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  "body": "\n## Proof\n\nWe find the inverse as the fixed point of a contraction, then show that it is differentiable, and finally of class $C^k$ by induction.\n\n1. Let $U$, $k$, $f$ and $a$ be as in the statement. Write $\\norm{x}$ for the Euclidean norm, and $\\norm{B}$ for the [[operator norm|0ZTEH055YF]] of a matrix $B$.\n2. Let $A = Df(a)$. Its [[Jacobian determinant|03W6Q1MFR5]] is not zero, so [[$A$ is invertible|03Q7RE2HXR]]. Let $c = \\norm{A^{-1}} \u003e 0$ and $\\lambda = 1/(2c)$.\n3. Since $f$ is of [[class C^1|0TCBZMEYGN]], it is [[differentiable|0KXSK8TEEB]] on $U$, and $Df(x)$ is the [[Jacobian matrix|0YF0VXKXGB]] of partial derivatives at $x$.\n4. For a matrix $B = (b_{ij})$ we have $\\norm{B} \\le (\\sum_{i,j} b_{ij}^2)^{1/2}$. This follows from the [[Cauchy–Schwarz inequality|0RFWW56Y22]] applied to each row of $B$.\n5. The partial derivatives of $f$ are continuous at $a$. By step 4, there is $r \u003e 0$ such that the closed ball $\\bar B = \\set{x : \\norm{x - a} \\le r}$ lies in $U$ and $\\norm{Df(x) - A} \\le \\lambda$ for all $x \\in \\bar B$.\n6. For $y \\in \\R^n$ and $x \\in U$, let $\\varphi_y(x) = x + A^{-1}(y - f(x))$. Since $A^{-1}$ is injective, $f(x) = y$ if and only if $\\varphi_y(x) = x$.\n7. The map $\\varphi_y$ is differentiable with $D\\varphi_y(x) = I - A^{-1} Df(x) = A^{-1}(A - Df(x))$. For $x \\in \\bar B$, step 5 gives $\\norm{D\\varphi_y(x)} \\le c \\lambda = \\tfrac{1}{2}$.\n8. The ball $\\bar B$ is [[convex|0K06X9QXDX]]. By the [[mean value inequality|02CYB184RC]] and step 7, $\\norm{\\varphi_y(x_1) - \\varphi_y(x_2)} \\le \\tfrac{1}{2} \\norm{x_1 - x_2}$ for all $x_1, x_2 \\in \\bar B$.\n9. We have $\\varphi_y(x_1) - \\varphi_y(x_2) = (x_1 - x_2) - A^{-1}(f(x_1) - f(x_2))$. So step 8 and the triangle inequality give $\\norm{x_1 - x_2} \\le \\tfrac{1}{2} \\norm{x_1 - x_2} + c \\norm{f(x_1) - f(x_2)}$.\n10. Hence $\\norm{x_1 - x_2} \\le 2 c \\norm{f(x_1) - f(x_2)}$ for all $x_1, x_2 \\in \\bar B$. In particular, $f$ is injective on $\\bar B$.\n11. Let $V = \\set{x : \\norm{x - a} \u003c r}$. It is open, $a \\in V$ and $V \\subseteq U$. Let $W = f(V)$, so $f(a) \\in W$.\n12. We show that $W$ is open. Let $y_0 \\in W$, with $y_0 = f(x_0)$ and $x_0 \\in V$. Choose $\\rho \u003e 0$ with $K = \\set{x : \\norm{x - x_0} \\le \\rho} \\subseteq V$. Let $y \\in \\R^n$ with $\\norm{y - y_0} \u003c \\lambda \\rho$.\n13. For $x \\in K$, step 8 gives $\\norm{\\varphi_y(x) - \\varphi_y(x_0)} \\le \\rho / 2$. Also $\\varphi_y(x_0) - x_0 = A^{-1}(y - y_0)$, whose norm is less than $c \\lambda \\rho = \\rho / 2$. So $\\norm{\\varphi_y(x) - x_0} \u003c \\rho$, and $\\varphi_y$ maps $K$ into $K$.\n14. The set $K$ is [[closed|0YS40B38JD]] in $\\R^n$, which is [[complete|0ABP2F2ZP7]]. So $K$ is a [[complete|0FBXJZX3MK]] nonempty metric space. By step 8, $\\varphi_y$ is a [[contraction|0MT2S0ZX4Y]] of $K$ with constant $\\tfrac{1}{2}$.\n15. By the [[Banach fixed point theorem|0VK6VEH345]], $\\varphi_y$ has a fixed point $x \\in K \\subseteq V$. By step 6, $f(x) = y$, so $y \\in W$. Thus $W$ contains the ball of radius $\\lambda \\rho$ around $y_0$, and $W$ is open.\n16. By steps 10 and 11, $f$ restricted to $V$ is a bijection from $V$ onto $W$. Let $g \\colon W \\to V$ be its inverse.\n17. For $x \\in V$, the matrix $Df(x)$ is invertible. Indeed, if $Df(x) h = 0$, then $\\norm{h} = \\norm{A^{-1} (A - Df(x)) h} \\le c \\lambda \\norm{h} = \\tfrac{1}{2} \\norm{h}$, so $h = 0$.\n18. Step 10 gives $\\norm{g(y_1) - g(y_2)} \\le 2c \\norm{y_1 - y_2}$ for $y_1, y_2 \\in W$. So $g$ is [[continuous|0S9EK5MQPE]].