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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $k \\ge 1$, and let $f \\colon U \\to \\mathbb{R}^n$ be of [[class C^k|0TCBZMEYGN]] on $U$.\nLet $a \\in U$, and suppose that the derivative $Df(a)$ is invertible, that is, the [[Jacobian determinant|03W6Q1MFR5]] of $f$ at $a$ is not zero.\nThen there are open sets $V \\subseteq U$ with $a \\in V$ and $W \\subseteq \\mathbb{R}^n$ with $f(a) \\in W$ such that $f$ restricted to $V$ is a bijection from $V$ onto $W$.\nIts inverse $f^{-1} \\colon W \\to V$ is of class $C^k$, and $D(f^{-1})(f(x)) = Df(x)^{-1}$ for each $x \\in V$.\n\n## Remarks\n\nThe theorem is local. The map $f(x, y) = (e^x \\cos y, e^x \\sin y)$ has invertible derivative at each point of $\\mathbb{R}^2$, but it is not injective, since $f(x, y + 2\\pi) = f(x, y)$.\nContinuity of the derivative matters. The function $g(x) = x + 2 x^2 \\sin(1/x)$, $g(0) = 0$, has $g'(0) = 1$, but $g$ is not injective on any interval around $0$.\nA standard proof applies the contraction mapping principle to $x \\mapsto x + Df(a)^{-1}(y - f(x))$ near $a$.\nThe formula for the derivative of the inverse is the [[derivative of the inverse function|0G65751VSW]].\n\n## Explanations\n\n### L1 human\n\nIf the linear approximation of $f$ at $a$ is invertible, then $f$ itself is invertible near $a$.\nSo one can solve the system $f(x) = y$ for $x$ when $y$ is close to $f(a)$, and the solution depends smoothly on $y$.\n",
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    "name": "Inverse function theorem",
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  "markdown": "https://math2.tamnd.com/node/0K9HSJFCMJ.md",
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  "name": "Inverse function theorem",
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