Proof of Inverse Function Theorem for $C^1$ Maps on Euclidean Open Sets

Reduction to the normalised case. Let $A=J_f(a)$. Since $\det J_f(a)\ne 0$, the matrix $A$ is invertible. The map $\tilde{f}(x)=A^{-1}(f(x+a)-f(a))$ is $C^1$ by the chain rule and linearity of $A^{-1}$, and satisfies $\tilde{f}(0)=0$ and $J_{\tilde{f}}(0)=A^{-1}A=I_n$. Since $f=f(a)+A\circ\tilde{f}(\cdot-a)$ and composition with the invertible affine map $y\mapsto f(a)+Ay$ preserves the local-$C^1$-diffeomorphism property, it suffices to prove the theorem under the additional assumptions $a=0$, $f(0)=0$, $J_f(0)=I_n$.

Step 1: Finding the contractive radius. Define $g:U\to\mathbb{R}^n$ by $g(x)=x-f(x)$. Then $g$ is $C^1$ and, since $J_f(0)=I_n$, we have $J_g(0)=I_n-I_n=0$. Because the entries of $J_g$ are continuous and vanish at $0$, there exists $r>0$ with $\overline{B}(0,r)\subseteq U$ such that

Step 2: Mean-value bound and injectivity. By the chain rule and the Fundamental Theorem of Calculus applied to $t\mapsto g_j(x'+t(x-x'))$, for any $x,x'\in\overline{B}(0,r)$ and each index $j$:

By the Cauchy--Schwarz inequality and the bound from Step 1,

Hence $\|f(x)-f(x')\|_2\ge\|x-x'\|_2-\|g(x)-g(x')\|_2\ge\tfrac{1}{2}\|x-x'\|_2$, so $f$ is injective on $\overline{B}(0,r)$.

Step 3: Existence of a local inverse via CMT. Set $W=\{y\in\mathbb{R}^n:\|y\|_2<r/2\}$. For each $y\in W$, define $T_y:\overline{B}(0,r)\to\mathbb{R}^n$ by $T_y(x)=g(x)+y$.

$T_y$ is a contraction: $\|T_y(x)-T_y(x')\|_2=\|g(x)-g(x')\|_2\le\tfrac{1}{2}\|x-x'\|_2$.

$T_y$ maps $\overline{B}(0,r)$ into itself: since $g(0)=0-f(0)=0$, for any $x\in\overline{B}(0,r)$,

The space $(\overline{B}(0,r),d_E)$ is complete, since $\overline{B}(0,r)$ is a closed subset of the complete Euclidean space $(\mathbb{R}^n,d_E)$.

By the Contraction Mapping Theorem, each $T_y$ has a unique fixed point $x^*(y)\in\overline{B}(0,r)$. The fixed point equation $T_y(x^*(y))=x^*(y)$ gives $f(x^*(y))=y$. Injectivity from Step 2 makes $x^*:W\to\overline{B}(0,r)$ the unique local inverse of $f$ on $W$.

Step 4: The local inverse is $C^1$. Let $V=f^{-1}(W)\cap B(0,r)$. Since $f$ is continuous and $W$, $B(0,r)$ are open, $V$ is open, and $f|_V:V\to W$ is a bijection with inverse $x^*$.

From the contraction estimate $\|x^*(y)-x^*(y')\|_2\le 2\|y-y'\|_2$, the inverse $x^*$ is Lipschitz, hence continuous. Since $\det J_f$ is continuous and nonzero at $0$, shrinking $r$ if necessary ensures $J_f(x)$ is invertible for all $x\in V$. Differentiating the identity $f(x^*(y))=y$ by the chain rule gives

The right-hand side is continuous in $y$ (composition of continuous $x^*$ with smooth $J_f$ and matrix inversion), so $x^*$ is $C^1$. Hence $f|_V:V\to W$ is a local $C^1$ diffeomorphism at $a=0$.