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Let $A$ be a differential graded algebra, $S\subset H^*(A)$. I would like to 'kill $S$ in a canonical way'. Is it possible to do it as follows: consider the $A_\infty$-algebra structure on $H^\ast(A)$, and factorize $H^*(A)$ by the '$A_\infty$-ideal' generated by $S$ (i.e. kill in $H^\ast(A)$ all sums of elements that can be obtained via $m_i$ taking elements of $S$ as one of the arguments)? Does this quotient possess any nice 'universal' properties? What other reasonable definitions of $H^\ast(A)/\langle S\rangle$ can one consider here?

I would like to concider the (triangulated?) category of $A_\infty$-modules over $H^\ast(A)/\langle S\rangle$; is there a way to describe it 'explicitly'? What is the relation between $A$-modules and $H^\ast(A)/\langle S\rangle$-ones? Can one define a functor by $A-\mod\cong H^\ast(A)-\mod \stackrel{\otimes_ {H^\ast(A)}H^\ast(A)/\langle S\rangle}{\to} H^\ast(A)/\langle S\rangle-mod$; does it possess any nice universal properties?

What are the 'canonical' references for these matters (in particular, for factorizing $A_\infty$-algebras modulo ideals)?

Upd. Unfortunately, I would like to consider an algebra that is not skew-commutative; it is closely related with a 'complicated' DG-category.

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Are you looking for the resulting map on cohomology to be some kind of quotient map? Are you looking for the quotient to have some kind of universal property? Are you looking for something else? For many definitions of "quotient" there is almost always a nontrivial moduli space of them. –  Tyler Lawson Jan 26 '13 at 4:56
    
I would expect the quotient to have some kind of universal property; yet I am not sure that this is possible. I don't know much about these things; any hints or references would be very welcome! What is 'the best' quotient in your opinion? –  Mikhail Bondarko Jan 26 '13 at 5:02
    
Also, I am rather interested in the category of modules over the algebra obtained than in the algebra itself. –  Mikhail Bondarko Jan 26 '13 at 5:06
    
Mikhail: The shortest quotient would be to take some kind of homotopy pushout; the category of modules would then be some kind of category of dg $A$-modules $M$ equipped with, for each $s \in S$, a chain homotopy from multiplication-by-$s$ to zero. But this only has a "versal" property, and it depends on the actual set of generators (e.g. if you use different generators or include redundant generators in $S$) you get a different and likely inequivalent algebra. This has kind of lousy properties. My own feeling is that quotients are just hard in the derived setting. –  Tyler Lawson Jan 26 '13 at 5:22
    
If you are in a good setting (maybe you have something commutative enough, with an ideal given by a regular sequence) then you can often construct a quotient that looks nice on the cohomology level, but there may multiple inequivalent ways to do it. –  Tyler Lawson Jan 26 '13 at 5:28
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