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  "body": "\n## Definition\n\nLet $\\mathcal{C}$ be a [[category|A2DE66937D]] and let $f \\colon X \\to Y$ be a [[morphism|859F0TFVA3]] of $\\mathcal{C}$.\nA *kernel pair* of $f$ is an object $K$ with morphisms $p_1, p_2 \\colon K \\to X$ such that $(K, p_1, p_2)$ is a [[pullback|Q05K04QJ71]] of $f$ and $f$.\nThat is:\n\n1. $f \\circ p_1 = f \\circ p_2$.\n2. For every object $W$ and morphisms $q_1, q_2 \\colon W \\to X$ with $f \\circ q_1 = f \\circ q_2$, there is a unique $u \\colon W \\to K$ with $p_1 \\circ u = q_1$ and $p_2 \\circ u = q_2$.\n\n## Remarks\n\nExample: in the category of sets, $K = \\set{(x, x') \\in X \\times X : f(x) = f(x')}$ with the two projections is a kernel pair.\nClause 1 holds by the definition of $K$, and for clause 2 the map $u$ is $w \\mapsto (q_1(w), q_2(w))$.\nFor a group homomorphism $f$ with kernel $N$, this $K$ is the subgroup of pairs $(x, x')$ with $x^{-1} x' \\in N$.\n\nNon-example: in the category of sets, if $f$ is not injective, then $(X, \\id_X, \\id_X)$ satisfies clause 1 but not clause 2.\nTake $W$ a point and $q_1, q_2$ picking $x \\neq x'$ with $f(x) = f(x')$; no $u$ has $u = q_1$ and $u = q_2$.\n\nA morphism $f$ is a monomorphism if and only if $(X, \\id_X, \\id_X)$ is a kernel pair of $f$.\n",
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