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  "body": "\n## Statement\n\nLet $\\mathcal{C}$ be a [[Cartesian closed category|BFBNQ36QS8]] with [[terminal object|EK85VSSRMR]] $1$.\nA *point* of an object $X$ is a morphism $1 \\to X$.\nLet $A$ and $B$ be objects, and let $B^A$ be the [[exponential object|RHWD8DY41R]].\nLet $\\phi \\colon A \\to B^A$ be point-surjective: for every point $q \\colon 1 \\to B^A$ there is a point $p \\colon 1 \\to A$ with $\\phi \\circ p = q$.\nThen every morphism $f \\colon B \\to B$ has a fixed point: there is a point $s \\colon 1 \\to B$ with $f \\circ s = s$.\n\n## Remarks\n\nProof idea: let $\\mathrm{ev} \\colon B^A \\times A \\to B$ be evaluation and $\\Delta \\colon A \\to A \\times A$ the diagonal, built from the [[product|S5744QZV4N]].\nPut $g = f \\circ \\mathrm{ev} \\circ (\\phi \\times \\id_A) \\circ \\Delta \\colon A \\to B$, so that informally $g(a) = f(\\phi(a)(a))$.\nThe transpose of $g$ is a point of $B^A$, so it equals $\\phi \\circ p$ for some point $p$ of $A$.\nThen $s = g \\circ p$ satisfies $s = f(\\phi(p)(p)) = f(g(p)) = f \\circ s$.\n\nThe contrapositive is the form used most.\nIn $\\mathbf{Set}$, take $B = \\set{0, 1}$ and $f$ the swap of $0$ and $1$, which has no fixed point.\nSo no map $A \\to \\set{0,1}^A$ is surjective, which is [[Cantor's theorem|0WTEMGYV2C]] that no set maps onto its power set.\nThe proof is an abstract form of [[Cantor's diagonal argument|0XCH8QVDCQ]].\n\n## Explanations\n\n### L1 human\n\nSuppose the elements of a set $A$ can serve as codes for all functions from $A$ to $B$.\nThen every map from $B$ to itself must leave some value unchanged.\nFlipping a bit leaves no value unchanged, so no set can code all bit-valued functions on itself.\nThis one short argument is behind Cantor's theorem and Russell's paradox.\n\n### L2 human\n\nThe theorem needs only finite products and exponentials, so it applies in any Cartesian closed category, including toposes and models of the typed lambda calculus.\nRead positively, it explains fixed-point combinators: if $A$ maps point-surjectively onto $A^A$, then every endomorphism of $A$ has a fixed point.\nRead negatively, it unifies diagonal arguments: Cantor's theorem, Russell's paradox and Tarski's undefinability of truth all come from an $f$ without fixed point that forbids a point-surjection.\n",
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