Is a differentiable (but not $C^1$) function $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ with invertible derivative everywhere an open map? I know that if we assume the function is $C^1$, then this is a consequence of the inverse function theorem (or a step on the way to proving the inverse function theorem). I've convinced myself no counter-example exists if $n = 1$, but I haven't been able to prove it or come up with a counter-example in the case $n = 2$. I've spent more time trying to come up with a counter-example, mostly trying variants involving $x^2 \sin(1/x)$.


The result holds in every dimension.

For the sake of simplicity, we assume that $f(0)=0$ and $f’(0)=Id$.

There is a continuous function (with vector values) $\epsilon$ such that for each $x$, $f(x)=x+|x|\epsilon(x)$.

Let $R >0$ be such that if $|x| \leq R$, $|\epsilon(x)| \leq 0.5$.

Let $0<r<R$, let $z$ be such that $|z| < r/4$.

It is easy to prove that for any $x$ with $|x|=r$, $|f(x)-z| > |f(0)-z|$.

So consider $g(x)=f(x)-z$ and $h(x)=|g(x)|$ for $|x| \leq r$.

Assume that $g$ does not vanish, then $h$ is differentiable, without critical points, and does not reach its minimum on the border of the disc, which is impossible.

So $g$ does vanish, ie, for all $z$ with $|z|$ small enough, there is some $y$ with $|y| \leq 5|z|$ sich that $f(y)=z$.

In other words: if $V$ is a neighborhood of $0$, so is $f(V)$, which ends the proof.

  • $\begingroup$ Thank you very much for the answer. Unfortunately I only had time to skim what you wrote now. May I ask, how do you know that h has no critical points? But I might be able to figure this out for myself in a few hours when I come back to the computer. Thanks again. $\endgroup$ – CJD May 8 at 19:47
  • $\begingroup$ Since $|\cdot|$ is a submersion from $\mathbb{R}^n\backslash \{0\}$ to $\mathbb{R}^{+*}$, if $\nabla h(x)=0$, then $g’(x)=0$, ie $f’(x)=0$, a contradiction. $\endgroup$ – Mindlack May 8 at 19:53
  • 1
    $\begingroup$ Perhaps it's easier to understand the argument if expressed this way: consider $h(x) = |g(x)|^2$. It reaches its minimum somewhere on the disk $B(0, r)$. It can't be at the boundary, so its gradient must vanish at some point. Computing the gradient, this shows that either $g$ has a zero at that point, or that the differential of $g$ is singular. The latter does not happen by assumption. $\endgroup$ – punctured dusk May 8 at 20:07
  • $\begingroup$ I do like your formulation better, it is much nicer. Do you mind if I edit it in (and credit you :) )? $\endgroup$ – Mindlack May 8 at 20:14
  • $\begingroup$ Thank you again! As far as I can tell, this is very similar to part of the proof of the inverse function theorem in Munkres's "Analysis on Manifolds". I was suspicious at first because that proof requires $C^1$, but as far as I can tell, your use of the function $\varepsilon(x)$ replaces the part of Munkres's proof where $C^1$ is used. $\endgroup$ – CJD May 8 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.