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Under what circumstances is a quasi-isomorphism between two complexes necessarily a homotopy equivalence? For instance, this is true for chain complexes over a field (which are all homotopy equivalent to their homology). It's also true in an $\mathcal{A}_\infty$ setting.

Is it true for chain complexes of free Abelian groups? The case I'm particularly interested in is chain complexes of free $(\mathbb{Z}/2\mathbb{Z})[U]$ modules or free $\mathbb{Z}[U]$ modules, but I'm also interested in general statements.

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Equivalent reformulation, considering the cone of the quasi-isomorphism: under what circumstances is an acyclic complex a split acyclic complex (i.e. spliced together from split short exact sequences)? True for complexes of projectives bounded to the right and, dually, for complexes of injectives bounded to the left. In free Z/4-modules, the unbounded complex .. -> Z/4 -2-> Z/4 -2-> Z/4 -> ... is acyclic, but not split acyclic. –  Matthias Künzer Mar 24 '11 at 6:27

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If your complexes are bounded, this is always true for any ring more generally replacing free modules with projectives. The statement is that D^b(A-mod) is equivalent to Ho(Proj-A mod) and you can find it in Weibel Chapter 10.4. If your complexes are unbounded things are more tricky. Then your statement is true in over any ring of finite homological dimension. Basically you have two notions K-projective(which have the property that you want) and complexes of projectives. Bounded complexes of projectives are K-projective, but unbounded ones are not unless you have the finiteness hypothesis(see Matthias' answer). See this post for the injective version of this story Question about unbounded derived categories of quasicoherent sheaves. In the cases you are interested in there is no problem.

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