## When is a quasi-isomorphism necessarily a homotopy equivalence?

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 2011 at 6:27