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  "body": "\n## Statement\n\nLet $X$ and $Y$ be [[compact|0TVJZYB6SH]] [[Hausdorff spaces|0S1FJ8VEX9]].\nLet $C(X)$ and $C(Y)$ be the spaces of continuous real functions with the [[supremum norm|0PARCKE8WP]].\nLet $T \\colon C(X) \\to C(Y)$ be a surjective [[linear isometry|07SJXTWVQW]].\nThen there are a [[homeomorphism|0QT00PB5G2]] $\\varphi \\colon Y \\to X$ and a continuous function $h \\colon Y \\to \\set{-1, 1}$ such that\n$$(T f)(y) = h(y) \\, f(\\varphi(y)) \\quad \\text{for all } f \\in C(X), \\ y \\in Y.$$\n\n## Remarks\n\nThe idea of the proof: the extreme points of the unit ball of the dual of $C(X)$ are the functionals $\\pm \\delta_x$, with $\\delta_x(f) = f(x)$.\nThe adjoint of $T$ is an isometry of the duals and maps extreme points to extreme points.\nSo it sends each $\\delta_y$ to $h(y) \\delta_{\\varphi(y)}$, and the weak-* topology makes $\\varphi$ and $h$ continuous.\n\nExample: on $X = Y = [0, 1]$, the map $(T f)(t) = -f(1 - t)$ is a surjective linear isometry.\nHere $h = -1$ and $\\varphi(t) = 1 - t$.\n\nFor complex scalars the same holds with $h$ a continuous function of absolute value $1$.\nAs a consequence, $X$ and $Y$ are homeomorphic if and only if $C(X)$ and $C(Y)$ are isometrically isomorphic.\nIsometry is needed.\nLet $S = \\set{0} \\cup \\set{1/n : n \\ge 1}$, a convergent sequence with its limit.\nThen $C(S)$ and $C(S \\times \\set{1, 2})$ are linearly isomorphic, but $S$ has one limit point and $S \\times \\set{1, 2}$ has two.\n\n## Explanations\n\n### L1 human\n\nThe norm of $C(X)$ remembers the space $X$.\nEvery way to match $C(X)$ with $C(Y)$ that keeps distances comes from matching the points of $Y$ with the points of $X$, up to a change of sign.\n\n### L2 human\n\nThe theorem shows that the Banach space $C(X)$, with its norm, determines the compact space $X$ up to homeomorphism.\nGelfand–Naimark theory gives a similar result for $C(X)$ as an algebra, and the Banach–Stone theorem gets it from the metric structure alone.\nWithout the isometry condition the result fails badly: by a theorem of Milutin, $C(X)$ is isomorphic to $C([0, 1])$ for every uncountable compact metric space $X$.\n",
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