---
tag: 7E7W0K3N8W
kind: definition
name: Quasi-isomorphism
msc: [18G35, 18G80]
defines: C9JEC2WQKV
uses: [Y9CZ3MTNF1, WQ82C0M3QB, HTK9WQZJ1G, 2J4JJJYP22, 4EBHAWJJ8C]
sources:
  - url: https://en.wikipedia.org/wiki/Quasi-isomorphism
---

## Definition

Let $\mathcal{A}$ be an [[abelian category|Y9CZ3MTNF1]] and let $f \colon C_\bullet \to D_\bullet$ be a [[chain map|HTK9WQZJ1G]] of [[chain complexes|WQ82C0M3QB]] in $\mathcal{A}$.
The map $f$ is a *quasi-isomorphism* if the induced morphism on [[homology|2J4JJJYP22]]
$$H_n(f) \colon H_n(C) \to H_n(D)$$
is an isomorphism for every integer $n$.

## Remarks

Every chain homotopy equivalence is a quasi-isomorphism, because [[chain homotopic|4EBHAWJJ8C]] maps induce the same maps on homology.
Quasi-isomorphisms are closed under composition, but in general a quasi-isomorphism has no chain map inverse up to homotopy.

Example: 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 $0$.
Let $f_0 \colon \Z \to \Z/2\Z$ be the reduction and $f_1 = 0$; this is a chain map since $f_0(2x) = 0$.
Both 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.
It 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 $\Z$.

Non-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 $\Z \to \Z$.