# Homological Algebra/Sequences

Lemma: In an $Ab$ -enriched category, if $f$ is a kernel, $g$ is a cokernel of $f$ , then $f$ is a kernel of $g$ .

Proof: Let $f$ be a kernel of $h$ . Since $hf=0$ , and $g$ is a cokernel of $f$ , there exists $k$ such that $h=kg$ . For all $l$ such that $gl=0$ , $hl=kgl=k0=0$ . Since $f$ is a kernel of $h$ , $l$ factorizes uniquely through $f$ . $\Box$ Corollary: In an abelian category, consider a sequence $a{\xrightarrow {f}}b{\xrightarrow {g}}c$ . The following conditions are equivalent:

1. $g$ is a cokernel of $f$ and $f$ is a kernel of $g$ .
2. $f$ is a monomorphism and $g$ is a cokernel of $f$ .
3. $g$ is an epimorphism and $f$ is a kernel of $g$ .

Definition (short exact sequence):

We call $\cdot {\xrightarrow {f}}\cdot {\xrightarrow {g}}\cdot$ a short exact sequence if it satisfies any of the equivalent conditions above.

Proposition (splitting lemma):

Let ${\mathcal {C}}$ be an abelian category, and suppose that

$\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0$

is a short exact sequence. Then the following are equivalent:

1. There exists a morphism $l:B\to A$ so that $l\circ f=\operatorname {id} _{A}$ 2. There exists a morphism $r:C\to A$ so that $g\circ r=\operatorname {id} _{B}$ 3. The short exact sequence $\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0$ is isomorphic to the short exact sequence $\displaystyle 0 \longrightarrow A \xlongrightarrow{\iota_A} A \oplus C \xlongrightarrow{\pi_C} C \longrightarrow 0$

Proof: Suppose first that $\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0$ is isomorphic to $\displaystyle 0 \longrightarrow A \xlongrightarrow{\iota_A} A \oplus C \xlongrightarrow{\pi_C} C \longrightarrow 0$ via isomorphisms $\varphi :A\to A$ , $\psi :B\to A\oplus C$ and $\chi :C\to C$ . Then there exists $l$ as in 1., since we may just define

$l:=\varphi ^{-1}\circ \pi _{A}\circ \psi$ ;

$l$ has the required property since by definition of morphisms of chain complexes, $\psi \circ f=\iota _{A}\circ \phi$ , and further $\pi _{A}\circ \iota _{A}=\operatorname {id} _{A}$ . Suppose now that there does exist $l:B\to A$ so that $l\circ f=\operatorname {id} _{A}$ . Since $A\oplus C$ is a biproduct, it is in particular a product, so that $l$ and $g$ define a unique morphism $\Phi :B\to A\oplus B$ such that

$l=\pi _{A}\circ \Phi$ and $g=\pi _{B}\circ \Phi$ .

Then, $\Phi$ , together with the identities on $A$ and $C$ , 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. $\Box$ 