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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 $\\mathcal{C}$.\nAn *equalizer* of $f$ and $g$ is an object $E$ with a morphism $e \\colon E \\to X$ such that:\n\n1. $f \\circ e = g \\circ e$;\n2. for every object $Z$ and every morphism $k \\colon Z \\to X$ with $f \\circ k = g \\circ k$, there is exactly one morphism $u \\colon Z \\to E$ with $e \\circ u = k$.\n\n## Remarks\n\nAn equalizer is the [[limit|CHS5AFJGWE]] of the diagram $X \\rightrightarrows Y$ formed by $f$ and $g$, so it is unique up to unique isomorphism.\nIt is a monomorphism: if $e \\circ a = e \\circ b$, then $a$ and $b$ both solve clause 2 for $k = e \\circ a$, so $a = b$.\n\nExample: in the category of sets, let $E = \\set{x \\in X : f(x) = g(x)}$ and let $e$ be the inclusion.\nClause 1 holds by the choice of $E$.\nIf $f \\circ k = g \\circ k$, every $k(z)$ lies in $E$, so $u(z) = k(z)$ works, and it is the only choice since $e$ is injective.\nIn the category of abelian groups the equalizer of $f$ and $g$ is the [[kernel|0W88CEGQP1]] of $f - g$ with its inclusion.\n\nNon-example: for $f, g \\colon \\R \\to \\R$ with $f(x) = x^2$ and $g(x) = x$, the inclusion $\\{0\\} \\to \\R$ satisfies clause 1 but not clause 2.\nThe map $k \\colon \\{*\\} \\to \\R$ with $k(*) = 1$ has $f \\circ k = g \\circ k$ but does not factor through $\\{0\\}$.\n",
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