Y$.
	\item  By the argument in \cite[Theorem 5.22]{DGMS75}, if $Y \to X$ is a holomorphic map of compact complex manifolds of non-zero degree, and $Y$ satisfies the $\partialdelbar$-lemma, then $X$ also satisfies the $\partialdelbar$-lemma; recall from loc. cit. that a manifold satisfying the $\partialdelbar$-lemma is formal.

		\item Suppose we have a non-zero degree map $Y \xrightarrow{f} X$ of rational Poincar\'e duality spaces of dimension $\leq 5n+2$, where $X$ is rationally $n$--connected. Then $Y$ being formal implies that $X$ is formal. Indeed, by \cite{CN20}, formality of $X$ is equivalent to the Bianchi--Massey tensor vanishing; a non-trivial Bianchi--Massey tensor pulls back non-trivially, obstructing formality of $Y$.
	\end{itemize}
	
	We add to this by giving an alternative proof of the aforementioned result from \cite{Ta10}, as well as by showing that formality is preserved under non-zero degree maps of closed manifolds of dimension $\leq 4$ when the domain has trivial rational cup product $H^1 \otimes H^1 \to H^2$.
	
	
	\iffalse
    proving the following:
	\begin{thmintro}\label{1.1}

Let $Y\xrightarrow{f} X$ be a non-zero degree map of orientable closed $n$-manifolds (or more generally Poincar\'e duality spaces) where $n\leq 4$ and the cup product $H^1(Y) \otimes H^1(Y) \rightarrow H^2(Y)$ is trivial. If $Y$ is formal then so is $X$.
	\end{thmintro}
	\fi
	
	

	
	We then show that Taylor's result on triple Massey products pulling back non-trivially does not generalize to the case of quadruple Massey products:
	
	\begin{thmintro}\label{thmintro}
	There exists a non-zero degree map $Y\xrightarrow{f} X$ of rational Poincar\'e duality spaces such that $X$ has a non-trivial quadruple Massey product which pulls back to be trivial on $Y$.
	\end{thmintro}
	
	The strategy is to start on the algebraic level and construct a suitable map between commutative differential graded algebras (cdga's) whose cohomology satisfies Poincaré duality. This can then be realized geometrically by using results from \cite{Su77}. To construct our algebraic example, we develop a procedure to modify any cdga (satisfying suitable finiteness conditions) into one whose cohomology satisfies Poincaré duality and study naturality properties of the construction. We refer to this procedure as \emph{Poincaré dualization}; the construction depends on a fixed integer $n$. We believe this might prove to be a useful technical tool to construct manifold examples from general cdga phenomena, somewhat akin to taking the boundary or double of a thickening of a cell complex in Euclidean space.
		
	\begin{thmintro}
	There is an association of cdga's $A\mapsto P_n A$ such that
	
	\begin{enumerate}
	    \item $A$ is naturally a sub-cdga of $P_n A$ and the inclusion induces an injection in cohomology.
	    \item If $A$ is cohomologically connected and has finite dimensional cohomology concentrated in degrees between $0$ and $n$, then $P_nA$ is a Poincar\'e duality cdga.
	    \item $P_n$ is functorial for pairs $(f,r)$ where $f:A\to B$ is a map of cdga's and $r:B\to A$ is a map of dg-$A$-modules.
	    \item If $A$ admits a non-vanishing Massey product, the induced Massey product in $P_n A$ is non-trivial.
	    \item If $A$ is a Sullivan cdga that is cohomologically connected, has finite dimensional cohomology concentrated in degrees between $0$ and $n/2$ and is formal, then $P_n A$ is also formal.
	\end{enumerate}
	\end{thmintro}
	
In view of the above, the following questions are natural to ask, and \mathbb{C}ref{thmintro} indicates that the answers might not be so straightforward:

\begin{questionin}\label{formalityquestion} Let $Y \to X$ be a non-zero degree map of rational Poincar\'e duality spaces, and suppose $Y$ is formal. Does it follow that $X$ is formal? If $m$ is some non-vanishing obstruction to formality on $X$ that can be pulled back to $Y$, under what conditions is the pullback $f^*m$ non-vanishing? \end{questionin}

The article is organized as follows: in \mathbb{C}ref{Preliminaries} we review the necessary concepts from (rational) homotopy theory we discuss. In \mathbb{C}ref{positiveresults} we give the aforementioned alternative argument to Taylor's theorem on triple Massey products pulling back non-trivially, and prove the mentioned results on manifolds of dimension $\leq 4$. In \mathbb{C}ref{poincaredualization} we introduce the Poincar\'e dualization construction and prove some basic properties. We then use this machinery, together with classical rational homotopy theory, in \mathbb{C}ref{counterexample} to construct the example in \mathbb{C}ref{thmintro}.

\mathfrak{su}bsection*{Acknowledgements} 

The authors would like to thank Michael Albanese, Manuel Amann, Joana Cirici, Pavel H\'ajek, Dieter Kotschick,  Christoforos Neofytidis, Dennis Sullivan, Peter Teichner, Shmuel Weinberger and Scott Wilson for illuminating conversations and helpful comments. We are very grateful to the referee for their comments which led us to discover a mistake in a previous version of this article.

Part of this research was supported through a ``Research in Pairs'' program visit by A.M. and J.S. at the Mathematisches Forschungsinstitut Oberwolfach; they thank the MFO for the excellent working conditions provided there. They are likewise grateful to the City University of New York Graduate Center for their hospitality. A.M. would also thank the Institut Mittag-Leffler in Djursholm for its generous hospitality during a visit to the ``Higher algebraic structures in algebra, topology and geometry'' program; he thanks \"Ozg\"ur Bay\i nd\i r, Alexander Berglund, and Andrea Bianchi for interesting discussions there.

J.S. and L.Z. thank the Ludwig-Maximilians-Universit\"at M\"unchen, and A.M. thanks the Max-Planck-Institut f\"ur Mathematik, for their enduring support.

\section{Preliminaries}\label{Preliminaries}

We recall the concepts used in the introduction and throughout, and set notation. Let $K$ be a field. We will consider graded commutative algebras $(A,d)$ over $K$, where we allow entries in negative degrees, i.e. $A=\bigoplus_{k\in\mathbb{Z}}A^k$, but most of the time we restrict to \emph{cohomologically connected} cdga's, i.e. those where $H^k(A)=0$ for $k<0$ and  $H^0(A)=K$. If $A^k=0$ for $k<0$ and $A^0=K$, we say $A$ is connected. We call $A$ a \emph{$K$--Poincar\'e duality cdga} if its cohomology satisfies Poincar\'e duality. That is, there is an index $n$ such that $H^n(A,d) \cong K$ and the pairing $$H^k(A,d)\otimes H^{n-k}(A,d) \to H^n(A,d) \cong K$$ given by $$\alpha \otimes \beta \mapsto \alpha \beta$$ is non-degenerate.

Restricting to the case of $K = \mathbb{Q}Q$, a commutative differential graded algebra (cdga) is said to be \emph{formal} if there is a chain of quasi-isomorphisms of cdga's

$$
\begin{tikzcd}[row sep=small]
        & {(B_1,d)} \arrow[ld] \arrow[rd] &           & \cdots \arrow[ld] \arrow[rd] &           & {(H,0)} \arrow[ld] \\
{(A,d)} &                                 & {(B_2,d)} &                              & {(B_r,d)} &                   
\end{tikzcd}
$$
connecting $(A,d)$ to a cdga with trivial differential. For $A$ cohomologically connected,  one may pick a Sullivan model $(\Lambda V,d)\to A$, i.e. a connected cdga that is free as an algebra, satisfying a nilpotence condition (c.f. \cite{FHT12}), with a quasi-isomorphism to $A$. One may even pick $(\Lambda V,d)$ to be minimal, i.e. $d(\Lambda V)\mathfrak{su}bseteq \Lambda^{\geq 2}V$; 
\cite[p.191]{FHT12}.  In terms of such a model, formality of $A$ is equivalent to the existence of a quasi-isomorphism $(\Lambda V,d)\to (H(A),0)$. (That is, we may replace the chain of quasi-isomorphisms by a single `roof'.) This reformulation of formality can be further rephrased as saying that one may pick a complement $N$ to the space of closed generators $C\mathfrak{su}bseteq V$ such that the ideal generated by $N$ in $\Lambda V$ is acyclic.
We will also consider the following weakening of the notion of formality (c.f. \cite{FM05}): A minimal Sullivan cdga $(\Lambda V, d)$ is called $s$--formal, where $s\in\mathbb{Z}_{\geq 1}$ if for each $i\leq s$ the closed generators $C^i\mathfrak{su}bseteq V^i$ have a complement $N^i$ such that  any closed element in the ideal $I\mathfrak{su}bseteq \Lambda^{\leq s}V$ generated by $\bigoplus_{i\leq s}N^i$ is exact in $\Lambda V$. Then an arbitrary (cohomologically connected) cdga is called $s$-formal if its minimal Sullivan model is $s$--formal.
\newline

