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  "body": "\n## Example\n\nWork in the [[concrete category|ZD0GTPSJS1]] $\\mathbf{Ring}$ of [[rings|0B1261B5ED]] with identity and [[ring homomorphisms|0GGE0Z0JQ1]] that preserve $1$.\nThe inclusion $i \\colon \\Z \\to \\Q$ of the [[integers|0C4JS9RF5M]] into the [[rationals|0NCJ5WXN6Z]] is an [[epimorphism|RVXDSY04KG]] in $\\mathbf{Ring}$, but it is not surjective.\n\nCheck: let $R$ be a ring and let $f, g \\colon \\Q \\to R$ be ring homomorphisms with $f \\circ i = g \\circ i$, so $f(n) = g(n)$ for every integer $n$.\nFor $n \\ne 0$, $f(1/n) \\, f(n) = f(1) = 1$ and $f(n) \\, f(1/n) = 1$, so $f(n)$ is a unit of $R$ with inverse $f(1/n)$.\nIn the same way $g(1/n)$ is the inverse of $g(n) = f(n)$.\nInverses are unique, so $f(1/n) = g(1/n)$, and then $f(m/n) = f(m) f(1/n) = g(m) g(1/n) = g(m/n)$ for all $m, n$.\nHence $f = g$, so $i$ is right-cancellable, that is, an epimorphism.\nIt is not surjective, since $1/2 \\notin \\Z$.\n\n## Remarks\n\nIn a concrete category of algebras every surjective homomorphism is an epimorphism, but the converse can fail.\nIt holds in the categories of groups and of abelian groups, where every epimorphism is surjective.\nOther non-surjective epimorphisms are the inclusion $(\\N, +) \\to (\\Z, +)$ in the category of monoids, and more generally every localization map $R \\to S^{-1} R$ of commutative rings.\n\n## Explanations\n\n### L1 human\n\nIn category theory, an epimorphism is an arrow that can be cancelled on the right, which for functions means being onto.\nFor rings this goes wrong.\nOnce a homomorphism out of $\\Q$ is known on the integers, it is already known on all fractions, because $1/n$ must go to the inverse of the image of $n$.\nSo the integers \"control\" the rationals, even though they are a small part of them.\n",
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  "msc": [
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