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homotopy category of chain complexes

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homological algebra

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Definition

Given an abelian category 𝒜\mathcal{A}, write Ch (𝒜)Ch_\bullet(\mathcal{A}) for its category of chain complexes.

Proposition

The relation of chain homotopy is an equivalence relation on the hom sets of this category.

Definition

Write

𝒦(𝒜)Ch (𝒜) \mathcal{K}(\mathcal{A}) \coloneqq Ch_\bullet(\mathcal{A})

for the category whose objects are chain complexes, and whose morphisms between objects C ,D C_\bullet, D_\bullet are chain homotopy-equivalence classes

𝒦(𝒜)(C ,D )Ch (𝒜)(C ,D ) ch. \mathcal{K}(\mathcal{A})(C_\bullet, D_\bullet) \coloneqq Ch_\bullet(\mathcal{A})(C_\bullet, D_\bullet)_{\sim_{ch}} \,.
Definition

Analogous to the variants discussed at category of chain complexes one considers also the full subcategories

K +,,b(𝒜)𝒦(𝒜) K^{+,-,b}(\mathcal{A}) \hookrightarrow \mathcal{K}(\mathcal{A})

on the chain complexes which are bounded above or bounded below or bounded, respectively.

Remark

The category 𝒦(𝒜)\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 (𝒜))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 𝒟(𝒜)\mathcal{D}(\mathcal{A}) of 𝒜\mathcal{A}:

𝒟(𝒜)Ho(Ch (𝒜)). \mathcal{D}(\mathcal{A}) \coloneqq Ho(Ch_\bullet(\mathcal{A})) \,.
Definition

By construction there is a canonical functor

Ch (𝒜)𝒦(𝒜) Ch_\bullet(\mathcal{A}) \to \mathcal{K}(\mathcal{A})

which is the identity on objects and the quotient projection on hom-sets.

Properties

Additive category structure

Proposition

Under componentwise addition of chain maps, 𝒦(𝒜)\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:

Definition

For C ,D Ch (𝒜)C_\bullet, D_\bullet \in Ch_\bullet(\mathcal{A}), let

NHt(C ,D )Ch (C ,D ) NHt(C_\bullet, D_\bullet) \hookrightarrow Ch_\bullet(C_\bullet, D_\bullet)

be the abelian subgroup on those chain maps which are null homotopic, hence for which there is a chain homotopy to the zero morphism.

Proposition

For C ,D Ch (𝒜)C_\bullet, D_\bullet \in Ch_\bullet(\mathcal{A}), there is an natural isomorphism of abelian groups

𝒦(𝒜)(C ,D )Ch (𝒜)/NHt(C ,D ) \mathcal{K}(\mathcal{A})(C_\bullet, D_\bullet) \simeq Ch_\bullet(\mathcal{A})/NHt(C_\bullet, D_\bullet)

of the hom-objects in 𝒦(𝒜)\mathcal{K}(\mathcal{A}) with the quotient group of all chain maps by those whose are null homotopic.

Triangulated category structure

Definition

A distinguished triangle in 𝒦(𝒜)\mathcal{K}(\mathcal{A}) is a sequence of morphisms of the form

Y (XY) X[1] Y[1] , Y_\bullet \to (X\sslash Y)_\bullet \to X[1]_\bullet \stackrel{}{\to} Y[1]_\bullet \,,

where C[1] 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 (𝒜)Ch_\bullet(\mathcal{A})

Y cone(f) X[1] f[1] Y[1] Y_\bullet \stackrel{}{\hookrightarrow} cone(f)_\bullet \stackrel{}{\to} X[1]_\bullet \stackrel{f[1]_\bullet}{\to} Y[1]_\bullet

of a chain map f :X Y f_\bullet : X_\bullet \to Y_\bullet in Ch (𝒜)Ch_\bullet(\mathcal{A}).

Theorem

With

the category 𝒦(𝒜)\mathcal{K}(\mathcal{A}) is a triangulated category.

Relation to the derived category

See at derived category in the section derived category - Properties.

Examples

Chain homotopies that ought to exist but do not

We discuss some basic examples of chain maps that ought to be identified in homotopy theory, but which are not yet identified in 𝒦(𝒜)\mathcal{K}(\mathcal{A}), but only in the derived category 𝒟(𝒜)\mathcal{D}(\mathcal{A}).

Example

In Ch (𝒜)Ch_\bullet(\mathcal{A}) for 𝒜=\mathcal{A} = Ab consider the chain map

0 0 0 2 id 0 2 mod2 2. \array{ \cdots &\to& 0 &\to& 0 &\to& 0 &\to& \mathbb{Z}_2 \\ && \downarrow && \downarrow && \downarrow && \downarrow^{\mathrlap{id}} \\ \cdots &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} &\stackrel{mod\,2}{\to}& \mathbb{Z}_2 } \,.

The codomain of this map is an exact sequence, hence is quasi-isomorphic to the 0-chain complex. Thereofore 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 2\mathbb{Z}_2 \to \mathbb{Z}, and this must be the 0-morphism, because \mathbb{Z} is a free group, but 2\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

0 0 2 mod2 0 0 0 2, \array{ \cdots &\to& 0 &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} \\ && \downarrow && \downarrow && \downarrow && \downarrow^{\mathrlap{mod\,2}} \\ \cdots &\to& 0 &\to& 0 &\to& 0 &\to& \mathbb{Z}_2 } \,,

where now the domain complex consists entirely of free groups. The composite of this with the original chain map is now

0 0 2 0 mod2 0 2 mod2 2. \array{ \cdots &\to& 0 &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} \\ && \downarrow && \downarrow && \downarrow^{0} && \downarrow^{\mathrlap{mod\,2}} \\ \cdots &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} &\stackrel{mod\,2}{\to}& \mathbb{Z}_2 } \,.

This is the corresponding resolution of the original chain map. And this indeed has a null homotopy:

0 0 2 id 0 id mod2 0 2 mod2 2. \array{ \cdots &\to& 0 &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} \\ && \downarrow &\swarrow& \downarrow &\swarrow_{-id}& \downarrow^{0} &\swarrow_{\mathrlap{id}}& \downarrow^{\mathrlap{mod\,2}} \\ \cdots &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} &\stackrel{mod\,2}{\to}& \mathbb{Z}_2 } \,.

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.

References

Lecture notes include

section 3.1 and 7.1 of

Revised on September 25, 2012 12:21:09 by Urs Schreiber (131.174.191.188)