Proof of Smooth Inverse Function Theorem on Euclidean Open Sets

Since every smooth map is $C^1$, the Inverse Function Theorem applies: there exist open sets $V\ni a$ and $W\ni f(a)$ with $V\subseteq U$ such that $h=(f|_V)^{-1}:W\to V$ is well-defined and $C^1$.

We show $h$ is smooth. By the chain rule applied to the identity $f(h(y))=y$,

so $J_h(y)=\bigl(J_f(h(y))\bigr)^{-1}$, where $(\cdot)^{-1}$ denotes the matrix inverse. The map $M\mapsto M^{-1}$ is smooth on the open set of invertible matrices: by Cramer's rule, each entry of $M^{-1}$ is a rational function of the entries of $M$, and such rational functions are smooth wherever the denominator $\det M$ is nonzero, by the product rule for smooth maps.

We now argue directly that $h$ is $C^k$ for every $k\ge 1$. Since $h$ is $C^1$ and $J_f$ is smooth, the composition $J_f\circ h$ has $C^1$ entries; since matrix inversion is smooth, $J_h=\bigl(J_f\circ h\bigr)^{-1}$ also has $C^1$ entries, so $h$ is $C^2$. Since $h$ is now $C^2$, the composition $J_f\circ h$ has $C^2$ entries, so $J_h$ has $C^2$ entries and $h$ is $C^3$. Repeating this argument shows $h$ is $C^k$ for every $k\ge 1$, hence smooth.