I want to find some method to do the following: given an DGA-module $M$ over some commutative ring $k$, positively graded ($M_i=0$ if $i<0$), where each component $M_i$ has a free action by the symmetric group $\mathbb{S}_i$, and $k$-free, then there is an acyclic DGA-module $N$ extending $M$ ($M\subset N$), with $N_0=M_0$ and also with free actions by the symmetric groups and $k$-free in each degree. I think it is possible and probably someone already do it, but i can't find articles or books talking about it. I know there are methods to construct free resolutions of modules(which correspond to the case when $M_i=0$ for $i>0$). Thanks!
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$\begingroup$ What's wrong with the cone? $\endgroup$– Fernando MuroCommented Jun 11, 2016 at 8:03
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$\begingroup$ Thanks for the hint, but i don't see the relation. $\endgroup$– emmagvrCommented Jun 11, 2016 at 13:13
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$\begingroup$ I mean the standard inclusion into the cone $M\subset CM$. What fails with this? Am I missing something? $\endgroup$– Fernando MuroCommented Jun 13, 2016 at 21:08
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$\begingroup$ The problem is that $CM$ is not acyclic, it is free and $H_*(CM)=0$, but $N$ needs to have the homology of a point. $\endgroup$– emmagvrCommented Jun 17, 2016 at 11:02
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1$\begingroup$ For me, acyclic means trivial homology. $\endgroup$– Fernando MuroCommented Jun 17, 2016 at 14:17
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