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  "body": "\n## Definition\n\nLet $\\mathcal{A}$ be an [[abelian category|Y9CZ3MTNF1]] and let $f \\colon C_\\bullet \\to D_\\bullet$ be a [[chain map|HTK9WQZJ1G]] of [[chain complexes|WQ82C0M3QB]] in $\\mathcal{A}$.\nThe map $f$ is a *quasi-isomorphism* if the induced morphism on [[homology|2J4JJJYP22]]\n$$H_n(f) \\colon H_n(C) \\to H_n(D)$$\nis an isomorphism for every integer $n$.\n\n## Remarks\n\nEvery chain homotopy equivalence is a quasi-isomorphism, because [[chain homotopic|4EBHAWJJ8C]] maps induce the same maps on homology.\nQuasi-isomorphisms are closed under composition, but in general a quasi-isomorphism has no chain map inverse up to homotopy.\n\nExample: let $C_\\bullet$ have $C_1 = C_0 = \\Z$ with $d_1$ multiplication by $2$, and let $D_\\bullet$ be $\\Z/2\\Z$ in degree $0$.\nLet $f_0 \\colon \\Z \\to \\Z/2\\Z$ be the reduction and $f_1 = 0$; this is a chain map since $f_0(2x) = 0$.\nBoth complexes have homology $\\Z/2\\Z$ in degree $0$ and $0$ elsewhere, and $H_0(f)$ sends the class of $1$ to $1$, so $f$ is a quasi-isomorphism.\nIt is not a chain homotopy equivalence: every chain map $D_\\bullet \\to C_\\bullet$ is $0$, because $\\Z/2\\Z$ has no nonzero map to $\\Z$.\n\nNon-example: the zero chain map from $\\Z$ in degree $0$ to itself is not a quasi-isomorphism, since $H_0$ of it is the zero map $\\Z \\to \\Z$.\n",
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