# Homological Algebra/Printable version

Homological Algebra

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# Defintion of abelian category

Definition (${\displaystyle Ab}$-enriched category):

An ${\displaystyle Ab}$-enriched category is a category ${\displaystyle {\mathcal {C}}}$ such that:

1. ${\displaystyle \forall a,b\in \operatorname {Obj} ({\mathcal {C}})}$, ${\displaystyle {\mathcal {C}}(a,b)}$ is an abelian group.
2. ${\displaystyle \forall a,b,c\in \operatorname {Obj} ({\mathcal {C}})}$, ${\displaystyle \circ _{a,b,c}:{\mathcal {C}}(b,c)\times {\mathcal {C}}(a,b)\to {\mathcal {C}}(a,c)}$ is bilinear.

Definition (zero object):

A zero object is an object in an ${\displaystyle Ab}$-enriched category that is both initial and terminal. We usually denote it by ${\displaystyle \mathbf {0} }$.

Definition (biproduct):

Given an ${\displaystyle Ab}$-enriched category ${\displaystyle {\mathcal {C}}}$, a biproduct of ${\displaystyle a,b\in \operatorname {Obj} ({\mathcal {C}})}$ is a tuple ${\displaystyle (c,i_{1}:a\to c,p_{1}:c\to a,i_{2}:b\to c,p_{2}:c\to b)}$ such that:

1. ${\displaystyle p_{1}\circ i_{1}=1_{a}.}$
2. ${\displaystyle p_{2}\circ i_{1}=0_{a,b}.}$
3. ${\displaystyle p_{1}\circ i_{2}=0_{b,a}.}$
4. ${\displaystyle p_{2}\circ i_{2}=1_{b}.}$
5. ${\displaystyle i_{1}\circ p_{1}+i_{2}\circ p_{2}=1_{c}.}$

We usually denote ${\displaystyle c}$ by ${\displaystyle a\oplus b}$.

An additive category is an ${\displaystyle Ab}$-enriched category ${\displaystyle {\mathcal {C}}}$ such that:

1. There is a zero product in ${\displaystyle {\mathcal {C}}}$.
2. Every ${\displaystyle a,b\in \operatorname {Obj} ({\mathcal {C}})}$ has a biproduct.

Definition ((co-)kernel):

Given ${\displaystyle f:a\to b}$ in an ${\displaystyle Ab}$-enriched category. A (co-)kernel of ${\displaystyle f}$ is a (co-)equalizer of ${\displaystyle f}$ and ${\displaystyle 0_{a,b}}$.

Definition (abelian category):

An abelian category is an additive category where:

1. Every morphism has a kernel and cokernel.
2. Every monomorphism is a kernel and every epimorphism is a cokernel.

Example:

The category of all left ${\displaystyle R}$-modules of a ring ${\displaystyle R}$ is an abelian category.

## Exercises

1. Given ${\displaystyle f:a\to b}$ in an ${\displaystyle Ab}$-enriched category with zero object. Prove that ${\displaystyle f=0_{a,b}}$ iff ${\displaystyle f}$ factors through ${\displaystyle \mathbf {0} }$.
1. Given a biproduct ${\displaystyle (a\oplus b,i_{1},p_{1},i_{2},p_{2})}$ of ${\displaystyle a}$ and ${\displaystyle b}$. Prove that ${\displaystyle (a\oplus b,i_{1},i_{2})}$ is a coproduct of ${\displaystyle a}$ and ${\displaystyle b}$ and ${\displaystyle (a\oplus b,p_{1},p_{2})}$ is a product of ${\displaystyle a}$ and ${\displaystyle b}$.
1. In an ${\displaystyle Ab}$-enriched category with zero object, a kernel of ${\displaystyle f:a\to b}$ can be equivalently be characterized as a pullback of ${\displaystyle \mathbf {0} \to b}$ along ${\displaystyle f}$.

# 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

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

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 $\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0$ 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 ${\displaystyle \varphi :A\to A}$, ${\displaystyle \psi :B\to A\oplus C}$ 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. ${\displaystyle \Box }$