Computable obstructions to formality are given by (ad hoc) Massey products \cite[Section 2]{M58}. Given three pure-degree classes $[x],[y],[z]\in H(A)$ such that $xy=da$, $yz=db$, the element $az-(-1)^{|x|}xb$ is closed and therefore gives rise to a cohomology class. Modulo the ideal generated by $[x]$ and $[y]$, this class is well-defined and independent of the choices of representatives and primitives. It is called the triple Massey product and denoted $\langle [x],[y],[z] \rangle :=[az-(-1)^{|x|}xb]\in H(A)/([x],[y])$. 

Quadruple Massey products are defined similarly: Given four classes $[w],[x],[y],[z]\in H(A)$, which for simplicity we assume to have pure even degree (which will be the case for us below), a defining system for the quadruple product $\langle [w],[x],[y],[z]\rangle\mathfrak{su}bseteq H(A)$ is a collection of pure-degree elements $a,b,c,f,g$ such that $da=wx$, $db=xy$, $dc=yz$ and $df=ay-wb$ and $dg=bz-xc$. For any such defining system, one obtains a cohomology class $[wg+ac+zf]\in H(A)$. The quadruple Massey product $\langle [w],[x],[y],[z]\rangle\mathfrak{su}bseteq H(A)$ is then defined to be the collection of classes obtained from all such defining systems. Again, this collection is independent of the chosen representatives for the classes. The quadruple product is said to be trivial (or vanish) if $0\in\langle [w],[x],[y],[z]\rangle$. The definitions for quintuple and higher products are similar; we refer the reader to \cite{K66}.

Massey products are invariants of the quasi-isomorphism type of a cdga, and on formal cdga's all Massey products vanish.

To a topological space $X$ we can associate its connected cdga $A_{PL}(X)$ of rational piecewise-linear forms \cite{Su77}, \cite{DGMS75}; this cdga computes the rational cohomology of $X$ (see e.g. \cite[Theorem 2.1]{DGMS75}, \cite[Theorem 1.21]{H07}). We say the space $X$ is formal if $A_{PL}(X)$ is formal as a cdga \cite[p.260]{DGMS75}, \cite[Definition 2.1]{H07}. A space is a \emph{rational Poincar\'e duality space} if its rational cohomology satisfies Poincar\'e duality.

	\section{Some positive results}\label{positiveresults}
	
	In this section, $K=\mathbb{Q}$ and $H(X)$ denotes rational cohomology. First let us give an alternative proof of Taylor's theorem \cite{Ta10} mentioned above.
	
	\begin{prop}\label{prop:Taylor} Let $Y\rightarrow X$ be a non-zero degree map between rational Poincar\'e duality spaces and let $a,b,c\in H(X)$ with $ab=bc=0$. If \[m:=\langle a,b,c\rangle\neq 0\in \frac{H(X)}{a\cup H(X)+H(X)\cup c},\]
		then also \[
		f^*(m)\neq 0\in \frac{H(Y)}{f^*a\cup H(Y)+H(Y)\cup f^*c}.
		\]
	\end{prop}
	
	\begin{proof}
		The map $f^*: H(X)\to H(Y)$ has a one-sided inverse given by $f_*':=\frac{1}{\deg f} f_*$, i.e. $f_*'f^*=Id_{H(X)}$. Here $f_*$ denotes the pushforward, determined by $f^*$ and Poincar\'e duality. This yields a splitting
		\[
		H(Y)\overset{(f_*',\operatorname{pr})}{\longrightarrow} H(X)\oplus H(Y)/f^*H(X).
		\]
		By the projection formula
		\[
		f_*(f^*x\cup y)=x\cup f_*y \quad\text{for }x\in H(X), y\in H(Y),\]
		this (additive) splitting is compatible with the natural $H(X)$-module structures on both sides (given by $f^*$ on the left and $(Id, f^*)$ on the right). Therefore, writing $K:=H(Y)/f^*H(X)$, the domain of definition of the Massey product decomposes as
		
		\[
		\frac{H(Y)}{f^*a\cup H(Y)+H(Y)\cup f^*c}\overset{\sim}{\longrightarrow}\frac{H(X)}{a\cup H(X)+H(X)\cup c}\oplus\frac{K}{f^*a \cup K+K\cup f^*c},
		\]
		and under this splitting, we have $f^*(m)=(m,0)$.
	\end{proof}
	
	\begin{rem} The above proof works for maps of Poincar\'e duality cdga's over any field, as long as we interpret $\deg f\neq 0$ to mean that $\deg f$ is invertible. 
	
	Often in examples one has a subalgebra of invariant forms, satisfying Poincar\'e duality, and containing an invariant volume form, in the algebra of all forms $A_X$ on a closed manifold $X$. By duality, this algebraic map is injective on cohomology. Then the same argument as above applies to show that a non-vanishing triple Massey product calculated on the invariant subalgebra continues to be non-vanishing on $A_X$; that is, if there is some defining system for the triple product on $A_X$ making it trivial, then there was a defining system on the subalgebra making it trivial to begin with.\end{rem}
	

	\begin{rem}\label{noPD} Without the Poincar\'e duality assumption, it is generally easy to find examples of cohomologically injective maps of cdga's such that a non-vanishing triple Massey product in the domain vanishes in the target. For example, consider the inclusion of cdga's $$A:=\left( \Lambda(x,y,z), dz = xy \right)\hookrightarrow B:=\left( \Lambda(x,y,z,u,v), dz = xy, dv = xz-yu \right),$$ where all generators are in degree 1. This induces an inclusion $A':=A/A^{\geq 3}\hookrightarrow B':=B/B^{\geq 3}$ which is injective on cohomology. Now, $\langle x, x, y \rangle$ is a non-vanishing triple product in $A'$, while in $B'$ it is represented by $[xz] = [yu]$, which lies in the indeterminacy. \end{rem}
	
    We also record the following result in which formality is preserved under a non-zero degree map.

\begin{prop}\label{1formalityprop}
Let $Y \xrightarrow{f} X$ be a cohomologically injective map of topological spaces,  where the cup product $H^1(Y) \otimes H^1(Y) \rightarrow H^2(Y)$ is trivial. If $Y$ is 1--formal, then $X$ is 1--formal.
\end{prop}
By \cite[Theorem 3.1]{FM05}, a closed manifold in dimensions $\leq 4$ is formal if and only if it is 1--formal. This directly implies the following:
\begin{cor}
Let $Y\xrightarrow{f} X$ be a non-zero degree map of orientable closed $n$-manifolds (or more generally rational Poincar\'e duality spaces) where $n\leq 4$ and the cup product $H^1(Y) \otimes H^1(Y) \rightarrow H^2(Y)$ is trivial. If $Y$ is formal then so is $X$.
\end{cor}

	
	
	Notice that our assumptions imply that $H^1(X) \otimes H^1(X) \to H^2(X)$ is trivial as well. Indeed, suppose $ab \neq 0$ for some $a,b \in H^1(X)$. Then we would have $(f^*a)(f^*b) = f^*(ab) \neq 0$ by the injectivity of $f^*$. 
    
	
\begin{rem}(The condition of having trivial $H^1(Y) \otimes H^1(Y) \xrightarrow{\cup} H^2(Y)$ on 3--manifolds)

There exist both formal and non-formal closed 3--manifolds satisfying this condition, e.g. connected sums of $S^1\times S^2$, resp. Heisenberg manifolds.

