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(This question was posed to me by a colleague; I was unable to answer it, so am posing it here instead.)

Let $f: {\bf R}^n \to {\bf R}^n$ be an everywhere differentiable map, and suppose that at each point $x_0 \in {\bf R}^n$, the derivative $Df(x_0)$ is nonsingular (i.e. has non-zero determinant). Does it follow that $f$ is locally injective, i.e. for every $x_0 \in {\bf R}^n$ is there a neighbourhood $U$ of $x_0$ on which $f$ is injective?

If $f$ is continuously differentiable, then the claim is immediate from the inverse function theorem. But if one relaxes continuous differentiability to everywhere differentiability, the situation seems to be much more subtle:

  1. In one dimension, the answer is "Yes"; this is the contrapositive of Rolle's theorem, which works in the everywhere differentiable category. (The claim is of course false in weaker categories such as the Lipschitz (and hence almost everywhere differentiable) category, as one can see from a sawtooth function.)
  2. The Brouwer fixed point theorem gives local surjectivity, and degree theory gives local injectivity if $\det Df(x_0)$ never changes sign. (This gives another proof in the case when $f$ is continuously differentiable, since $\det Df$ is then continuous.)
  3. On the other hand, if one could find an everywhere differentiable map $f: B \to B$ on a ball $B$ that was equal to the identity near the boundary of $B$, whose derivative was always non-singular, but for which $f$ was not injective, then one could paste infinitely many rescaled copies of this function $f$ together to produce a counterexample. The degree theory argument shows that such a map does not exist in the orientation-preserving case, but maybe there is some exotic way to avoid the degree obstruction in the everywhere differentiable category?

It seems to me that a counterexample, if one exists, should look something like a Weierstrass function (i.e. a lacunary trigonometric series), as one needs rather dramatic failure of continuity of the derivative to eliminate the degree obstruction. To try to prove the answer is yes, one thought I had was to try to use Henstock-Kurzweil integration (which is well suited to the everywhere differentiable category) and combine it somehow with degree theory, but this integral seems rather unpleasant to use in higher dimensions.

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according to your second item(2): Is there a differentiable function for which $det Df(x)$ changes sign, but does not vanish? –  Ali Taghavi Jan 2 at 22:18

4 Answers 4

up vote 42 down vote accepted

The usual reference to the proof is A. V. Cernavskii in "Finite-to-one open mappings of manifolds", Mat. Sb. (N.S.), 65(107) (1964), 357–369 and "Addendum to the paper "Finite-to-one open mappings of manifolds"", Mat. Sb. (N.S.), 66(108) (1965), 471–472. If I remember it correctly, he does not state it explicitly, but it follows from what is there.

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4  
Welcome to MO, David! –  Gil Kalai Sep 10 '11 at 21:49
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Thanks David! (Now I just need to get a physical copy of the paper and learn some Russian...) –  Terry Tao Sep 11 '11 at 0:30
14  
(Ah, there is a translated copy at American Mathematical Society Translations, Series 2. Vol. 100: Fourteen papers on logic, geometry, topology and algebra. American Mathematical Society, Providence, R.I., 1972. iv+316 pp. Still need to get to the library, though. ) –  Terry Tao Sep 11 '11 at 0:44

For another proof that your question has a positive answer, you may look at the folowing paper by Jean Saint Raymond: http://www.math.jussieu.fr/~raymond/preprints/inversion.dvi

It seems that he was not aware of the reference given by David Preiss above - his proof is in English so at least it should avoid you having to learn Russian just for that...

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7  
Thanks! It is a nice argument; I decided to write it up (so that I could properly understand it) at terrytao.wordpress.com/2011/09/12/… –  Terry Tao Sep 13 '11 at 0:10
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You're more than welcome. I'm amazed how quickly you read the paper and wrote about it... –  Julien Melleray Sep 13 '11 at 14:13

From http://arxiv.org/abs/1011.1288 by Ivar Ekeland:

I present an inverse function theorem for differentiable maps between Frechet spaces which contains the classical theorem of Nash and Moser as a particular case. In contrast to the latter, the proof does not rely on the Newton iteration procedure, but on Lebesgue's dominated convergence theorem and Ekeland's variational principle. As a consequence, the assumptions are substantially weakened: the map F to be inverted is not required to be C^2, or even C^1, or even Frechet-differentiable.

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5  
Thanks for the interesting reference! Reading it, though, it seems that this paper is focused on the infinite dimensional case, and only establishes local surjectivity rather than local injectivity (it seems oriented towards the task of proving existence (but not uniqueness) for PDE. It also has an additional hypothesis that $Df(x_0)^{-1}$ is locally bounded. –  Terry Tao Sep 10 '11 at 16:06

From this result one can obtain a differentiable but nonsmooth version of the Implicit Function Theorem by the usual argument.

There are two interesting closely related questions apparently still remaining:

  1. Is there a corresponding version of the continuous Implicit Function Theorem not requiring continuous differentiability in the variables being solved for?

  2. In the original inductive proof of the Implicit Function Theorem, continuous differentiability was needed to insure a decreasing chain of locally nonvanishing minors for the Jacobian determinant. Does there always exist such a chain if simple differentiability with nonzero Jacobian is assumed?

A positive answer to 2 gives a positive answer to 1 and an inductive proof of the differentiable nonsmooth Inverse Function Theorem.

For a more careful description of these problems, see the exposition of the Implicit Function Theorem in my real analysis manuscript on my website http://wolfweb.unr.edu/homepage/bruceb/ .

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This great question should be posted as a separate question. Here it is somehow lost and probably does not receive proper attention. –  András Bátkai Dec 4 '13 at 7:28
    
I agree. If you decide to write a separate question, it may help to add a link to this one (and perhaps write a comment on the question at the top, pointing to your question). –  S. Carnahan Dec 4 '13 at 11:29
    
Thanks, I did that; see mathoverflow.net/questions/150856/… –  Bruce Blackadar Dec 4 '13 at 22:48

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