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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 $x^2 \sin(1/x)$.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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