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I ask for help in understanding the following.

Let $\mathcal{D}$ be the derived category of $k$-vector spaces. Assume $M$ is an element of this category; then, we can choose a quasi isomorphism $$\Psi\colon \bigoplus_iH^i(M)[-i]\rightarrow M, $$ that is, in the derived category the complex $M$ is isomorphic (in a non canonical way) to its cohomology complex with zero differentials.

Assume now that our complex $M$ is equipped with a product structure, i.e. a chosen class $\mu\in\hom_{\mathcal{D}}(M\otimes M, M)$. It induces products $\mu_{i,j}\colon H^i(M)\otimes H^j(M)\rightarrow H^{i+j}(M)$.

My question is: is it possible to choose a quasi isomorphism $\Psi$ such that
$\Psi\circ (\oplus_{i,j}\mu_{i,j})=\mu\circ (\Psi\otimes\Psi) $ as elements of $\hom_{\mathcal{D}}(\bigoplus_{i}H^i(M)[-i]\otimes \bigoplus_{j}H^j(M)[-j], M)$ ?

A naive idea about to tackle this question is:

By the above we get a decomposition, depending on $\Psi$, \begin{align*} \hom_{\mathcal{D}}(M\otimes M,M) & \cong \hom_{\mathcal{D}}({\bigoplus_iH^i(M)[-i]\otimes \bigoplus_jH^j(M)[-j],\bigoplus_kH^k(M)[-k]})\\ &=\bigoplus_{i,j,k}\hom_{\mathcal{D}}(H^i(M)[-i]\otimes H^j(M)[-j], H^k(M)[-k]) \\ & =\bigoplus_k \bigoplus_{i+j=k}\hom_{\mathcal{D}}(H^i(M)\otimes H^j(M),H^k(M)). \end{align*} The last equality is due to the vanishing of higher Ext in the category of vector spaces. If we require $\Psi$ to induce the identity in cohomology, which may be easily achieved, then, necessarily, $\mu$ corresponds to $\oplus_i\oplus_{i+j=k} \mu_{j,k}$. Isn't this saying that we have always a positive answer to the question above ?

Side question: how many quasi isomorphisms $\Psi$ are there? Given one of them, we get a decomposition \begin{align*} \hom_{\mathcal{D}}(M,M) & \cong \hom_{\mathcal{D}}({\bigoplus_iH^i(M)[-i],\bigoplus_jH^j(M)[-j]})\\ &=\bigoplus_{i.j}\hom_{\mathcal{D}}(H^i(M)[-i], H^j(M)[-j]) \\ & =\bigoplus_i \hom_{\mathcal{D}}(H^i(M),H^i(M)), \end{align*} the last equality again standing because there are no higher Ext between $k$-vector spaces. In Deligne's old paper it is proved (Corollary 1.12) that any such $\Psi$ is uniquely determined by pairwise orthogonal projectors $Z_i\colon M\rightarrow M$, each $Z_i$ acting in cohomology as the identity in degree $i$ and as zero in the other degrees. By the decomposition above, these are uniquely determined.

Comments: I think that there must be some mistake in my reasoning. Wouldn't my answer to the first question prove that any dg-algebra over $k$ is formal? I am pretty confused at the moment about this, I hope somebody could clearify the situation for me.

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    Thank you for your comment, but I am note sure of what you exactly mean.2017-02-24
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    Thank you for your comment. What do you exactly mean with this? My point is that if the multiplication $\mu$ is only defined in the derived category (so, up to quasi-isomorphisms), then it is determined by the induced multiplication on cohmology.2017-02-24
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    And now I see the diference between asking for formality of a dg-algebra: there the multiplication is already defined on the complex, therefore one needs to work within the derived category of algebras.2017-02-24
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    Sorry, I misunderstood your question. It is simply not the case that $\Psi$ inducing an isomorphism on cohomology implies that it sends the multiplication on cohomology to the multiplication on $M$, and your argument simply does not establish this (as you say, if it did then every dg algebra would be formal).2017-02-24
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    Well, I am not sure about this. for a dg-algebra (i.e. a complex with a graded algebra structure), formality means to have a chain of quasi isomorphism of dg algebras, not only a quasi isomorphism of complexes preserving the product. The first problem involves a much more rigid structure2017-02-24

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