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Mapping cone (homological algebra)

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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.

1 Definition
2 Properties
3 Mapping cylinder
4 Topological inspiration
5 References

Definition

The cone may be defined in the category of chain 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 <math>A, B</math> be two complexes, with differentials <math>d_A, d_B;</math> i.e.,

<math>A = \dots \to A^{n - 1} \xrightarrow{d_A^{n - 1}} A^n \xrightarrow{d_A^n} A^{n + 1} \to \cdots</math>

and likewise for <math>B.</math> For a map of complexes <math>f : A \to B,</math> we define the cone, denoted by either <math>\operatorname{Cone}(f)</math> or <math>C(f),</math> or a nearby variant, to be the following complex:

<math>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</math> on terms,

with differential

<math>d_{C(f)} = \begin{pmatrix} d_{A[1]} & 0 \\ f[1] & d_B \end{pmatrix}</math> (acting as though on column vectors).

Note that this is different from the natural differential on <math>A[1] \oplus B.</math> Thus, if for example our complexes are of abelian groups, the differential would act as

<math>\begin{array}{ccl}

d^n_{C(f)}(a^{n + 1}, b^n) &=& \begin{pmatrix} d^n_{A[1]} & 0 \\ f[1]^n & d^n_B \end{pmatrix} \begin{pmatrix} a^{n + 1} \\ b^n \end{pmatrix} \\

 &=& \begin{pmatrix} (-1)^n d^{n + 1}_A & 0 \\ f^{n + 1} & d^n_B \end{pmatrix} \begin{pmatrix} a^{n + 1} \\ b^n \end{pmatrix} \\
 &=& \begin{pmatrix} (-1)^n d^{n + 1}_A (a^{n + 1}) \\ f^{n + 1}(a^{n + 1}) + d^n_B(b^n) \end{pmatrix}\\
 &=& \left((-1)^n d^{n + 1}_A (a^{n + 1}), f^{n + 1}(a^{n + 1}) + d^n_B(b^n)\right).

\end{array} </math>

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

<math>A \xrightarrow{f} B \to C(f) \to</math>

where the maps <math>B \to C(f), C(f) \to A[1]</math> are the projections onto 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:

<math>\dots \to H^{i - 1}(C(f)) \to H^i(A) \xrightarrow{f^*} H^i(B) \to H^i(C(f)) \to \cdots</math>

and if <math>C(f)</math> is acyclic then by definition, the outer terms above are zero. Since the sequence is exact, this means that <math>f^*</math> 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 <math>A, B</math> have only one nonzero term in degree 0:

<math>A = \dots \to 0 \to A^0 \to 0 \to \cdots,</math>

<math>B = \dots \to 0 \to B^0 \to 0 \to \cdots,</math>

and therefore <math>f \colon A \to B</math> is just <math>f^0 \colon A^0 \to B^0</math> (as a map of objects of the underlying abelian category). Then the cone is just

<math>C(f) = \dots \to 0 \to \underset{[-1]}{A^0} \xrightarrow{f^0} \underset{[0]}{B^0} \to 0 \to \cdots.</math>

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

<math>H^{-1}(C(f)) = \operatorname{ker}(f^0),</math>

<math>H^0(C(f)) = \operatorname{coker}(f^0),</math>

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 of a continuous map of topological spaces <math>\phi : X \rightarrow Y</math>: the complex of singular chains of the topological cone <math>cone(\phi)</math> 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.