# Homological Algebra/Sequences

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

Proof: Let ${\displaystyle f}$ be a kernel of ${\displaystyle h}$. Since ${\displaystyle hf=0}$, and ${\displaystyle g}$ is a cokernel of ${\displaystyle f}$, there exists ${\displaystyle k}$ such that ${\displaystyle h=kg}$. For all ${\displaystyle l}$ such that ${\displaystyle gl=0}$, ${\displaystyle hl=kgl=k0=0}$. Since ${\displaystyle f}$ is a kernel of ${\displaystyle h}$, ${\displaystyle l}$ factorizes uniquely through ${\displaystyle f}$. ${\displaystyle \Box }$

Corollary: In an abelian category, consider a sequence ${\displaystyle a{\xrightarrow {f}}b{\xrightarrow {g}}c}$. The following conditions are equivalent:

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

Definition (short exact sequence):

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

Proposition (splitting lemma):

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

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

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

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

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

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

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

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

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