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

$0\longrightarrow A\xrightarrow {\qquad f\qquad } B\xrightarrow {\qquad g\qquad } C\longrightarrow 0$

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 $0\longrightarrow A\xrightarrow {\qquad f\qquad } B\xrightarrow {\qquad g\qquad } C\longrightarrow 0$ is isomorphic to the short exact sequence $0\longrightarrow A\xrightarrow {\qquad \iota _{A}\qquad } A\oplus C\xrightarrow {\qquad \pi _{C}\qquad } C\longrightarrow 0$

Proof: Suppose first that $0\longrightarrow A\xrightarrow {\qquad f\qquad } B\xrightarrow {\qquad g\qquad } C\longrightarrow 0$ is isomorphic to $0\longrightarrow A\xrightarrow {\qquad \iota _{A}\qquad } A\oplus C\xrightarrow {\qquad pi_{C}\qquad } C\longrightarrow 0$ 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. $\Box$