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  "body": "\n## Definition\n\nLet $\\mathcal{C}$ be a finitely complete category and $X$ an object of $\\mathcal{C}$.\nAn *internal equivalence relation* on $X$ is a pair $r_1, r_2 \\colon R \\to X$ with this property.\nFor every object $T$, the set $\\set{(r_1 \\circ t, r_2 \\circ t) : t \\colon T \\to R}$ is an [[equivalence relation|0G8VEZ3WTK]] on the set $\\mathcal{C}(T, X)$, and $t$ is determined by this pair.\nThe relation is *effective* if it is the [[kernel pair|E70DQGM8X5]] of some morphism $f \\colon X \\to Y$.\nThe category $\\mathcal{C}$ is *exact in the sense of Barr* if both conditions hold.\n\n1. $\\mathcal{C}$ is a [[regular category|2SM3JRJXA4]].\n2. Every internal equivalence relation in $\\mathcal{C}$ is effective.\n\n## Remarks\n\nThe second part of the first clause says that $(r_1, r_2) \\colon R \\to X \\times X$ is a [[monomorphism|Y2NP1JFR86]].\nThe kernel pair of any morphism is always an internal equivalence relation, so clause 2 asks for the converse.\n\nExample: the category of sets is Barr-exact.\nAn equivalence relation $R \\subseteq X \\times X$ is the kernel pair of the quotient map $X \\to X / R$.\n\nNon-example: the category of torsion-free abelian groups.\nIn $\\Z$, the subgroup $R = \\set{(a, b) : a \\equiv b \\pmod 2}$ of $\\Z \\times \\Z$ is torsion-free and is an internal equivalence relation.\nLet $f \\colon \\Z \\to Y$ be a homomorphism into a torsion-free group with $f(2) = 0$.\nThen $2 f(1) = 0$, so $f(1) = 0$ and the kernel pair of $f$ is all of $\\Z \\times \\Z$.\nSo $R$ is not effective and clause 2 fails.\n",
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