# Homological Algebra/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.