By a theorem of Sullivan \cite{Su75}, for any finite--dimensional rational vector space $H$ and any skew--symmetric trilinear form $H^{\otimes 3} \to \mathbb{Q}Q$, there is an oriented 3--manifold $M$ realizing this data as the trilinear form $H^1(M)^{\otimes 3} \to \mathbb{Q}Q$. In the case of 3--manifolds, our result applies to the case of the zero form.

 If there were $a, b \in H^1(M)$ such that $ab \neq 0$, then by Poincar\'e duality we could choose $c \in H^1(M)$, linearly independent of $a$ and $b$, such that $abc \neq 0$. Choosing integral representatives of non-zero multiples of $a,b,c$ hence gives us a non-zero degree map $M \to T^3$. 

	
\end{rem}
	
	
	
	\begin{proof}[Proof of \mathbb{C}ref{1formalityprop}]
		
		
		Take minimal models $M_X = (\Lambda V^X, d)$ and $M_Y = (\Lambda V^Y, d)$ for $X$ and $Y$. We will argue that $M_X$ is 1--formal. Namely, we will show there is a splitting of the space of degree 1 generators $V_1^X = C_1^X \oplus N_1^X$, where $C_1^X = \ker d$, such that any closed element in $\Lambda(V_1^X)$ in the ideal generated by $N_1^X$ is exact in $M_X$. 
		
		Before doing that, let us show that $V_1^X$ injects into $V_1^Y$ under the induced map on models, which we also denote by $f^*$. Certainly $C_1^X$ injects (into $C_1^Y$) since $f^*$ is injective on cohomology. Consider the increasing filtration of $V_1^X$ given by $F^0 = C_1^X$ and $F^i=d^{-1}(\Lambda^2 F^{i-1})$ for $i\geq 1$. This filtration is exhaustive and preserves the differential by nilpotency. Now take a non-closed element in $F^1$ (if it is closed, it pulls back non-trivially); then the image of its differential under $f^*$ is non-zero by freeness, and hence the image of the element is non-zero. Inductively we obtain our claim. In fact, we have shown that the cdga $\left(\Lambda V_1^X, d\right)$ injects into $M_Y$.
		
		Now we observe that, in general, on a 1--formal minimal cdga $(\Lambda V, d)$ with trivial cup product $H^1 \otimes H^1 \to H^2$, we can choose \textit{any} complement to the closed elements in degree 1 in the definition of 1--formality. Indeed, by 1--formality we have some splitting $V_1 = C_1 \oplus N_1$ with the desired properties. Choose another complement $N_1'$ to $C_1$, and take a closed element in $\Lambda V_1$ in the ideal generated by $N_1'$. We can write this element as a sum of an element in the ideal of $N_1$ and a product of elements in $C_1$. Since the latter is closed, the former is closed as well, and hence it is exact in $(\Lambda V, d)$ by assumption. The product of elements in $C_1$ is also exact in $(\Lambda V, d)$ by the assumption on the cup product.
		
		Now choose any complement $N_1^X$ to $C_1^X$ in $V_1^X$. Mapping over (injectively) to $M_Y$ via $f^*$, we can complete a basis of $C_1^X$ to a basis of $C_1^Y$, and a basis of $N_1^X$ to a basis for a complement to $C_1^Y$ in $V_1^Y$. Suppose now that we have a closed element in $\Lambda(V_1^X)$ in the ideal generated by $N_1^X$. Pulling back to $M_Y$, this is now a closed element in $\Lambda(V_1^Y)$ in the ideal generated by $N_1^Y$. By 1--formality of $M_Y$, it is exact. Hence, by the injectivity of $f^*$ on cohomology, the element must have been exact in $M_X$ to begin with.
\end{proof}

\section{Poincar\'e dualization}\label{poincaredualization}

Motivated by the example in \mathbb{C}ref{noPD}, we detail a construction that ``completes'' any cohomologically connected cdga to one satisfying Poincar\'e duality on its rational cohomology, which in certain cases is functorial.

Fix a natural number $n$ and a field $K$. Let $(A,d)$ be a complex of $K$-vector spaces. We define the ($n$-th) \textbf{dual complex} $D_nA$ by $(D_nA)^k:=(A^{n-k})^\vee$ with differential $(D_nA)^k \to (D_nA)^{k+1}$ given on pure-degree elements $\varphi\in D_n A$ by $d(\varphi)(a):=(-1)^{|\varphi|-1}\varphi(da)$ for any $a\in A$. Clearly, $D_n$ is a contravariant functor (given by $\left((D_n r)(\varphi)\right) (b) = \varphi(r(b))$ for a map of complexes $B \xrightarrow{r} A$) and
\begin{equation*}
    H^k(D_nA)=(H^{n-k}(A))^\vee.
\end{equation*}

Now let us assume $A$ carries in addition the structure of a graded-commutative algebra, such that $d$ is a derivation (i.e. $A$ is a cdga). 

\begin{defi}
Let $(A,\wedge, d)$ be as above. The $n$-th \textbf{Poincar\'e dualization} of $A$ is given, as a complex, by
\[
P_nA:= A\oplus D_nA,
\]
with multiplication (extending that on $A$) defined on pure-degree elements $a\in A$, $\varphi\in D_nA$ by the dual complex element given by
\begin{align*}
(a\wedge\varphi)(b)&:=(-1)^{|a||\varphi|}\varphi(a\wedge b),\\
(\varphi\wedge a)(b)&:=\varphi(a\wedge b),
\end{align*}
and setting $\varphi\wedge\psi=0$ for $\varphi,\psi\in D_n A$. 
\end{defi}

\begin{lem}
The Poincar\'e dualization $P_nA$ is indeed a cdga. 
\end{lem}
\begin{proof}
Graded commutativity of the multiplication holds by definition. For associativity, we only need to check the case $\varphi\in D_n A$ and $a,b \in A$ as all other combinations of products of three elements are either zero or entirely in $A$, where associativity holds since $A$ is a cdga. We compute, for $c \in A$:
\begin{align*}
    ((\varphi\wedge a)\wedge b)(c)&=(\varphi\wedge a)(b\wedge c)\\
    &=\varphi(a\wedge b\wedge c)\\
    &=(\varphi\wedge(a\wedge b))(c).
\end{align*}

That $d$ is a derivation again only has to be checked on products of the form $\varphi\wedge a$ with $\varphi\in D_n A$ and $a\in A$. In this case, we compute:
\begin{align*}
    (d\varphi\wedge a)(b)&=d\varphi(a\wedge b)\\
    &=(-1)^{|\varphi|-1}\varphi(d(a\wedge b))\\
    &=(-1)^{|\varphi|-1}\varphi(da\wedge b)+(-1)^{|\varphi|-1+|a|}\varphi(a\wedge db)\\
    &=(-1)^{|\varphi|-1}(\varphi\wedge da)(b) + (-1)^{|\varphi\wedge a|-1}(\varphi\wedge a)(db)\\
    &= (-1)^{|\varphi|-1}(\varphi\wedge da)(b)+d(\varphi\wedge a)(b). \qedhere
\end{align*}
\end{proof}

For simplicity, let us now further assume that $A$ is cohomologically connected, $H(A)$ is finite-dimensional, and $H^k(A)\neq 0$ at most for $0\leq k< n/2$.
\begin{lem}

The cohomology $H(P_nA)$ is finite dimensional and concentrated in degrees $0,...,n$. Further, $P_nA$ is a Poincar\'e duality cdga, i.e. for any integer $k$, the pairing
\[
H^k(P_n A)\times H^{n-k}(P_nA)\overset{\wedge}{\longrightarrow} H^n(P_nA)\cong K 
\]
is non-degenerate.
\end{lem}
\begin{proof}
For simplicity of notation, let $k<n/2$. By construction, $H^k(P_nA)=H^k(A)$ and $H^{n-k}(P_nA)= H^{n-k}(D_nA)=(H^k(A))^\vee$,
and the pairing is given (up to a non-zero scalar) by evaluation. \qedhere
\end{proof}

\begin{rem} In fact, $P_n(A)$ is an oriented differential Poincar\'e duality algebra in the sense of \cite{LS08} (though without the assumption of no elements in negative degree as therein), i.e. it satisfies Poincar\'e duality already on the chain level and there is a canonical orientation (i.e. an isomorphism $(P_nA)^n\cong K$) induced by the dual of the unit map $K\to A$ sending $1\mapsto 1$. However, we will not make use of this additional structure here. \end{rem}

\begin{ex} \,
\begin{itemize}
\item Let $A=\left( \Lambda(x)/x^2, d = 0 \right)$ with $|x|\geq 1$ and let $n>|x|$. Then \[P_n(A)\cong \left( \Lambda(x,y)/(x^2,y^2), d = 0 \right)\] with $|y|=n-|x|$, i.e. we obtain the cohomology algebra of the product of spheres $S^{|x|}\times S^{|y|}$.

