Massey products and $A_{\infty}$ structures

Question

I know the general theorem of Kadeishvili which says that, for a DGA $C$, when $H^{i}(C)$, $i\geq 0$, is free, $H(C)$ can be made into an $A_{\infty}$ algebra. If my understanding is correct, the proof essentially uses the freeness of $H^{i}(C)$ to produce a section $$s:H(C)\to C$$ of the projection $p:C\to H(C)$. Then a choice of chain homotopy between the identity and $sp$ can be used to define the maps $$m_{i}:\otimes^{i}H(C)\to H(C).$$ It is not hard to see that these give you an element of the higher Massey products (when defined).

Here is my question: When $H^{i}(C)$ is not free, is there any good way to define an $A_{\infty}$ structure on $H(C)$? It seems that the fact that one has an actual chain homotopy is crucial to the construction and in general $p$ may only be a quasi-iso. I hope this question is not too elementary (I am still a lowly graduate student). It just seems odd to me that the higher Massey products can be defined in general, even for torsion $H(C)$ (as long as the lower ones vanish) but one may not be able to identify these with a full $A_{\infty}$ structure.

Answer 1

Let $C = \Bbb Z[x,y] \otimes \Lambda[u,v]$, with $x$ and $y$ in (homological) degree 2 and $u$ and $v$ in degree 3, with $dx = dy = 0$ and $du = 2x$, $dv = 2y$. Then $H_5(C)$ is $\Bbb Z/2$, generated by the Massey product $x v - u y = \langle x,2,y\rangle$ (with no indeterminacy in this case).

This structure does not come from the homology equipped with an $A_\infty$ structure, because then we would have $$\langle x,2,y \rangle = m_3(x \otimes 2 \otimes y) = 2 \cdot m_3(x \otimes 1 \otimes y) = 0.$$

The usual generalization clarifies that the original could be viewed as, instead of a theorem about homology, a theorem about chain equivalence: if $C \to D$ is a chain homotopy equivalence and $C$ is a DGA, then $D$ gets an $A_\infty$ structure.