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  "body": "\n## Definition\n\nAn *abelian category* is an [[additive category|B2VP96NJSK]] $\\mathcal{A}$ such that:\n\n1. every morphism of $\\mathcal{A}$ has a [[kernel|S417YNEJ6M]] and a [[cokernel|H8W38933NC]];\n2. every [[monomorphism|Y2NP1JFR86]] of $\\mathcal{A}$ is the kernel of some morphism;\n3. every [[epimorphism|RVXDSY04KG]] of $\\mathcal{A}$ is the cokernel of some morphism.\n\n## Remarks\n\nClauses 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 $\\operatorname{im} f = \\ker(\\operatorname{coker} f)$.\nIn an abelian category, a morphism that is both a monomorphism and an epimorphism is an isomorphism.\n\nExample: the category of left [[modules|0NZ8JA25WK]] over a ring $R$ is abelian.\nA monomorphism $m \\colon M \\to N$ is injective, and it is the kernel of the quotient map $N \\to N / m(M)$.\nAn epimorphism $e \\colon N \\to P$ is surjective, and it is the cokernel of the inclusion of its kernel.\n\nNon-example: the category of [[free abelian groups|0WJX8MGG37]] is additive but not abelian.\nMultiplication 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 $u = v$.\nIt 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 $2$.\nSo clause 3 fails.\n",
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