\item Take now, for example, $A$ to be a minimal model for $S^2$, i.e. $A = \left( \Lambda(x, y), dy = x^2 \right)$. Then $P_n(A)$ is generated as an algebra by $x, y,$ and a dual basis $\{\widehat{x^k}, \widehat{x^k y}\}_{k\geq 0}$ for $\{x^k, x^ky\}$, in degrees $n - 2k$ and $n - 2k - 3$ respectively. satisfying $$\widehat{x^k y} \wedge x^l y = \widehat{x^{k-l}}, \ \  \widehat{x^k y} \wedge x^l = \widehat{x^{k-l}y}, \ \  \widehat{x^k} \wedge x^l y = 0, \ \ \widehat{x^k} \wedge x^l = \widehat{x^{k-l}}$$ for $k\geq l$. All other products of dual elements with basis elements from $A$ are zero. The differential is determined by $d(\widehat{x})=0$ and $d(\widehat{x^k}) = - \widehat{x^{k-2}y}$ for $k\geq 2$. A quasi-isomorphism from $P_n(A)$ to $\left(H(S^2 \times S^{n-2}), d=0\right)$ is given by sending $x$ and $\widehat{x}$ to the volume classes of $S^2$ and $S^{n-2}$ respectively (and 1 to 1), and all other generators to zero. 
\item For a non-manifold example, consider the formal space $S^2 \vee S^3$, with model $$\left( \Lambda(x, y)/(x^2, xy), d = 0\right)$$ with $|x|=2$, $|y|=3$. Then for $n\geq 4$, its $n$-th Poincar\'e dualization is the cohomology ring of the manifold $\left( S^2 \times S^{n-2} \right) \# \left( S^3 \times S^{n-3} \right) $ equipped with trivial differential. Note that for large enough $n$, the boundary of a thickening of $S^2 \vee S^3$ in $\mathbb{R}^n$ is $\left( S^2 \times S^{n-2} \right) \# \left( S^3 \times S^{n-3} \right) $. \\ \end{itemize}

The above examples suggest that there is a certain quasi-isomorphism invariance in the Poincar\'e dualization construction, and a geometric construction mirrored by it. It would be interesting, but not necessary for the present purposes, to pursue these points further.

\end{ex}

Now consider a cdga $B$, satisfying the same finiteness and connectedness conditions as $A$, and a map of cdga's $f:A\to B$. In general, it is not true that this can be extended to a map $P_n A\to P_n B$. However, one has:
\begin{lem}\label{lem:naturality}
Given a map $r:B\to A$ of dg-$A$-modules, i.e.\ a map of complexes satisfying $r(f(a)\wedge b)=a\wedge r(b)$, the map
\[
f\oplus D_n r: P_n A\to P_n B
\]
is a map of cdga's.  When $r(1)=1$ (or equivalently $r\circ f=\operatorname{id}$), the map $f\oplus D_n r$ has non-zero degree.
\end{lem}
\begin{proof}
Because both $f: A\to B$ and $D_n r: D_n A \to D_n B$ are maps of complexes, so is $f\oplus D_n r$, where $f$ and $D_n r$ are extended trivially to all of $P_n A$. It thus remains to show that the map is compatible with the product. If both factors are in $A\mathfrak{su}bseteq P_n A$, this is true since $f$ is an algebra map. If both entries are in $D_n A$, their product is zero, and so is the product of their images under $D_n r$. The remaining case, $a\in A, \varphi\in D_n A$, follows from: 
\begin{align*}
    \Bigl( (f\oplus D_nr)(\varphi\wedge a) \Bigr) (b)&=\Bigl( D_n r(\varphi\wedge a) \Bigr) (b)\\
    &=(\varphi\wedge a)(r(b))\\
    &=\varphi(a\wedge r (b))\\
    &=\varphi(r(f(a)\wedge b))\\
    &= \left(D_nr(\varphi) \right) (f(a)\wedge b)\\
    &=\Bigl( D_nr(\varphi)\wedge f(a) \Bigr) (b)\\
    &=\Bigl( (f\oplus D_n r)(\varphi)\wedge(f\oplus D_n r)(a)\Bigr) (b).
\end{align*}
The statement about the degree follows since the top-degree cohomology in $P_n A$ and $P_n B$ is generated by any class that evaluates non-trivially on $1$.
\end{proof}
\begin{rem}
Consider the category whose objects are cdga's satisfying the conditions of $A$ above and morphisms $A\to B$ given by pairs $(f,r)$ as above, with composition $(g,s)\circ (f,r)=(g\circ f, r\circ s)$. Then Poincar\'e dualization $P_n$ defines a functor from this category to that of Poincar\'e duality cdga's. Restricted to degree-wise finite dimensional cdga's, it is fully faithful.
\end{rem}

From now on, we will take our ground field $K$ to be the rationals $\mathbb{Q}Q$.

\begin{prop}\label{prop:PDagainFormal}
Let $(\Lambda V,d)$ be a formal Sullivan cdga. Assume that $H(\Lambda V, d)$ is finite dimensional and concentrated in degree $\leq k$ with $2k < n$. Then $P_n(\Lambda V,d)$ is formal.
\end{prop}

\begin{proof}
We assume first that $(\Lambda V,d)$ is minimal.
Consider the canonical map $\varphi\colon (\Lambda V,d)\rightarrow P_n(\Lambda V,d)$ and note that $\varphi$ induces an isomorphism in cohomology up until degree $s:=\lfloor n/2\rfloor$. Following the usual algorithm for the construction of a minimal model $(\Lambda W,d)\xrightarrow{\simeq} P_n(\Lambda V,d)$ (see \cite[Section 5]{Su77}), we may arrange that $W^{\leq s}=V^{\leq s}$ and that $\varphi$ factors as
\[(\Lambda V,d)\xrightarrow{\psi} (\Lambda W,d)\rightarrow P_n(\Lambda V,d)
\]
with $\psi$ being the identity in degrees $\leq s$. Note that though $P_n(\Lambda V, d)$ may contain elements in negative degree, due to it being cohomologically connected, the construction in \cite[Section 5]{Su77} carries through verbatim.

Now since $(\Lambda V,d)$ is formal we may write $V=C\oplus N$ such that $d(C)=0$, $d|_N$ is injective, and every closed element in the ideal $N\cdot \Lambda V$ is exact \cite[Theorem 4.1]{DGMS75}. We claim that the decomposition of $W^{\leq s}=V^{\leq s}$ into $C^{\leq s}\oplus N^{\leq s}$ satisfies the condition of $s$-formality for $(\Lambda W,d)$. To check this we only need to verify that any closed element in $N^{\leq s}\cdot \Lambda V^{\leq s}$ becomes exact in $(\Lambda W,d)$. But this holds since such an element lies in the image of $\psi$ and is already exact in $(\Lambda V,d)$. Now by \cite[Theorem 3.1]{FM05}, $s$-formality of $(\Lambda W,d)$ already implies formality (note that the theorem therein is stated for closed manifolds, but the proof holds for rational Poincar\'e duality cdga's). 

