Morphisms between formal dg-algebras

Suppose we are given a map $f:A \rightarrow B$ between two dg-algebras which are formal. Is the map $f$ also "formal" in some sense?

More precisely can we find isomorphisms $\phi_A:A\rightarrow H^\bullet(A)$ and $\phi_B:B\rightarrow H^\bullet(B)$ in the derived category of dg-algebras such that

$$H^\bullet(f) \circ \phi_A= \phi_B \circ f$$ holds in the derived category of dg-algebras?

I fear that this may be wrong in general. What would be a counter example? Are there criteria which ensure that $f$ is "formal" in this sense?

Let me give you an example which has a topological flavour. Let us consider, the De-Rham complex of differential forms on a sphere $S^n$, we denote it $A^*(S^n)$, it is a formal commutative differential graded algebra. Let us look at the morphisms of commutative differential graded algebras between $A(S^2)$ and $A(S^3)$ into the derived category.

A good way to compute them is to take a cofibrant replacement of $A(S^2)$, such a cofibrant resolution is given by the algebra $(\mathbb{R}[x_2]\otimes\Lambda(x_3),D)$ where the differential satisfies $D(x_2)=0$ and $D(x_3)=x_2^2$.

And as $A(S^3)$ is formal you can replace it by its cohomology algebra which is the exterior algebra $\Lambda(y_3)$. It is easy to see that a morphism of algebra $\phi$ between these two algebras is completely determined by the image of $x_3$ thus by a real number $\lambda$ such that $\phi(x_3)=\lambda.x_3$. We have proved: $$[A(S^2),A(S^3)]_{CDGA}\cong\mathbb{R},$$
while $H(\phi)=0$.

This example has a topological flavour because it is a way (not the best of course) to show that $\pi_3(S^2)\otimes \mathbb{R}\cong \mathbb{R}$.

People in rational homotopy theory have studied the concept of formality for morphisms:

• Vigué-Poirrier, Micheline Formalité d'une application continue. C. R. Acad. Sci. Paris Sér. A-B 289 (1979), no. 16

• Félix, Yves; Tanré, Daniel Sur la formalité des applications. Publ. U.E.R. Math. Pures Appl. IRMA 3 (1981), no. 2, exp. no. 1, 45 pp.

• Oprea, John F. DGA homology decompositions and a condition for formality. Illinois J. Math. 30 (1986), no. 1, 122–137.

Edit: if you want to play with dgas, then the only thing that you have to change in the example above is the cofibration resolution of the algebra $A(S^2)$. You start with $(T(x_2,x_3),D)$ the tensor algeba on $2$ generators with differential $D(x_2)=0$ and $D(x_3)=x_2\otimes x_2$ it has the same cohomology as $A(S^2)$ until the differential degree $5$ where you have created a new cycle :$[x_3,x_2]=x_3\otimes x_2-x_2\otimes x_3$ which is not a boundary. Then you have to add a generator $x_4$ to kill this cycle i.e. you put $D(x_4)=[x_3,x_2]$. And so on and so forth, it is easy to check that all the generators you add to $(T(x_2,x_3),D)$ in order to build a cofibrant resolution of $A(S^2)$ will have degree at least $4$.

With this resolution in hand you check that a homotopy class of morphism of dgas is also completely determined by the image of $x_3$.

• Thanks for this nice example! Its only slight disadvantage is that it lives in the category of commutative dgas. Do you know a simple counterexample in the category of dgas? – Jan Weidner Nov 28 '12 at 15:40
• I have edited my answer to take into account your remark. – David C Nov 28 '12 at 16:11