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  "body": "\n## Statement\n\nLet $M$ and $N$ be [[smooth manifolds|F45VQ47GPV]] of dimensions $m$ and $n$, and let $f \\colon M \\to N$ be a smooth map.\nA point $y \\in N$ is a *regular value* of $f$ if the [[differential|6P9PXQDSEN]] $df_x \\colon T_xM \\to T_yN$ is surjective for every $x \\in f^{-1}(y)$.\nLet $y$ be a regular value of $f$ with $f^{-1}(y)$ not empty.\nThen the [[level set|0B2RZ0778N]] $f^{-1}(y)$ is an [[embedded submanifold|VKFW27WXEZ]] of $M$ of dimension $m - n$.\nIts tangent space at each point $x \\in f^{-1}(y)$ is $\\ker df_x$.\n\n## Remarks\n\nThe set of points where $df_x$ is surjective is open, so $f$ is a [[submersion|BMY55EW3RF]] on a neighbourhood of $f^{-1}(y)$.\nIn charts the claim becomes the [[implicit function theorem|0BXBADR94C]].\nNear each $x$ there are coordinates in which $f$ is the projection onto the first $n$ coordinates, and the fibre is then a coordinate slice.\nExample: $f \\colon \\R^{n+1} \\to \\R$, $f(x) = \\abs{x}^2$, has $df_x(v) = 2 \\langle x, v \\rangle$, which is onto for $x \\neq 0$.\nSo $1$ is a regular value and the unit sphere $S^n = f^{-1}(1)$ is a submanifold of dimension $n$.\nThe hypothesis is needed: $0$ is not a regular value of $g(x, y) = xy$, and $g^{-1}(0)$ is the union of the two axes, which is not a submanifold.\n\n## Explanations\n\n### L1 human\n\nSolve an equation $f(x) = y$ on a smooth space.\nIf the equation is \"non-degenerate\" at every solution, the solutions form a smooth space again, of dimension equal to the number of unknowns minus the number of equations.\nThis is how one shows that spheres, many matrix groups and other familiar shapes are manifolds.\n\n### L2 human\n\nThe theorem is the main tool for producing examples of manifolds.\nFor instance, the orthogonal group $O(n)$ is the fibre over the identity of the map $A \\mapsto A^T A$ into symmetric matrices.\nThe identity is a regular value, so $O(n)$ is a manifold of dimension $n(n-1)/2$.\nBy [[Sard's theorem|Z7R8J71SJN]] almost every value is regular, so the theorem applies to almost every fibre.\nIt is the special case of transversality to a point, and the same proof gives that the preimage of a submanifold under a transverse map is a submanifold.\n",
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