Now, in general, if $(\Lambda V,d)$ is only a Sullivan cdga, it decomposes (uniquely) as a tensor product $(\Lambda V,d)\cong(\Lambda U,d)\otimes (\Lambda(C\oplus dC),d)$ of a minimal cdga with a contractible one \cite[Theorem 2.2]{Su77}. Note that the inclusion $\iota\colon (\Lambda U,d)\rightarrow (\Lambda U,d)\otimes (\Lambda(C\oplus dC),d)$ admits a retraction $r$ which is the identity on $U$ and sends $C\oplus dC$ to $0$. The pair $\iota,r$, satisfies the requirements of \mathbb{C}ref{lem:naturality} and thus induces a quasi--isomorphism $P_n(\Lambda U,d)\rightarrow P_n(\Lambda V,d)$.\end{proof}

\begin{prop}\label{prop:PDnotFormal}
Let $(A,d)$ be a cdga which admits a non-trivial Massey product. Then also $P_n(A,d)$ has a non-trivial Massey product. Namely, the image of a non-trivial Massey product $m$ in $(A,d)$ by the inclusion into the Poincar\'e dualization is also non-trivial.
\end{prop}

\begin{proof}
By construction, $P_nA=A\oplus D_n A$ with $A$ a subalgebra and $D_nA$ a differential ideal. Thus, the inclusion of cdga's $i:A\to P_nA$ admits a one-sided inverse map of cdga's  $r:P_n A\to A$ with $r\circ i=\operatorname{id}$. Now for any non-trivial Massey product $m\in H(A)$, $i(m)$ is non-trivial as $(r\circ i) (m)\mathfrak{su}bseteq m$. Recall, we treat a Massey product as the set of cohomology classes obtained via any possible defining system (see \cite{K66}), with the Massey product being trivial if the zero class is contained in this set.
\end{proof}

\begin{rem}\label{trivialMassey} Since we will implicitly use it in the next section, we remark on the following obvious property: if a Massey product $m$ is defined and trivial in $A$, then the same holds in $P_n(A)$ for any $n$, as $A$ embeds into $P_n(A)$. \end{rem}

\section{A quadruple Massey product pulling back trivially}\label{counterexample}
Our goal is to prove the following statement:

\begin{thm}\label{thm: counterex} There is a non-zero degree map of cohomologically connected rational Poincar\'e duality cdga's $P_1 \xrightarrow{f} P_2$ such that $P_1$ carries a non-trivial quadruple Massey product, which becomes trivial in $P_2$. That is, there are cohomology classes $[w],[x],[y],[z] \in H(P_1)$ such that the Massey product $\langle [w],[x],[y],[z] \rangle$ is defined and does not contain zero, and $0 \in \langle [f(w)], [f(x)], [f(y)], [f(z)] \rangle$. \end{thm}

We first construct cdga's with finite-dimensional cohomology $A,B$ (not satisfying Poincar\'e duality), together with a cdga morphism $f\colon A\rightarrow B$ and a differential graded $A$-module homomorphism $r\colon B\rightarrow A$ sending $1\mapsto 1$, such that $A$ carries a non-trivial quadruple Massey product which $f$ sends to a trivial one. Then Poincar\'e dualizing for large enough $n$ and applying \mathbb{C}ref{prop:PDnotFormal} (and the subsequent remark) will yield a map $P_n(A) \to P_n(B)$ with the desired properties.

\begin{rem}
In view of Taylor's result, see Proposition \ref{prop:Taylor}, the above datum $f,r\colon A\leftrightarrows B$ with $A$ carrying a non-trivial triple Massey product which becomes trivial in $B$ can not exist (in particular, there is no such $r$ in \mathbb{C}ref{noPD}). 

Indeed, consider a triple Massey product $\langle [x],[y],[z]\rangle$ in $A$. Then, as the Massey product $\langle [f(x)],[f(y)],[f(z)]\rangle$ vanishes in $B$, we can find a defining system $a,b\in B$ with $da= f(x)f(y)$, $db=f(y)f(z)$ such that $af(z)-(-1)^{|x|}f(x)b$ is exact. But then using that $r$ is a dg-$A$-module morphism we find that $r(a),r(b)$ is a defining system for $\langle [x],[y],[z]\rangle$ and the representing cocycle
\[r(a)z-(-1)^{|x|} x r(b)=r\left( af(z)-(-1)^{|x|}f(x)b\right) \] is exact.
This shows triviality of the Massey product $\langle [x],[y],[z]\rangle$. 

In order to motivate what is happening in the example we will construct for \mathbb{C}ref{thm: counterex}, it is rather instructive to check where the above argument fails for quadruple Massey products. To this end consider a quadruple Massey product $\langle [w],[x],[y],[z]\rangle$ in $A$. (For simplicity of notation, assume these classes are in even degrees.) As before, choose a defining system $a,b,c,g,h$ for $\langle [f(w)],[f(x)],[f(y)],[f(z)]\rangle$ such that $da=f(w)f(x)$, $db=f(x)f(y)$, $dc=f(y)f(z)$, $dg=af(y)-f(w)b$ and $dh=bf(z)-f(x)c$. While it still holds that $r(a),r(b),r(c),r(g),r(h)$ is a defining system for $\langle [w],[x],[y],[z]\rangle$ it is in general no longer true that the cocycles representing the Massey products get mapped to one another, i.e.\ we might have
\[ wr(h)+r(a)r(c)+zr(g)\neq r(f(w)h+ac+f(z)g)\]
if $r(ac)\neq r(a)r(c)$, which can happen since $r$ is not fully multiplicative. In particular the right hand side being exact does not force the left hand side to be so. In other words: while a non-trivial triple Massey product would obstruct the construction of the module retract $r$ in the counterexample below, the freedom of choosing $r(ac)$ will allow us to construct $r$ even in the presence of a non-trivial quadruple Massey product. \qed
\end{rem}

We begin with the construction of $A$. Set $(A,d):=(\Lambda (V^{\leq 5})\otimes \Lambda(V^{\geq 6}),d)$, where  $V^{\leq 5}=\langle X,Y,a,b,c,e,f,h,i\rangle$ with

\begin{center}
\begin{tabular}{ c | c | c }
degree & generators & differential \\ \hline
$2$ & $X,Y$ & $X,Y\mapsto 0$ \\
$3$ & $a,b,c$ & $a\mapsto X^2 \quad b\mapsto XY \quad c\mapsto Y^2$\\
$4$ & $e,f$ & $e\mapsto Ya-Xb\quad f\mapsto Yb-Xc$\\
$5$ & $h,i$ & $h\mapsto Xe+ab\quad i\mapsto Yf+bc$

\end{tabular}
\end{center}
and $V^{\geq 6}$ is a vector space  which we construct inductively in order to eliminate all cohomology in degrees $\geq 7$. To be precise, we first choose cycles representing a basis for degree $7$ cohomology. Then for each these element introduce a generator in $V^6$ and map it to the chosen cycle under the differential. The resulting algebra will have trivial degree $7$ cohomology while cohomology in degrees $\leq 6$ remains unchanged. Now repeat this process inductively for all higher degrees.

\begin{lem}\label{lem:generatorsinA}
The cohomology of $(A,d)$ is generated by the linearly independent cohomology classes of the cocycles $1,X,Y,m$, where $m=Ye+ac+Xf$. Furthermore the Massey product $\langle [X],[X],[Y],[Y]\rangle$ is non-trivial and represented by $[m]$.\end{lem}

\begin{proof}
Clearly $1,X,Y$ generate cohomology in degrees $\leq 2$. Furthermore $A^3=\langle a,b,c\rangle$ maps isomorphically onto $\Lambda^2 (X,Y)$ so there is no cohomology in degree $3,4$ in $\Lambda (X,Y,a,b,c)$. This changes in degree $5$, where the $\ker d$ is generated by $Ya-Xb$, $Yb-Xc$. Note that the corresponding cohomology classes do indeed form a basis of $H^5(\Lambda(X,Y,a,b,c),d)$, since $d$ vanishes on the degree $4$ span of the above generators. Thus after introducing $e,f$ we obtain $H^5(\Lambda (X,Y,a,b,c,e,f),d)=0=H^4(\Lambda (X,Y,a,b,c,e,f),d)$. At this point we compute that the degree $6$ part of $\ker d$ is $\langle Xe+ab,m,Yf+bc\rangle\oplus \Lambda^3(X,Y)$. The differential maps the degree $5$ span of the above generators onto $\Lambda^3(X,Y)$ so the cocycles in the left hand factor yield a basis for the cohomology at this stage, after introducing $h,i,V$ only the class of $m$ remains, generating $H^6(A)$. This proves the first part of the lemma. The reader can also verify this with the ``Commutative Differential Graded Algebras'' module in \cite{Sage}: 

