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Is there any function $f: \mathbb{R}^2 \to \mathbb{R}^2$ which is $C^1$ and has an invertible derivative matrix at all points, but is not 1-1?

Thanks!

Arnold
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  • The condition $| F(u)-F(v)| \geq c|u -v|$ (with $c > 0$) already proves $F$ is $1-1$ without any assumptions about being $C^1$ or having invertible derivative. To see this, assume $F$ is a (set theoretic) function satisfying $F(u) = F(v)$. Then $0 = | F(u) - F(v)| \geq c|u-v|$ which implies $|u-v| = 0$, so $u=v$. – Jason DeVito - on hiatus Apr 03 '13 at 19:12
  • True, that part of my question was irrelevant so I removed it. I was getting confused, you could use that condition + C1 to show that the function is actually onto, which is a more substantial argument. – Arnold Apr 03 '13 at 19:14

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Let $F(x,y)=(e^x\cos y,e^x\sin y)=(u,v)$. Then $F(x,y)=F(x,y+2\,\pi)$, so that $F$ is not $1$ to $1$. But $$ \frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}e^x\cos y &- e^x\sin y\\ e^x\sin y & e^x\cos y\end{vmatrix}=e^x\ne0 \quad\forall(x,y), $$ so that it is locally invertible.

In case you wonder how I came up with such example, $u$ and $v$ are the real and imaginary part of $e^ {x+iy}$.

Julián Aguirre
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