I'm trying to understand the step highlighted in this demonstration:

enter image description here

from Zorich, Mathematical Analysis I, sec. 8.6, pag 510. What I know is that if $f'(x_0)$ is invertible ($f:G\subset\mathbb{R}^m\to\mathbb{R}^m$, $f\in\mathcal{C}^p(G)$), then f is locally a diffeomorphism (of smoothness $p$). In the step highlighted, instead, the opposite is stated, that is the other-sense implication ($\Leftarrow$). Is it always true? How would you prove it?

Thanks in advance.


The map $f$ is not only differentiable. It is assumed that $f$ is a diffeomorphism. This means that $f$ is bijective, and that the inverse map $g:=f^{-1}$ is differentiable as well. Since $g\bigl(f(x)\bigr)\equiv x$ the chain rule implies that $$g'\bigl(f(x)\bigr)\circ f'(x)={\rm id}\ ,$$ hence $f'(x)$ has to be regular.

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  • $\begingroup$ Thank you very much :) $\endgroup$ – Nameless Jan 19 at 11:45

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