{\tiny \begin{align*} &T.<X,Y,a,b,c,e,f,h,i> = \textrm{GradedCommutativeAlgebra(QQ, degrees = (2,2,3,3,3,4,4,5,5))} \\
&A = T.\textrm{cdg}\_{\textrm{algebra}}(\{a: X*X, b: X*Y, c: Y*Y, e: Y*a - X*b, f: Y*b - X*c, h: X*e + a*b, i: Y*f + b*c\}) \\
&[A.\textrm{cohomology}(i) \textrm{ for } i \textrm{ in } [1..7]] \end{align*} }

When writing down a defining system for the Massey product $\langle [X],[X],[Y],[Y]\rangle$, there is no choice for the primitives of $X^2,XY,Y^2$, except for $a,b,c$. When choosing primitives $p_1,p_2$ for the cocycles $Ya-Xb$ and $Yb-Xc$, we get $p_1=e+\alpha_1$, $p_2=f+\alpha_2$ for some $\alpha_i\in (\ker d)^4=\Lambda^2(X,Y)$. Then the resulting representative of $\langle [X],[X],[Y],[Y]\rangle$ is $m+Y\alpha_1+X\alpha_2$. Independent of the choice of $\alpha_i$, this is cohomologous to $m$. Hence we get a unique non-trivial cohomology class representing $\langle [X],[X],[Y],[Y]\rangle$.
\end{proof}

Now we come to the construction of $B$. Consider first the algebras $H_1=\Lambda( x,y)/\Lambda ^{\geq 2}(x,y)$ with $|x|=|y|=2$ and $H_2=\Lambda(\Tilde{a},\Tilde{c})$ with $|\Tilde{a}|=|\Tilde{c}|=3$. In low degrees, our algebra $B$ will be quasi-isomorphic to the product\footnote{i.e. the pullback in the category of augmented cdga's of the diagram $(H_1,0) \rightarrow (\mathbb{Q}Q,0) \leftarrow (H_2,0)$ of augmentation maps} of the augmented cdga's $H_1$ and $H_2$, i.e. $(H,d)=(H_1\oplus_\mathbb{Q} H_2,0)$, where $(H_1\oplus_\mathbb{Q}H_2)^k=H_1^k\oplus H_2^k$ for $k\geq 1$ and $(H_1\oplus_\mathbb{Q} H_2)^0=\langle (1,1)\rangle$, with multiplication $(n_1,n_2)\wedge(m_1,m_2)=(n_1m_1,n_2m_2)$. 

Consider $f_1\colon A\rightarrow H_1$ which sends $X\mapsto x$, $Y\mapsto y$ and all other generators to $0$. In order to check that this morphism is compatible with the differential we only need to check that for every generator $v$, we have $f_1(dv)=df_1(v)=0$, which is clearly satisfied (note that while we have not explicitly specified the differential on $V$, $d(V)$ lies in degrees $\geq 7$ and thus maps to $0$). In the same fashion we check that the algebra morphism $f_2\colon A\rightarrow H_2$ which sends $a\mapsto \Tilde{a}$, $c\mapsto \Tilde{c}$, and all other generators to $0$ is a morphism of cdga's. These two morphisms piece together to form the components of a cdga morphism $f=(f_1,f_2)\colon A\rightarrow H$, where e.g. $f(X) = (x, 0)$ and $f(a) = (0,\Tilde{a})$. We consider the relative minimal model of $f$, i.e. we build a commutative diagram

\[\xymatrix{ (A,d)\ar[r]^f\ar[dr]_\iota & (H,0)\\ & (A\otimes \Lambda W',d)\ar[u]_{\varphi}}
\]
with $\varphi$ inducing an isomorphism in cohomology and $\iota$ being the standard inclusion. Since $A^1=H^1=0$ there is an explicit degreewise inductive procedure (see \cite[§5]{Su77}) to compute $\Lambda W'$, $\varphi$, and the differential in terms of $A$ and $f$. We will need to do so up until degree $4$. The result is shown in the following table.

\begin{center}
\begin{tabular}{ c | c | c | c }
degree & generators & differential & image under $\varphi$ \\ \hline
$3$ & $\alpha,\gamma$ & $\alpha,\gamma\mapsto 0$ & $\alpha\mapsto\Tilde{a}\quad \gamma\mapsto\Tilde{c}$\\
$4$ & $s_{X\alpha},s_{Y\alpha},s_{X\gamma},s_{Y\gamma}$ & $s_{*}\mapsto *$ & $s_*\mapsto 0$\\

\end{tabular}
\end{center}

Let us briefly reason that this is indeed the beginning of a relative minimal model of $f$. We add new generators by induction over degree turning $\varphi$ into a quasi isomorphism. The original map $f$ is cohomologically injective as can be seen by applying it to the generators of $H(A)$ described in \mathbb{C}ref{lem:generatorsinA} (note that the element $m$ maps to $\Tilde{a}\Tilde{c}$). The only degree in which $f^*$ is not an isomorphism is in degree $3$ where $\operatorname{coker}(f^*)=\langle \Tilde{a},\Tilde{c}\rangle$ (as the differential on $A$ is injective on $\langle a,c \rangle$). Thus we introduce the generators $\alpha,\gamma$ to add to cohomology and map them onto $\Tilde{a},\Tilde{c}$. At this point $\varphi$ is cohomologically surjective so it remains to eliminate $\ker \varphi^*$. The introduction of $\alpha,\gamma$ has created new unwanted cohomology in degree $5$ in form of the classes of $X\alpha$, $Y\alpha$, $X\gamma$, and $Y\gamma$. Thus we introduce the generators $s_{X\alpha}$, $s_{Y\alpha}$, $s_{X\gamma}$, $s_{Y\gamma}$ to make these exact. 

We check that at this stage a basis of
$H^6(A\otimes \Lambda (\alpha,\gamma,s_{X\alpha},s_{Y\alpha},s_{X\gamma},s_{Y\gamma}),d)$ is represented by $m$ and the cocycles

\begin{center}
\begin{tabular}{l l l }
$v_1=m-\alpha\gamma,$ & $v_2=Ys_{X\alpha}-Xs_{Y\alpha},$ & $v_3=Ys_{X\gamma}-Xs_{Y\gamma},$\\
 $v_4=a\alpha-Xs_{X\alpha},$ & $v_5=b\alpha-Xs_{Y\alpha},$ & $v_6=c\alpha-Ys_{Y\alpha},$\\
 $v_7=a\gamma-Xs_{X\gamma},$ & $v_8=b\gamma-Xs_{Y\gamma},$
& $v_9=c\gamma-Ys_{Y\gamma}.$
\end{tabular}
\end{center}

It is at this point that we deviate from the construction of the relative minimal model of $f$, setting $B=A\otimes \Lambda W$ with $W^{\leq 4}=(W')^{\leq 4}$ and the rest of $W$ defined as follows: we introduce new generators $W^5=\langle t_1,\ldots,t_9\rangle$ in degree $5$ with $dt_i=v_i$. Then, we carry on indefinitely to define $W^{\geq 6}$ so that $H^{\geq 7}(B)=0$. We will not need to describe this last step explicitly.

\begin{rem}
Note, $B$ differs from the relative minimal model (even though it has cohomology $H$). In fact, $v_6$ and $v_7$ do not lie in the kernel of $\varphi^*$ but are exact in $B$ which prevents $\varphi$ from extending to $B$. A choice of $v_i$ that span the kernel of $\varphi^*$ is incompatible with the definition of the module retract below.
\end{rem}

