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Mar 13, 2012 at 18:23 vote accept Jan Weidner
Mar 12, 2012 at 22:47 history edited David White CC BY-SA 3.0
Fixed typo
Mar 12, 2012 at 19:12 answer added Leonid Positselski timeline score: 9
Mar 12, 2012 at 18:18 comment added Fernando Muro There's a very basic lemma on homological algebra which says that projective resolutions are essentially unique, in the sense that two of them are homotopy equivalent, therefore their endomorphism DG-algebras become homotopy equivalent too (through a zig-zag of equivalences, though). The same holds for injective resolutions by the same reason that $Ext$ can be computed both kinds of resolutions.
Mar 12, 2012 at 18:04 comment added MTS Ah, I see. Thanks for the clarification. I don't know the answer, but it's an interesting question for sure.
Mar 12, 2012 at 17:24 comment added Jan Weidner I guess one can do both. I mean the total complex of the double complex $Hom(P,P)$. On $Hom(P,M)$ the dg-algebra structure is not so obvious to me, thats why I have choosen $Hom(P,P)$.
Mar 12, 2012 at 17:04 comment added MTS In the construction of the Ext-algebra, don't you form the Hom complex $Hom^\bullet (P,M)$ and then take homology? It seems like $Hom^\bullet (P,P)$ would be a double complex...
Mar 12, 2012 at 16:39 history asked Jan Weidner CC BY-SA 3.0