# 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.

## ExercisesEdit

- 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

**Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0}**

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
**Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0}**is isomorphic to the short exact sequence**Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{\iota_A} A \oplus C \xlongrightarrow{\pi_C} C \longrightarrow 0}**

**Proof:** Suppose first that **Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0}**
is isomorphic to **Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{\iota_A} A \oplus C \xlongrightarrow{\pi_C} C \longrightarrow 0}**
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.