Now we verify that the quadruple product $\langle [X], [X], [Y], [Y] \rangle$ becomes trivial in $B = A \otimes \Lambda W$ under the inclusion $A \xrightarrow{\iota} A \otimes W$:

\begin{lem} We have $0 \in \langle \iota [X], \iota [X], \iota [Y], \iota [Y] \rangle$. \end{lem}

\begin{proof} For simplicity we omit explicitly mentioning the inclusion map $\iota$. Choose $X, Y$ as representatives of $[X], [Y]$, and make the following choice of primitives: $d(a - \alpha) = X^2$, $db = XY$, $d(c+\gamma) = Y^2$. With these choices, the triple product $\langle [X], [X], [Y] \rangle$ is represented by $(a-\alpha)Y - Xb$, for which we make the choice of primitive $d(e-s_{Y\alpha}) = (a-\alpha)Y - Xb$. For the triple product $\langle [X], [Y], [Y] \rangle$, we have $d(f - s_{X\gamma}) = bY - X(c+\gamma)$ for the given representative. Hence the quadruple product, with these choices of primitives, is represented by $$ (e-s_{Y\alpha})Y + (a-\alpha)(c+\gamma) + X(f-s_{X\gamma}) = d(t_1 + t_6 + t_7). \qedhere $$
\end{proof}

It remains to construct a dg-$A$-module retract of the map $\iota\colon A\rightarrow A\otimes \Lambda W$. In order to do this we recall the following

\begin{defi}
Let $A$ be a dga and $(M,d)$ be a dg-$A$-module. Then a semi-free extension of $(M,d)$ is a dg-$A$-module of the form $(M\oplus (A\otimes V),d)$, where $V$ is a graded vector space and $d(1\otimes V)\mathfrak{su}bset M$.\end{defi}

For us this concept is helpful due to the following standard lemma. Part (1) is an immediate observation, while part (2) is a more explicit form of \cite[Lemma 14.1]{FHT12} which will prove useful when dealing with the explicit example.

\begin{lem}\label{lem:dgmodules}
\begin{enumerate}
    \item Let $f\colon M\rightarrow N$ a morphism of dg-$A$-modules and $(M\oplus (A\otimes V),d)$ a semi-free extension of $M$. Let $(v_i)_{i\in I}$ be a basis of $V$, and let $(\alpha_i)_{i\in I}$ be a collection of elements in $N$ with $d\alpha_i= f(d v_i))$. Then $f$ extends to a morphism of dg-$A$-modules $M\oplus (A\otimes V) \to N$ by setting $f(v_i)=\alpha_i$.
    \item Let $A\rightarrow A\otimes \Lambda W$ be a relative minimal cdga with $A^1=W^1=0$. For $0\leq j\leq i$, set $V_{(i,j)}=(\Lambda^{i-j}W)^{2i-j}$. Then $\Lambda W=\bigoplus_{0\leq j\leq i} V_{(i,j)}$ and for any $(i,j)$ as above the inclusion
    \[A\otimes \left(\bigoplus_{(k,l)<(i,j)} V_{(k,l)}\right)\rightarrow A\otimes \left(\bigoplus_{(k,l)\leq(i,j)} V_{(k,l)}\right)\]
    is a semi-free extension, where we use the lexicographical order on tuples.
\end{enumerate}
\end{lem}

\begin{proof}
Part (1) is straightforward verification. For the proof of part $(2)$ we observe that due to $W=W^{\geq 2}$ we indeed have \[\Lambda W=\bigoplus_{0\leq 2k\leq l} (\Lambda^k W)^{l}= \bigoplus_{0\leq j\leq i} (\Lambda^{i-j}W)^{2i-j}.\] 
It remains to check that $d(V_{(i,j)})\mathfrak{su}bset A\otimes \left(\bigoplus_{(k,l)<(i,j)} V_{(k,l)}\right)$. To see this, we investigate the differential with respect to its bidegree $A\otimes ((\Lambda^p W)^q)$, where $p$ is the wordlength degree in $W$ and $q$ is the cohomological degree in $\Lambda W$.
If $p$ does not increase then $q$ decreases by at least $1$ due to minimality and $A^1=0$. Furthermore $p$ can decrease by at most $1$ in which case $q$ decreases by $2$ since $d(W)\cap A$ lies in degrees $\geq 3$. Consequently
\begin{align*}
    d\left((\Lambda^{i-j} W)^{2i-j}\right)\mathfrak{su}bset A\otimes \left((\Lambda^{\geq i-j+1} W)^{\leq 2i-j+1}\oplus  (\Lambda^{i-j} W)^{\leq 2i-j-1}\oplus (\Lambda^{ i-j-1} W)^{\leq 2i-j-2}\right)
    \end{align*}
which proves the claim.
\end{proof}

Thus by this lemma, in order to define the retraction $r\colon A\otimes \Lambda W\rightarrow A$ it suffices to inductively specify images of a suitable basis of $\Lambda W$ and extend $A$-linearly. In fact, even less is sufficient by the following:
\begin{lem}\label{lem:extension} Any morphism
\[r\colon A\otimes \left(\bigoplus_{(i,j)\leq (4,3)} V_{(i,j)}\right)\rightarrow A\]
of dg-$A$-modules extends to $A\otimes \Lambda W$.
\end{lem}
 \begin{proof}
 Recall that by part (1) of \mathbb{C}ref{lem:dgmodules} the only obstruction to extend $r$ over a new generator $v$ is that the class $[r(dv)]\in H^{*}(A)$ has to vanish. By definition, for $(i,j)>(4,3)$ the space $V_{(i,j)}$ is concentrated in cohomological degrees $\geq 6$ (since $V_{(i,i)}=0 $ for $i\neq 0$) while $H(A)$ is concentrated in degrees $\leq 6$.
 \end{proof} 

Furthermore note that for $(i,j)\leq (4,3)$, we have $V_{(i,j)}\mathfrak{su}bset \Lambda (W^{\leq 5})$ which means we have already computed all the required algebra generators. We define $r$ according to the following table, where we list all non-trivial $V_{(i,j)}$ with $(i,j)\leq (4,3)$ in their order of occurrence.

\begin{center}
\begin{tabular}{ c | c | c }
extension & generators & image under $r$ \\ \hline
$V_{(0,0)}$ & $1$ & $1\mapsto 1$\\
$V_{(2,1)}$ & $\alpha,\gamma$ & $\alpha,\gamma\mapsto 0$\\
$V_{(3,1)}$ & $s_{X\alpha},s_{Y\alpha},s_{X\gamma},s_{Y\gamma}$ & $s_{*}\mapsto 0$\\
$V_{(4,2)}$ & $\alpha\gamma$ & $\alpha\gamma\mapsto m$
\\
$V_{(4,3)}$ &  $t_1,\ldots,t_9$ & $t_i\mapsto 0$
\end{tabular}
\end{center}

One checks that indeed for any of the generators $v$ above we have $r(dv)=dr(v)$. Then by \mathbb{C}ref{lem:dgmodules} and \mathbb{C}ref{lem:extension} we obtain the desired retraction $r\colon A\otimes \Lambda W\rightarrow A$.

In conclusion, applying the Poincar\'e dualization construction, together with \mathbb{C}ref{prop:PDnotFormal} and \mathbb{C}ref{trivialMassey}, we obtain a non-zero degree map of simply connected Poincar\'e duality algebras with the desired properties, as long as $n \geq 8$. This can then be realized by a map of (simply connected) spaces \cite{Su77} which satisfy rational Poincar\'e duality.

