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Let $f:ℝ^{r+1}→ℝ^{r+1}$ be a real analytic function. Assume that the determinant of its jacobian matrix is non zero for all $x∈ℝ^{r+1}$, i.e., for every point $x∈ℝ^{r+1}$, there exists a neighborhood about $x$ over which $f$ is locally invertible.

My question is: How I can prove that the jacobian matrix of the inverse function $f⁻¹$ is also non zero for all $x∈ℝ^{r+1}$?

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From the inverse function theorem we have that if $f$ is $C^1$ on some neighborhood $U$ of $\hat{x}$, and $Df(\hat{x})$ is invertible, then there exists some neighborhood $V$ of $f(\hat{x})$, and a $C^1$ function $\phi:V \to U$ such that $\phi(f(x)) = x$. Furthermore, $D \phi(f(\hat{x})) = (Df(\hat{x}))^{-1}$

If $f$ has an inverse, then we must have $f^{-1} = \phi$ on $V$, and hence $D(f^{-1})(f(\hat{x}) ) = (Df(\hat{x}))^{-1}$. Since $\det Df(\hat{x}) \neq 0$, it follows that $\frac{1}{\det Df(\hat{x})} \neq 0$, hence $\det D(f^{-1})(f(\hat{x}) ) \neq 0$.

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