# Mapping cone (homological algebra)

(Redirected from Mapping cone of complexes)

In homological algebra, the mapping cone is a construction on a map of chain complexes inspired by the analogous construction in topology. In the theory of triangulated categories it is a kind of combined kernel and cokernel: if the chain complexes take their terms in an abelian category, so that we can talk about cohomology, then the cone of a map f being acyclic means that the map is a quasi-isomorphism; if we pass to the derived category of complexes, this means that f is an isomorphism there, which recalls the familiar property of maps of groups, modules over a ring, or elements of an arbitrary abelian category that if the kernel and cokernel both vanish, then the map is an isomorphism. If we are working in a t-category, then in fact the cone furnishes both the kernel and cokernel of maps between objects of its core.

## Definition

The cone may be defined in the category of cochain complexes over any additive category (i.e., a category whose morphisms form abelian groups and in which we may construct a direct sum of any two objects). Let ${\displaystyle A,B}$ be two complexes, with differentials ${\displaystyle d_{A},d_{B};}$ i.e.,

${\displaystyle A=\dots \to A^{n-1}{\xrightarrow {d_{A}^{n-1}}}A^{n}{\xrightarrow {d_{A}^{n}}}A^{n+1}\to \cdots }$

and likewise for ${\displaystyle B.}$

For a map of complexes ${\displaystyle f:A\to B,}$ we define the cone, often denoted by ${\displaystyle \operatorname {Cone} (f)}$ or ${\displaystyle C(f),}$ to be the following complex:

${\displaystyle C(f)=A[1]\oplus B=\dots \to A^{n}\oplus B^{n-1}\to A^{n+1}\oplus B^{n}\to A^{n+2}\oplus B^{n+1}\to \cdots }$ on terms,

with differential

${\displaystyle d_{C(f)}={\begin{pmatrix}d_{A[1]}&0\\f[1]&d_{B}\end{pmatrix}}}$ (acting as though on column vectors).

Here ${\displaystyle A[1]}$ is the complex with ${\displaystyle A[1]^{n}=A^{n+1}}$ and ${\displaystyle d_{A[1]}^{n}=-d_{A}^{n+1}}$. Note that the differential on ${\displaystyle C(f)}$ is different from the natural differential on ${\displaystyle A[1]\oplus B}$, and that some authors use a different sign convention.

Thus, if for example our complexes are of abelian groups, the differential would act as

${\displaystyle {\begin{array}{ccl}d_{C(f)}^{n}(a^{n+1},b^{n})&=&{\begin{pmatrix}d_{A[1]}^{n}&0\\f[1]^{n}&d_{B}^{n}\end{pmatrix}}{\begin{pmatrix}a^{n+1}\\b^{n}\end{pmatrix}}\\&=&{\begin{pmatrix}-d_{A}^{n+1}&0\\f^{n+1}&d_{B}^{n}\end{pmatrix}}{\begin{pmatrix}a^{n+1}\\b^{n}\end{pmatrix}}\\&=&{\begin{pmatrix}-d_{A}^{n+1}(a^{n+1})\\f^{n+1}(a^{n+1})+d_{B}^{n}(b^{n})\end{pmatrix}}\\&=&\left(-d_{A}^{n+1}(a^{n+1}),f^{n+1}(a^{n+1})+d_{B}^{n}(b^{n})\right).\end{array}}}$

## Properties

Suppose now that we are working over an abelian category, so that the cohomology of a complex is defined. The main use of the cone is to identify quasi-isomorphisms: if the cone is acyclic, then the map is a quasi-isomorphism. To see this, we use the existence of a triangle

${\displaystyle A{\xrightarrow {f}}B\to C(f)\to }$

where the maps ${\displaystyle B\to C(f),C(f)\to A[1]}$ are given by the direct summands (see Homotopy category of chain complexes). Since this is a triangle, it gives rise to a long exact sequence on cohomology groups:

${\displaystyle \dots \to H^{i-1}(C(f))\to H^{i}(A){\xrightarrow {f^{*}}}H^{i}(B)\to H^{i}(C(f))\to \cdots }$

and if ${\displaystyle C(f)}$ is acyclic then by definition, the outer terms above are zero. Since the sequence is exact, this means that ${\displaystyle f^{*}}$ induces an isomorphism on all cohomology groups, and hence (again by definition) is a quasi-isomorphism.

This fact recalls the usual alternative characterization of isomorphisms in an abelian category as those maps whose kernel and cokernel both vanish. This appearance of a cone as a combined kernel and cokernel is not accidental; in fact, under certain circumstances the cone literally embodies both. Say for example that we are working over an abelian category and ${\displaystyle A,B}$ have only one nonzero term in degree 0:

${\displaystyle A=\dots \to 0\to A^{0}\to 0\to \cdots ,}$
${\displaystyle B=\dots \to 0\to B^{0}\to 0\to \cdots ,}$

and therefore ${\displaystyle f\colon A\to B}$ is just ${\displaystyle f^{0}\colon A^{0}\to B^{0}}$ (as a map of objects of the underlying abelian category). Then the cone is just

${\displaystyle C(f)=\dots \to 0\to {\underset {[-1]}{A^{0}}}{\xrightarrow {f^{0}}}{\underset {[0]}{B^{0}}}\to 0\to \cdots .}$

(Underset text indicates the degree of each term.) The cohomology of this complex is then

${\displaystyle H^{-1}(C(f))=\operatorname {ker} (f^{0}),}$
${\displaystyle H^{0}(C(f))=\operatorname {coker} (f^{0}),}$
${\displaystyle H^{i}(C(f))=0{\text{ for }}i\neq -1,0.\ }$

This is not an accident and in fact occurs in every t-category.

## Mapping cylinder

A related notion is the mapping cylinder: let f: A → B be a morphism of complexes, let further g : Cone(f)[-1] → A be the natural map. The mapping cylinder of f is by definition the mapping cone of g.

## Topological inspiration

This complex is called the cone in analogy to the mapping cone (topology) of a continuous map of topological spaces ${\displaystyle \phi :X\rightarrow Y}$: the complex of singular chains of the topological cone ${\displaystyle cone(\phi )}$ is homotopy equivalent to the cone (in the chain-complex-sense) of the induced map of singular chains of X to Y. The mapping cylinder of a map of complexes is similarly related to the mapping cylinder of continuous maps.