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I was reading up on the inverse function theorem and I was wondering if the following thought I had was correct:

If $f: \mathbb{R}^n \to \mathbb{R}^n$ $g: \mathbb{R}^n \to \mathbb{R}^n$ are eachothers inverse functions and $f,g$ are both continuously differentiable, does that imply that the the derivatives $f'$ and $g'$ are invertible themselves? And more specifically is $f'$ the inverse of $g'$ and vice versa?

My first reaction would be yes since it follows that the Jacobian of the composite functions is simply the identity. Therefore the Jacobians of $f$ and $g$ are inverses. Is my reasoning flawed?

github
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  • Try $n = 1$, $f = g = \mathsf{id}_{\Bbb{R}}$. – Rob Arthan Oct 09 '16 at 22:33
  • Perhaps it's badly worded, but if it asks what I think it asks, it is indeed true that if $f,g:\Bbb R^n\to\Bbb R^n$ are continuously differentiable at each point and $f\circ g=g\circ f=id_{\Bbb R^n}$, then $\nabla f(x)\cdot \nabla g(f(x))=\nabla g(f(x))\cdot \nabla f(x)=I_n$ for all $x\in\Bbb R^n$. It follows directly from the chain rule. –  Oct 09 '16 at 22:35

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By the chain rule $$ D (f \circ g) ({\bf x}) = Df(g({\bf x})) Dg({\bf x}) $$ and for the identity operator $$ DI({\bf x}) = I $$ Since $f \circ g = I$, what this tells you is that for any ${\bf x} \in \mathbb{R}^n$, the matrix $Df(g({\bf x}))$ and $Dg({\bf x})$ are inverses of each other and, by symmetry, the same is true for $Dg(f({\bf x}))$ and $Df({\bf x})$.

Tom
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