quasi-isomorphism

(also nonabelian homological algebra)

**homotopy theory, (∞,1)-category theory, homotopy type theory**

flavors: stable, equivariant, rational, p-adic, proper, geometric, cohesive, directed…

models: topological, simplicial, localic, …

see also **algebraic topology**

**Introductions**

**Definitions**

**Paths and cylinders**

**Homotopy groups**

**Basic facts**

**Theorems**

A *quasi-isomorphism* is a chain map that induces isomorphisms on all homology groups. These are the natural choice of weak equivalences between chain complexes in the context of (stable) homotopy theory.

The localization of a category of chain complexes at the quasi-isomorphisms is called the *derived category* of the underlying abelian category.

Under the relation between topological spaces and chain complexes established by forming singular simplicial complexes, quasi-isomorphism can be understod as the abelianization of weak homotopy equivalences (see the Hurewicz theorem for more on this).

Let $\mathcal{A}$ be an abelian category and write $Ch_\bullet(\mathcal{A})$ for its category of chain complexes.

A chain map $f_\bullet : C_\bullet \to D_\bullet$ in $Ch_\bullet(\mathcal{A})$ is called a **quasi-isomorphism** if for each $n \in \mathbb{N}$ the induced morphisms on chain homology groups

$H_n(f) \colon H_n(C) \to H_n(D)$

is an isomorphism.

Quasi-isomorphisms are also called, more descriptively, **homology isomorphisms** or **$H_\bullet$-isomorphisms**. See at *homology localization* for more on this.

The relation “There exists a quasi-isomorphism from $C_\bullet$ to $D_\bullet$.” is a reflexive and transitive relation, but it is not a symmetric relation.

Reflexivity and transitivity are evident. An explicit counter-example showing the non-symmetry is the chain map

$\array{
\cdots &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} &\to& 0 &\to& \cdots
\\
\cdots && \downarrow && \downarrow && \downarrow && \downarrow && \cdots
\\
\cdots &\to& 0 &\to& 0 &\to& \mathbb{Z}/2\mathbb{Z} &\to& 0 &\to& \cdots
}$

from the chain complex concentrated on the morphism of multiplication by 2 on integers, to the chain complex concentrated on the cyclic group of order 2.

This clearly induces an isomorphism on all homology groups. But there is not even a non-zero chain map in the other direction, since there is no non-zero group homomorphism $\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}$.

This is as for weak homotopy equivalences, see the discussion at *Relation to homotopy types* there.

A chain map is a quasi-isomorphism precisely if its homotopy cofiber in the (∞,1)-category of chain complexes has trivial homology groups.

By basic properties discussed at *truncated object in an (∞,1)-category*.

Concretely this means in particular the following.

A chain map $f_\bullet : C_\bullet \to D_\bullet$ is a quasi-isomorphism precisely if its mapping cone $cone(f)_\bullet \in Ch_\bullet(\mathcal{A})$ has all trivial chain homology groups.

This follows for instance from the homology long exact sequence

$\cdots
\to
H_{n+1}(c)\to H_{n+1}(D) \to H_{n+1}(cone(f))
\to
H_n(C) \to H_n(D) \to H_n(cone(f))
\to
H_{n-1}(C) \to H_{n-1}(D) \to H_{n-1}(cone(f))
\to
\cdots
\,.$

If here by assumption $H_n(cone(f)) = 0$ for all $n$, then this involves exact sequences of the form

$0 \to H_n(C) \stackrel{H_n(f)}{\to} H_n(D) \to 0$

for all $n$. But this says that the kernel and cokernel of $H_n(f)$ are trivial for all $n$, hence that $H_n(f)$ is an isomorphism for all $n$, hence that $f_\bullet$ is a quasi-isomorphism.

Quasi-isomorphisms are the weak equivalences in the most common model category structures on the category of chain complexes. See at *model structure on chain complexes* and *derived category*.

A basic introduction is around definition 1.1.2 in

A more systematic discussion is in section 12 of

Revised on January 17, 2015 14:55:56
by Adeel Khan
(77.9.198.13)