# Showing a vector valued function is differentiable

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? – user1551 Dec 11 '12 at 5:44
Doesn't the Inverse Function theorem say precisely this? – Javier Dec 11 '12 at 14:53
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. – Gyu Eun Lee Dec 11 '12 at 15:26
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. – Nick Strehlke Dec 12 '12 at 19:53
The inverse function theorem doesn't necessarily apply, because we haven't assumed that $df$ is invertible everywhere. – Nick Strehlke Dec 12 '12 at 19:54