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Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be continuously differentiable. Suppose there is a function $g:\mathbb{R}^n\to\mathbb{R}^n$ such that $fg=gf=1_{\mathbb{R}^n}$.

Does it follow that $g$ is differentiable?

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    What do you mean by $fg$? Is it the function composition of $f$ and $g$? And what is $1$? A vector of ones?2012-12-11
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    Doesn't the Inverse Function theorem say precisely this?2012-12-11
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    The inverse function theorem is the converse of this: if we assume $f$ is invertible then its inverse looks like this. Here we're not really assuming $f$ is invertible, we're just saying there's a function that acts like the inverse without actually saying it's the ivnerse.2012-12-11
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    The equality $fg=gf=1_{\mathbb R}$ means that $f$ is (globally) invertible with inverse $g$, and you supposed $f$ continuously differentiable. Hence, as pointed out by Javier Badia, you can use the inverse function theorem. http://en.wikipedia.org/wiki/Inverse_function_theorem2012-12-12
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    The answer should be no. Take $f(x) = x^3$ on $\mathbb R^1$. Then $g(x) = x^{1/3}$ satisfies your condition (if I understand your notation correctly). However, $g$ is not differentiable at the origin.2012-12-12
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    The inverse function theorem doesn't necessarily apply, because we haven't assumed that $df$ is invertible everywhere.2012-12-12
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    @NickStrehlke Yes, indeed you're right, I take back what I said in my previous comment.2012-12-12

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