\begin{rem}
If one could take $B$ in the above construction to be the relative model for $A\to H$, one would obtain a formal space dominating a non-formal one. For the purposes of \mathbb{C}ref{thm: counterex} as it is now, it may be more transparent, though somewhat more ad hoc, to consider the following variant of the above construction more in the spirit of \mathbb{C}ref{noPD}. Define
$A':=\Lambda(X,Y,a,b,c,e,f)$ with the following degrees and differential:
\begin{center}
\begin{tabular}{ c | c | c }
degree & generators & differential \\ \hline
$2$ & $X,Y$ & $X,Y\mapsto 0$ \\
$3$ & $a,b,c$ & $a\mapsto X^2 \quad b\mapsto XY \quad c\mapsto Y^2$\\
$4$ & $e,f$ & $e\mapsto Ya-Xb\quad f\mapsto Yb-Xc$.
\end{tabular}
\end{center}
 Denote by $m:=Xf+ac+Ye$ a representative for the quadruple Massey product $\langle [X],[X],[Y],[Y] \rangle$. Set $A:=A'/(A')^{\geq 7}$. Next, define $B'=A\otimes\Lambda W$ with $
W = \mathrm{span}\langle \alpha,\gamma,g,h,\Omega\rangle$
with the following degrees and differential:
\begin{center}
\begin{tabular}{ c | c | c }
degree & generators & differential \\ \hline
$3$ & $\alpha,\gamma$ & $\alpha,\gamma\mapsto 0$ \\
$4$ & $s_{Y\alpha},s_{X\gamma}$ & $s_{Y\alpha} \mapsto Y\alpha\quad s_{X\gamma} \mapsto X\gamma$\\
$5$ & $\Omega$ & $\Omega\mapsto m-Xs_{X\gamma}-Ys_{Y\alpha}+a\gamma-\alpha c-\alpha\gamma$
\end{tabular}
\end{center} 
Now, set $B:= B'/I$ where $I$ is the ideal generated by $(\Lambda W)^{\geq 7}$. Then, as an $A$-module, $B$ is free of rank $6$:
\[
B=A\oplus A\alpha\oplus A\gamma\oplus A s_{Y\alpha} \oplus A s_{X\gamma} \oplus A\Omega\oplus A\alpha\gamma .
\]
We can define a module retract $r:B\to A$ by sending all but the first and last summand to zero, sending $1\mapsto 1$ and $\alpha\gamma\mapsto m$. This map of modules is compatible with the differential, i.e. $rd=dr$. Indeed, this is clear on the first three summands. On the last four, for any $u\in A$, one has
\begin{align*}
    rd(u\otimes s_{Y\alpha})&=r(du\otimes s_{Y\alpha} + (-1)^{|u|} uY\otimes \alpha)=0=dr(u\otimes s_{Y\alpha}),\\
    rd(u\otimes s_{X\gamma})&=r(du\otimes s_{X\gamma} + (-1)^{|u|} uX\otimes \gamma)=0=dr(u\otimes s_{X\gamma}),\\
    rd(u\otimes \Omega)&=r\left( du\otimes \Omega + (-1)^{|u|} (um\otimes 1 - uX\otimes s_{X\gamma} - uY\otimes s_{Y\alpha} + ua\otimes \gamma + uc\otimes \alpha - u\otimes\alpha\gamma)\right) \\ &= (-1)^{|u|}(um - um) = 0=dr(u\otimes \Omega),\\
    rd(u\otimes \alpha\gamma)&=r(du\otimes\alpha\gamma)=(du) m=dr(u\otimes\alpha\gamma).
\end{align*}

The Massey product $\langle [X], [X], [Y], [Y] \rangle$ vanishes on $B$. Indeed, choose primitives $d(a-\alpha) = X^2, db = XY, d(c+\gamma) = Y^2$. Then the triple product $\langle [X], [X], [Y] \rangle$ is represented by $aY - Xb - \alpha Y = d(e-s_{Y\alpha})$, and $\langle [X], [Y], [Y] \rangle$ is represented by $bY - Xc - X\gamma = d(f-s_{X\gamma})$. With these choices of primitives, the quadruple product is represented by $X(f-s_{X\gamma}) + (a-\alpha)(c+\gamma) + (e-s_{Y\alpha})Y = d\Omega$.

Poincar\'e dualizing the pair $(A,B)$ with respect to the inclusion and $r$ gives another example as in \mathbb{C}ref{thm: counterex}.
\end{rem}

\begin{rem}
If one were to find a non-zero degree map of rational Poincar\'e duality cdga's $P_1 \to P_2$ with non-formal domain and formal target, then this could be upgraded to a non-zero degree map of closed manifolds with formal domain and non-formal target, at least if the formal dimension is not divisible by four, or if $P_1$ and $P_2$ are obtained by Poincar\'e dualization of some algebra. 

In fact, such a map of cdga's can be realized, up to a lift of a non-zero grading automorphism on the cohomology of $P_2$, by a non-zero degree smooth map of closed manifolds. First of all, there is a topological map of simply connected rational spaces $M_{\mathbb{Q}Q} \to N_{\mathbb{Q}Q}$ corresponding to this map of cdga's. For this, one checks that the algorithms for building a minimal model and the induced map between the models \cite[Section 5]{Su77}, \cite[Proposition 12.9]{FHT12}, usually stated for connected cdga's, work essentially verbatim for cohomologically connected cdga's (we already made use of the former in \mathbb{C}ref{prop:PDagainFormal}). Thus, we may replace the map between $P_1$ and $P_2$ by one between connected cdga's and hence by one between rational spaces. If the formal dimension is not divisible by four, we can apply the realization theorem \cite[Theorem 13.2]{Su77} to obtain simply connected closed smooth manifolds $M$ and $N$ with rationalization maps $M \to M_{\mathbb{Q}Q}$ and $N \to N_{\mathbb{Q}Q}$. If $P_1$ and $P_2$ are Poincar\'e dualizations, then one can apply the realization theorem in any dimension, by choosing all rational Pontryagin classes to vanish. Namely, the signature of a Poincar\'e dualization is zero and its intersection form (if the dimension is divisible by four) is of the form $\mathfrak{su}m_i \pm x_i^2$, as can be easily verified.

We are thus led to consider the following lifting problem:
$$
\begin{tikzcd}
M \arrow[d] \arrow[r, dashed] & N \arrow[d] \\
M_{\mathbb{Q}Q} \arrow[r]             & N_{\mathbb{Q}Q}    
\end{tikzcd}
$$

The obstructions to lifting (up to homotopy) the composite $M \to N_{\mathbb{Q}Q}$ lie in $H(M; \pi_{*-1}(F))$, where $F$ is the homotopy fiber of the map $N \to N_{\mathbb{Q}Q}$. Since this map is a rationalization, the homotopy groups of $F$ are torsion. Now, since $M$ has the homotopy type of a simply connected finite complex, and is formal, there are sufficiently many endomorphisms of $M$ that allow us to surpass the obstructions to lifting, upon precomposing with such an endomorphism \cite[Theorem 12.2]{Su77} (we remark that there is an implicit nilpotency assumption on the space therein). We refer to \cite[Corollary 4.3]{A15} for full details of the argument. To see that $M_{\mathbb{Q}Q}$ satisfies the hypothesis of \cite[Corollary 4.3]{A15}, i.e. that it admits a self-map with $k^{\textrm{th}}$ multiples for all large enough $k$, see the proof of \cite[Corollary 4.4]{A15}. Namely, the map in question is a lift of any grading automorphism on the cohomology; such a lift exists by formality \cite[Theorem 12.7]{Su77}.

Ultimately we obtain a non-zero degree endomorphism $f$ of $M$ and a homotopy commutative diagram

$$
\begin{tikzcd}
M \arrow[r, "f"'] \arrow[rr, bend left, shift left] \arrow[d] & M \arrow[d]       & N \arrow[d] \\
M_{\mathbb{Q}Q} \arrow[r]                                             & M_{\mathbb{Q}Q} \arrow[r] & N_{\mathbb{Q}Q}    
\end{tikzcd}
$$ giving us the desired map of closed manifolds.

\end{rem}

\iffalse

Some additional possibly relevant papers:

Reid, A.W. and Wang, S., 1999. Non-Haken 3-manifolds are not large with respect to mappings of non-zero degree. Communications in Analysis and Geometry, 7(1), pp.105-132.

https://edoc.ub.uni-muenchen.de/17204/1/Neofytidis_Christoforos.pdf  

Lafont, J.F. and Neofytidis, C., 2019. Steenrod problem and the domination relation. Topology and its Applications, 255, pp.32-40.

Connell, C. and Farb, B., 2003. The degree theorem in higher rank. Journal of Differential Geometry, 65(1), pp.19-59.

Kotschick, D., Löh, C. and Neofytidis, C., 2016. On stability of non-domination under taking products. Proceedings of the American Mathematical Society, 144(6), pp.2705-2710.

Zhang, W., 2017. Geometric structures, Gromov norm and Kodaira dimensions. Advances in Mathematics, 308, pp.1-35.

Neofytidis, C. and Zhang, W., 2021. Geometric structures, the Gromov order, Kodaira dimensions and simplicial volume. Pacific Journal of Mathematics, 315(1), pp.209-233.

\fi

\end{document}