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The question is whether there is a function from $f: \mathbb{R}^2 \to \mathbb{R}^2$ which is continuous but NOT one-to-one, and maps open sets to open sets.

I can show this for the $\mathbb{R} \to \mathbb{R}$ case, by saying that not one-to-one means there is an $x$ and a $y$ with $f(x) = f(y)$, and while $f([x,y])$ is a closed interval by the extreme and intermediate value theorems, removing a single point $f(x) = f(y)$ from a closed interval will not make it open.

But this argument doesn't really generalize in any way that I can see.

Thank you!

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  • $\begingroup$ Do you want $f$ to be surjective as well? $\endgroup$ – Asaf Karagila Apr 3 '13 at 13:51
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In complex analysis it is shown that every analytic function maps open sets to open sets. So if you take $f(z) = z^2$ it will satisfy the conditions you seek. Written out as a function on ${\mathbb R}^2$, you'd have $f(x,y) = (x^2 - y^2, 2xy)$.

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  • $\begingroup$ Thanks. So for the R->R case, there is no such function still am I right? $\endgroup$ – Kevin Apr 3 '13 at 18:08
  • $\begingroup$ How could I prove that this maps open sets to open sets without resorting to complex analysis? $\endgroup$ – Kevin May 1 '13 at 4:07
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Try $z\to e^z$, i.e. $(x,y)\to (e^x\cos y , e^x\sin y)$.

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