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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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    $\begingroup$ 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. $\endgroup$
    – Matthias Künzer
    Commented Mar 24, 2011 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 $\mathrm{D}^b(A\text{-}mod)$ is equivalent to $\mathrm{Ho}(Proj\text{-}A)$ 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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  • $\begingroup$ Following answer by Daniel, this two complexes here in the first answer math.stackexchange.com/questions/93273/… are bounded and quasi-isomorphic but not homotopy equivalent. Should there be further conditions? $\endgroup$
    – Arun
    Commented Sep 26, 2015 at 11:05
  • $\begingroup$ One of the complexes you mention is not made up of projective modules as is required above. The above chapter has all of the details and I recommend consulting it. $\endgroup$
    – Daniel Pomerleano
    Commented Sep 27, 2015 at 11:21

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