Let \(R\) be a ring and \(C_\bullet\), \(D_\bullet\) chain complexes of \(R\)-modules. A chain map \(f\colon C_\bullet\to D_\bullet\) is a quasi-isomorphism if the induced map \(H_n(f)\colon H_n(C_\bullet)\to H_n(D_\bullet)\) is an isomorphism for every \(n\in\ZZ\).
Every chain homotopy equivalence is a quasi-isomorphism.
A quasi-isomorphism need not be a chain homotopy equivalence. Let \(C_\bullet\) be the complex of abelian groups with \(C_1 = \ZZ\), \(C_0 = \ZZ\), differential \(C_1\to C_0\) multiplication by \(2\), and all other terms zero, and let \(D_\bullet\) be \(\ZZ/2\) concentrated in degree \(0\). The reduction map \(C_0 = \ZZ\to\ZZ/2 = D_0\) is a chain map inducing an isomorphism \(H_0(C_\bullet) = \ZZ/2\to H_0(D_\bullet)\), and all other homology groups vanish, so it is a quasi-isomorphism. Every chain map \(D_\bullet\to C_\bullet\) is zero, since \(\Hom(\ZZ/2, \ZZ) = 0\), so it induces the zero map on \(H_0\) and cannot be a homotopy inverse.
The derived category of \(R\)-modules is obtained from the category of chain complexes by formally inverting the quasi-isomorphisms.
[@Wei94].