Is $f$ bijective on$\mathbb{R}^2$?

Question

Let $f(x,y)=(x^2-y^2,2xy)$ a function from $\mathbb{R}^2\to\mathbb{R}^2$. Study if $f$ does have an inverse in whole $\mathbb{R}^2$?

My approach: Since $\det(Df(x,y))=(2x)(2x)-(-2y)(2y)=4x^2+4y^2\neq 0$ for $x,y\neq 0$ then $f$ is locally invertible, for any $(x,y)\neq (0,0)$. But how can I know if this inverse function is the same for all the points in $\mathbb{R}^2$?

Thanks!

$f(1,0)=f(-1,0)$, so $f$ is not injective. – Joffysloffy Jul 14 '15 at 20:44 — Joffysloffy, Jul 14 '15 at 20:44
It doesn't, $f(a,b)=f(-a,-b)$. – André Nicolas Jul 14 '15 at 20:45 — André Nicolas, Jul 14 '15 at 20:45
Thanks!! Now I see how to proceed! – apa Jul 14 '15 at 20:50 — apa, Jul 14 '15 at 20:50

score 5 · Answer 1 · answered Jul 14 '15 at 20:44

5

Hints:

Look for two points (perhaps with small integer coordinates) to show that it is not injective.
A deeper hint: do you know complex numbers? What is $(x+iy)^2$?

answered Jul 14 '15 at 20:44

Berci

90,745

score 3 · Answer 2 · answered Jul 14 '15 at 20:46

3

Hint: you have something squared, which often tends to not be injective. Try to find two points (there are some fairly easy ones) to disprove that it is injective.

answered Jul 14 '15 at 20:46

Hirshy

5,040

score 1 · Answer 3 · answered Jul 14 '15 at 20:58

1

The function $f(x,y)=(x^2-y^2, 2xy)$ is the Real version of the Complex function $f(z)=z^2$. Its inverse is only a local one, since $f(z)=f(-z)$, and this local inverse is the square root function, which is not 1-1.

answered Jul 14 '15 at 20:58

Gary.

2,432

Is $f$ bijective on$\mathbb{R}^2$?

3 Answers3