Abelian category

Remarks

Clauses 2 and 3 are equivalent, given clause 1, to the condition that for every morphism $f$ the canonical morphism $\operatorname{coim} f \to \operatorname{im} f$ is an isomorphism, where $\operatorname{coim} f = \operatorname{coker}(\ker f)$ and $\operatorname{im} f = \ker(\operatorname{coker} f)$. In an abelian category, a morphism that is both a monomorphism and an epimorphism is an isomorphism.

Example: the category of left modules over a ring $R$ is abelian. A monomorphism $m \colon M \to N$ is injective, and it is the kernel of the quotient map $N \to N / m(M)$. An epimorphism $e \colon N \to P$ is surjective, and it is the cokernel of the inclusion of its kernel.

Non-example: the category of free abelian groups is additive but not abelian. Multiplication by $2$ on $\Z$ is an epimorphism there: if $u \circ 2 = v \circ 2$ with torsion-free target, then $2(u - v) = 0$ forces $u = v$. It is not the cokernel of any morphism $k$: from $2 \circ k = 0$ follows $k = 0$, whose cokernel is the identity of $\Z$, not multiplication by $2$. So clause 3 fails.