and
nonabelian homological algebra
Given an abelian category $\mathcal{A}$, write $Ch_\bullet(\mathcal{A})$ for its category of chain complexes.
The relation of chain homotopy is an equivalence relation on the hom sets of this category.
Write $\mathcal{K}(\mathcal{A})$ for the category whose objects are chain complexes, and whose morphisms between objects $C_\bullet, D_\bullet$ are chain homotopy-equivalence classes
Analogous to the variants discussed at category of chain complexes one considers also the full subcategories
on the chain complexes which are bounded above or bounded below or bounded, respectively.
The category $\mathcal{K}(\mathcal{A})$ is sometimes called the “homotopy category of chain complexes”. But beware that there is another category that also and more properly deserves to be called by this name:
the homotopy category $Ho(Ch_\bullet(\mathcal{A}))$ in the sense of the result of localization at the chain maps that are quasi-isomorphisms is traditionally called the derived category $\mathcal{D}(\mathcal{A})$ of $\mathcal{A}$:
By construction there is a canonical functor
which is the identity on objects and the quotient projection on hom-sets.
Under componentwise addition of chain maps, $\mathcal{K}(\mathcal{A})$ is an additive category.
In fact using the abelian group-structure on chain maps, we can equivalently reformulate the quotient by chain homotopy as follows:
For $C_\bullet, D_\bullet \in Ch_\bullet(\mathcal{A})$, let
be the abelian subgroup on those chain maps which are null homotopic, hence for which there is a chain homotopy to the zero morphism.
For $C_\bullet, D_\bullet \in Ch_\bullet(\mathcal{A})$, there is an natural isomorphism of abelian groups
of the hom-objects in $\mathcal{K}(\mathcal{A})$ with the quotient group of all chain maps by those whose are null homotopic.
A distinguished triangle in $\mathcal{K}(\mathcal{A})$ is a sequence of morphisms of the form
where $C[1]_\bullet$ denotes the suspension of a chain complex, which is isomorphic to the image under the projection fuctor def. 3 of a mapping cone sequence in $Ch_\bullet(\mathcal{A})$
of a chain map $f_\bullet : X_\bullet \to Y_\bullet$ in $Ch_\bullet(\mathcal{A})$.
With
shift functor given by suspension of chain complexes;
distinguished triangles as in def. 5
the category $\mathcal{K}(\mathcal{A})$ is a triangulated category.
See at derived category in the section derived category - Properties.
We discuss some basic examples of chain maps that ought to be identified in homotopy theory, but which are not yet identified in $\mathcal{K}(\mathcal{A})$, but only in the derived category $\mathcal{D}(\mathcal{A})$.
In $Ch_\bullet(\mathcal{A})$ for $\mathcal{A} =$ Ab consider the chain map
The codomain of this map is an exact sequence, hence is quasi-isomorphic to the 0-chain complex. Therefore in homotopy theory it should behave entirely as the 0-complex itself. In particular, every chain map to it should be chain homotopic to the zero morphism (have a null homotopy).
But the above chain map is chain homotopic precisely only to itself. This is because the degree-0 component of any chain homotopy out of this has to be a homomorphism of abelian groups $\mathbb{Z}_2 \to \mathbb{Z}$, and this must be the 0-morphism, because $\mathbb{Z}$ is a free group, but $\mathbb{Z}_2$ is not.
This points to the problem: the components of the domain chain complex are not free enough to admit sufficiently many maps out of it.
Consider therefore a free resolution of the above domain complex by the quasi-isomorphism
where now the domain complex consists entirely of free groups. The composite of this with the original chain map is now
This is the corresponding resolution of the original chain map. And this indeed has a null homotopy:
Indeed, for this to happen it is sufficient that the resolution is by a degreewise projective complex. This is the statement of this lemma at projective resolution.
Lecture notes include
section 3.1 and 7.1 of