Is a differentiable (but not $C^1$) function $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ with invertible derivative everywhere an open map?

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 $$0, let $$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.

• 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. – CJD May 8 at 19:47
• 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. – Mindlack May 8 at 19:53
• 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. – punctured dusk May 8 at 20:07
• I do like your formulation better, it is much nicer. Do you mind if I edit it in (and credit you :) )? – Mindlack May 8 at 20:14
• 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. – CJD May 8 at 22:34