# Mapping cone (homological algebra)

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 $A,B$  be two complexes, with differentials $d_{A},d_{B};$  i.e.,

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

and likewise for $B.$

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

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

with differential

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

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

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

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

## Properties

Suppose now that we are working over an abelian category, so that the homology 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

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

where the maps $B\to C(f),C(f)\to A$  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 homology groups:

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

and if $C(f)$  is acyclic then by definition, the outer terms above are zero. Since the sequence is exact, this means that $f^{*}$  induces an isomorphism on all homology 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 $A,B$  have only one nonzero term in degree 0:

$A=\dots \to 0\to A_{0}\to 0\to \cdots ,$
$B=\dots \to 0\to B_{0}\to 0\to \cdots ,$

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

$C(f)=\dots \to 0\to {\underset {[-1]}{A_{0}}}{\xrightarrow {f_{0}}}{\underset {}{B_{0}}}\to 0\to \cdots .$

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

$H_{-1}(C(f))=\operatorname {ker} (f_{0}),$
$H_{0}(C(f))=\operatorname {coker} (f_{0}),$
$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 $f\colon A\to B$  be a morphism of chain complexes, let further $g\colon \operatorname {Cone} (f)[-1]\to A$  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 $\phi :X\rightarrow Y$ : the complex of singular chains of the topological cone $cone(\phi )$  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.