Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The following result can be found in the paper "Natural operations on differential forms" by Richard S. Palais:

Corollary 4.3: "Let $f_{1}, \ldots , f_{n}$ be $n$ real valued functions of $n$ real variables. Necessary and sufficient conditions that the mapping $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ defined by $f(x) = (f_{1}(x), \ldots , f_{n}(x))$ be a diffeomorphism of $\mathbb{R}^{n}$ onto itself are:
(1) $\det (df_{i}/dx_{j})$ never vanishes.
(2) $\lim_{||x||\rightarrow \infty} ||f(x)|| = \infty$."

If we assume that each of the $f_{i}$:s are smooth functions also of some other variables in $y \in \mathbb{R}^{m}$, is satisfaction of the above conditions for all $y \in \mathbb{R}^{m}$ necessary and sufficient for the $x_{i}$:s to be functions of $y$ in the entire $\mathbb{R}^{n}$ (intuitively the answer is yes I suppose)?

Also, if $f$ is smooth in $x$ and $y$, will $x$ be smooth as a function of $y$?

(I just realized my question is related to a previous one: How to "globalize" the inverse function theorem?)

Best regards

share|improve this question
Your questions are purely local in nature and therefore follow by the standard inverse function theorem. Voting to close. –  Deane Yang Jan 7 '12 at 9:53
Why do you think they are purely local? The second condition, that the norm of f tends to infinity, does not appear in the usual inverse function theorem –  user12400 Jan 8 '12 at 7:37
In case someone else comes across this. A motivation for why the second condition above is needed is of course given by considering the map $x \rightarrow e^{x}$. Obviously the derivative never vanishes, but $e^{x} = a$ has no solution for $a \leq 0$. –  user12400 Apr 12 '12 at 8:30
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.