\n19. Let $y \\in W$, $x = g(y)$ and $T = Df(x)^{-1}$. Let $k \\ne 0$ with $y + k \\in W$, and let $h = g(y + k) - x$. Then $f(x + h) = y + k \\ne y$, so $h \\ne 0$. By step 18, $\\norm{h} \\le 2c \\norm{k}$.\n20. Since $f(x + h) - f(x) = k$ and $T Df(x) h = h$, we get $g(y + k) - g(y) - T k = h - T k = -T \\big( f(x + h) - f(x) - Df(x) h \\big)$.\n21. So $\\dfrac{\\norm{g(y + k) - g(y) - T k}}{\\norm{k}} \\le 2 c \\norm{T} \\cdot \\dfrac{\\norm{f(x + h) - f(x) - Df(x) h}}{\\norm{h}}$.\n22. As $k \\to 0$, also $h \\to 0$ by step 19. Since $f$ is differentiable at $x$, the right side of step 21 tends to $0$. So $g$ is differentiable at $y$ with $Dg(y) = Df(g(y))^{-1}$. This is the formula $D(f^{-1})(f(x)) = Df(x)^{-1}$.\n23. Identify $n \\times n$ matrices with $\\R^{n^2}$. The determinant is a polynomial in the entries, so the set $\\mathcal{M}$ of invertible matrices is open. For $B \\in \\mathcal{M}$, each entry of $B^{-1}$ is an entry of the adjugate divided by $\\det B$ ([[inverse via the adjugate|0ENVM493VK]]; for $n = 1$, $B^{-1} = 1/B$).\n24. So each entry map $\\Psi_{il}(B) = (B^{-1})_{il}$ is a polynomial divided by a nonvanishing polynomial on $\\mathcal{M}$. Its partial derivatives of every order are again of this form, so $\\Psi_{il}$ is of class $C^j$ on $\\mathcal{M}$ for every $j$.\n25. Lemma. Let $j \\ge 1$, let $O \\subseteq \\R^N$ and $W' \\subseteq \\R^p$ be open, let $\\Phi \\colon O \\to \\R$ be of class $C^j$, and let $G \\colon W' \\to O$ be of class $C^j$. Then $\\Phi \\circ G$ is of class $C^j$. We prove this by induction on $j$.\n26. Both $\\Phi$ and $G$ are differentiable, so the [[chain rule|0EAAS2EHDR]] gives $\\partial_i (\\Phi \\circ G) = \\sum_l \\big( (\\partial_l \\Phi) \\circ G \\big) \\, \\partial_i G_l$. For $j = 1$, the right side is continuous, so $\\Phi \\circ G$ is of class $C^1$.\n27. Assume the lemma for $j$, and let $\\Phi$ and $G$ be of class $C^{j+1}$. In step 26, $(\\partial_l \\Phi) \\circ G$ is of class $C^j$ by the induction hypothesis, and so is $\\partial_i G_l$.\n28. A product $p q$ of two $C^j$ functions is $\\mu \\circ (p, q)$ with the polynomial $\\mu(s, t) = s t$, so it is of class $C^j$ by the induction hypothesis. Sums of $C^j$ functions are of class $C^j$.\n29. So every first partial derivative of $\\Phi \\circ G$ is of class $C^j$. Hence $\\Phi \\circ G$ is of class $C^{j+1}$, which proves the lemma.\n30. By step 22, the partial derivatives of $g$ are $\\partial_l g_i = \\Psi_{il} \\circ (Df \\circ g)$ on $W$. The entries of $Df \\circ g$ are the functions $(\\partial_l f_i) \\circ g$.\n31. The functions $\\partial_l f_i$ and $g$ are continuous, so $Df \\circ g$ is continuous, and so is each $\\partial_l g_i$. Hence $g$ is of class $C^1$.\n32. Let $1 \\le j \u003c k$, and assume that $g$ is of class $C^j$. Each $\\partial_l f_i$ is of class $C^{k-1}$, hence of class $C^j$. By the lemma, each $(\\partial_l f_i) \\circ g$ is of class $C^j$, so $Df \\circ g \\colon W \\to \\mathcal{M}$ is of class $C^j$.\n33. By step 24 and the lemma, each $\\partial_l g_i = \\Psi_{il} \\circ (Df \\circ g)$ is of class $C^j$. So $g$ is of class $C^{j+1}$.\n34. By induction, $g = f^{-1}$ is of class $C^k$ on $W$. With steps 11, 15, 16 and 22 this proves the theorem.\n\n## Idea\n\nTo solve $f(x) = y$ near $a$, replace $f$ by its linear part $A$ and correct: the map $x \\mapsto x + A^{-1}(y - f(x))$ has fixed points exactly at the solutions. Near $a$ its derivative is small, so it is a contraction and has a unique fixed point. The same estimate shows that the inverse is Lipschitz, which gives its derivative, and the formula $Dg = (Df \\circ g)^{-1}$ lifts smoothness from $C^j$ to $C^{j+1}$.\n",
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