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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 $C^k$.


Thank you.

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Welch ein schöner, romantischer Deckname! –  Georges Elencwajg Jan 27 '12 at 10:55
@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). –  Sam Jan 27 '12 at 17:34
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up vote 4 down vote accepted

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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Dear Robert and little son, I think you mean $g' = Inv \circ f' \circ g$ , not $g' = Inv \circ f \circ g$. –  Georges Elencwajg Jan 27 '12 at 11:01
Yes, thanks for catching the typo, –  Robert Israel Jan 27 '12 at 16:59
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