Homological Algebra/Printable version
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Defintion of abelian category
Definition (-enriched category):
An -enriched category is a category such that:
- , is an abelian group.
- , is bilinear.
Definition (zero object):
A zero object is an object in an -enriched category that is both initial and terminal. We usually denote it by .
Definition (biproduct):
Given an -enriched category , a biproduct of is a tuple such that:
We usually denote by .
Definition (additive category):
An additive category is an -enriched category such that:
- There is a zero product in .
- Every has a biproduct.
Definition ((co-)kernel):
Given in an -enriched category. A (co-)kernel of is a (co-)equalizer of and .
Definition (abelian category):
An abelian category is an additive category where:
- Every morphism has a kernel and cokernel.
- Every monomorphism is a kernel and every epimorphism is a cokernel.
Example:
The category of all left -modules of a ring is an abelian category.
Exercises
[edit | edit source]- Given in an -enriched category with zero object. Prove that iff factors through .
- Given a biproduct of and . Prove that is a coproduct of and and is a product of and .
- In an -enriched category with zero object, a kernel of can be equivalently be characterized as a pullback of along .
Sequences
Lemma: In an -enriched category, if is a kernel, is a cokernel of , then is a kernel of .
Proof: Let be a kernel of . Since , and is a cokernel of , there exists such that . For all such that , . Since is a kernel of , factorizes uniquely through .
Corollary: In an abelian category, consider a sequence . The following conditions are equivalent:
- is a cokernel of and is a kernel of .
- is a monomorphism and is a cokernel of .
- is an epimorphism and is a kernel of .
Definition (short exact sequence):
We call a short exact sequence if it satisfies any of the equivalent conditions above.
Proposition (splitting lemma):
Let be an abelian category, and suppose that
is a short exact sequence. Then the following are equivalent:
- There exists a morphism so that
- There exists a morphism so that
- The short exact sequence is isomorphic to the short exact sequence
Proof: Suppose first that is isomorphic to via isomorphisms , and . Then there exists as in 1., since we may just define
- ;
has the required property since by definition of morphisms of chain complexes, , and further . Suppose now that there does exist so that . Since is a biproduct, it is in particular a product, so that and define a unique morphism such that
- and .
Then, , together with the identities on and , yields an isomorphism of chain complexes. The equivalence of 2. and 3. is a dual statement to the equivalence of 1. and 3., whence it doesn't have to be proved separately.