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  "body": "\n## Definition\n\nLet $\\mathcal{C}$ be a [[category|A2DE66937D]] and let $f, g \\colon X \\to Y$ be two [[morphisms|859F0TFVA3]] of $\\mathcal{C}$.\nA *coequalizer* of $f$ and $g$ is an object $Q$ with a morphism $q \\colon Y \\to Q$ such that:\n\n1. $q \\circ f = q \\circ g$;\n2. for every object $Z$ and every morphism $k \\colon Y \\to Z$ with $k \\circ f = k \\circ g$, there is exactly one morphism $u \\colon Q \\to Z$ with $u \\circ q = k$.\n\n## Remarks\n\nA coequalizer is the [[colimit|BJS1CGTKQ5]] of the diagram formed by $f$ and $g$, and it is an [[equalizer|CTJ5AD8ST5]] in the opposite category.\nIt is an epimorphism, by the dual of the argument for equalizers.\n\nExample: in the category of sets, let $\\sim$ be the smallest [[equivalence relation|0G8VEZ3WTK]] on $Y$ with $f(x) \\sim g(x)$ for all $x \\in X$, let $Q = Y/{\\sim}$ and let $q$ be the quotient map.\nClause 1 holds by construction.\nIf $k \\circ f = k \\circ g$, the relation \"$k(y) = k(y')$\" is an equivalence relation containing all pairs $(f(x), g(x))$, so it contains $\\sim$, and $u([y]) = k(y)$ is well defined; it is unique because $q$ is surjective.\nIn the category of abelian groups the coequalizer of $f$ and $g$ is $Y / \\im(f - g)$ with the quotient map.\n\nNon-example: for $X = \\{*\\}$, $Y = \\{0, 1, 2\\}$, $f(*) = 0$ and $g(*) = 1$, the constant map $q \\colon Y \\to \\{*\\}$ satisfies clause 1 but not clause 2.\nThe map $k \\colon Y \\to \\{0, 2\\}$ with $k(0) = k(1) = 0$ and $k(2) = 2$ has $k \\circ f = k \\circ g$ but does not factor through a one-point set.\n",
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