# Is there a function from $\mathbb R^2 \to \mathbb R^2$ that is continuous, and *not* one to one, but still maps open sets to open sets?

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!

• Do you want $f$ to be surjective as well? – Asaf Karagila Apr 3 '13 at 13:51

## 2 Answers

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 $f(x,y) = (x^2 - y^2, 2xy)$.

• Thanks. So for the R->R case, there is no such function still am I right? – Kevin Apr 3 '13 at 18:08
• How could I prove that this maps open sets to open sets without resorting to complex analysis? – Kevin May 1 '13 at 4:07

Try $z\to e^z$, i.e. $(x,y)\to (e^x\cos y , e^x\sin y)$.