# $f(U)=U$ but $f$ is not injective.

I was given this exercise:

Let $U=\{(x,y): 1<x^2+y^2<2\}$ and $f:U\rightarrow \mathbb {R^2}$ defined by:

$$f(x,y)=\left(\frac {x^2-y^2}{r},\frac {2xy}{r}\right)$$

where $r=\sqrt {x^2+y^2}$.

Then I have to show that $f(U)=U$ but $f$ is not injective... Think that polar coordinates might help.. but how? I'm kind stuck.

Thanks for any help!

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$f$ is not injective because $f(-x,-y)=f(x,y)$, so for example the two points $(-1,-1)$ and $(1,1)$ are mapped to the same point $(0,\sqrt2)$. Polar coordinates are indeed a good start. Write $$x=r \cos(\phi)$$ $$y=r \sin(\phi)$$ Then $$f(x,y)=\left(\frac {r^2\cos^2(\phi)-r^2\sin^2(\phi)}{r},\frac {2r \cos(\phi)r \sin(\phi)}{r}\right)=\left(r(\cos^2(\phi)-r^2\sin^2(\phi)),r(2\sin(\phi)\cos\phi))\right)=\left(r\cos(2\phi),r\sin(2\phi)\right)$$ So, $f$ sends $(r,\phi)$ to $(r,2\phi)$. Since $U$ is just the annulus $1<r<\sqrt 2$ and $r$ is left invariant under $f$, $f$ sends $U$ to $U$ and every $(r,\phi) \in U$ has exactly two preimages $(r,\phi/2)$ and $(r,\phi/2+180°)$

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Thanks. And how do I show that $f(U)=U$ ? – Haytham Nov 26 '12 at 19:00
@Melvin: One way is to show that for all $u \in U$, there is a $z=(x,y)$ such that $f(z)=u$. DonAntonio's formulation of $f$ is a good step in that direction. – Ross Millikan Nov 26 '12 at 19:06
@Melvin See my answer below. – Fly by Night Nov 26 '12 at 19:18

With polar coordinates:

$$x=r\cos t\,\,,\,\,y=r\sin t\,\,,\,\,r\geq 0\,\,,\,0\leq t\leq 2\pi\Longrightarrow$$

$$U=\{(r,t)\;;\;\;1<r^2<2\}\,\,,\,f(r,t)=\left(r\cos 2t\,,\,r\sin 2t\right)$$

Thus, $\,f(r,t)=f(r,t+\pi)\,$

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Use complex numbers!

Let $z = x+iy$, then $|z| = \sqrt{x^2+y^2}$ and $z^2 = (x^2-y^2)+2ixy$. Thus, the complex function: $f : z \mapsto z^2/|z|$ can be identified with your function by taking the real part as the first coordinate, and the imaginary part as the second coordinate.

You hav $U = \{z \in \mathbb{C} : 1 < |z|^2 < 2 \}.$ Let $z \in U$, and consider

$$|f(z)| = \left|\frac{z^2}{|z|}\right| = \frac{|z|^2}{|z|} = |z|$$

It follows that $|z| = |f(z)|.$ Thus, $z \in U \iff f(z) \in U.$ Consider the two facts: $z \in U \implies f(z) \in U$ tells us that $f(U) \subseteq U,$ while $f(z) \in U \implies z \in U$ tells us that $U \subseteq f(U).$ It follows that $U = f(U).$

The function is clearly not injective since $f(\pm 1) = 1.$ In my notation $1 = 1 + 0i \in \mathbb{C},$ which corresponds to $(1,0) \in \mathbb{R}^2$ in your notation.

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Hint:

Given some $(x',y')$ s.t $$1<x'^{2}+y'^{2}<2$$ solve $$x'=\frac{x^{2}-y^{2}}{r},y'=\frac{2xy}{r}$$ for $x,y$ and show that $1<x^{2}+y^{2}<2$.

That proves that $U\subseteq f(U)$.

To prove $f(U)\subseteq U$ that $x,y$ s.t $$1<x^{2}+y^{2}<2$$ and prove that $$1<\left(\frac{x^{2}-y^{2}}{r}\right)^{2}+\left(\frac{2xy}{r}\right)^{2}<2$$

Both are done with relatively simple algebra.

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