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I don't understand the reason for the conclusion, written in boldface at the end, in the following argument (taken from Elon LIMA, Curso De Análise, Vol .2).

If $f:U\longrightarrow V\subset \mathbb{R}^m$ is a diffemorphism which is $C^{k}$, then $g=f^{-1}$ is $C^k$ as well: indeed, by the chain rule, g'(y)=(Inv\circ f'\circ g) (y), where $Inv$ is a $C^{\infty}$ map from $GL(\mathbb{R}^n)$ onto itself. And, since $f$ is $C^k$, it follows, from these facts, that $g$ is $C^k$.

Why?

Thank you.

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    @GeorgesElencwajg: Good effort, but 'Deckname' is sort of reserved for spies in German ;). Maybe one should rather go with 'Pseudonym' here (although there might be a better word which doesn't come to mind right now).2012-01-27

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The composition of $C^j$ maps is $C^j$. Knowing that $g$ is continuous, g' = Inv \circ f' \circ g is continuous so $g$ is $C^1$. If $g$ is $C^1$ and $f$ is $C^2$, then f' is $C^1$ so g' is $C_1$ and $g$ is $C^2$. Induct...

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    Yes, thanks for catching the typo,2012-01-27