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Mapping cone (homological algebra) : ウィキペディア英語版
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^ \xrightarrow A^ \to \cdots
and likewise for B.
For a map of complexes f : A \to B, we define the cone, often denoted by \operatorname(f) or C(f), to be the following complex:
:C(f) = A() \oplus B = \dots \to A^n \oplus B^ \to A^ \oplus B^n \to A^ \oplus B^ \to \cdots on terms,
with differential
:d_ = \begin d_ & 0 \\ f() & d_B \end (acting as though on column vectors).
Here A() is the complex with A()^n=A^ and d^n_=-d^_.
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
d^n_(a^, b^n) &=& \begin d^n_ & 0 \\ f()^n & d^n_B \end \begin a^ \\ b^n \end \\
&=& \begin - d^_A & 0 \\ f^ & d^n_B \end \begin a^ \\ b^n \end \\
&=& \begin - d^_A (a^) \\ f^(a^) + d^n_B(b^n) \end\\
&=& \left(- d^_A (a^), f^(a^) + d^n_B(b^n)\right).
\end


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