Showing onto and not 1-1 in multi variables

Question

Let $f : \mathbb {R^2 } \to \mathbb {R^2 }$ be given by $f(x, y) = (x^ 2 − y^ 2 , 2xy) $

Show that $f(x_1, y_1) = f(x_2, y_2) $ if and only if $(x_1, y_1) = ±(x_2, y_2)$

Seems pretty obvious that this an onto function if we hold y at 1 and let x range through all values we get everything on the second coordinate and most of the positive stuff. Holding x at 1 and ranging y from 1 to infinity gives us the rest of the first coordinate.

How do i show explicitly that f is onto and is only more than onto only when the signs flip?

Answer 1

Perhaps a bit beyond intentions with the exercise, but you could consider $f$ as a map of complex numbers : $$\widehat{f} : z=x+iy \mapsto z^2= (x^2-y^2) + 2i xy.$$ Then $z^2=w^2$ iff $(z-w)(z+w)=0$ iff $z=\pm w$. So the map is 2 to 1 except when $z=0$.

To show that it is 'onto' you may note that in polar coordinates the map reads $(r,\phi)\mapsto (r^2,2\phi)$ (mod $2\pi$ in angles) and this is clearly onto. But agree, this uses some theory possibly beyond what may be used in your exercise.

Answer 2

Choose $(a,b)\in \mathbb{R}^2$. Consider the equations $x^2-y^2=a$ and $2xy=b$. If $b=0$, we either have that $a$ is positive or negative. If $a$ is positive, we can take $x=\sqrt{a}$ and $y=0$. Otherwise take $y=\sqrt{-a}$ and $x=0$, then $f(x,y)=(a,0)$ in both cases.

Now assume that $b\neq 0$. Then we know that $x$ and $y$ cannot be zero. Hence $2xy=b$ implies that $y=\frac{b}{2x}$. Hence the first equation becomes $x^2-y^2=x^2-\frac{b^2}{4x^2}=a$. Thus $4x^4-4ax^2-b^2=0$. Renaming $x^2=z$, this yields $4z^2-4az-b^2=0$, which is a quadratic equation. Since it's discriminant is $16a^2+16b^2>0$, this equation has solutions. In fact, $z=\frac{4a+\sqrt{16a^2+16b^2}}{8}=\frac{a+\sqrt{a^2+b^2}}{2}$ is a positive solution. Now take $\sqrt{z}=x$. Then $f(x,y)=(a,b)$.

This shows that $f$ is onto.

Answer 3

$f(x_1,y_1)=f(x_2,y_2) \implies x_1^2-y_1^2=x_2^2-y_2^2$ and $2x_1y_1=2x_2y_2$. Multiplying the first equation by $y_1^2y_2^2$,

$$ x_1^2y_1^2y_2^2-y_1^4y_2^2=x_2^2y_1^2y_2^2-y_2^4y_1^2 $$

Rearranging and using the second equation,

$$ x_1^2y_1^2(y_2^2-y_1^2)+y_1^2y_2^2(y_2^2-y_1^2)=0$$

So

$$ (y_2^2-y_1^2)(x_1^2y_1^2+y_1^2y_2^2)=(y_2^2-y_1^2)y_1^2(x_1^2+y_2^2)=0 $$

There are three cases; $y_1=\pm y_2$ leads to $x_1=\pm x_2$

$y_1=0$ leads to $x_2=x_1=y_2=0$ or $y_2=0 ,x_1=\pm x_2$

$x_1^2+y_1^2=0$ leads to $x_1=y_1=x_2=y_2=0$

Overall, summing them up we get $(x_1,y_1)=\pm(x_2,y_2)$

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For surjectivity, if $f(x,y)=(a,b)$ then $x^2-y^2=a$ and $2xy=b$.

You can proceed as is suggested by the other answer; but you can sort of see it immediately via graphs

if $b=0$ then one of $x$ or $y$ is $0$, so that case is simple (depending on sign of $a$ you can choose either $x$ or $y$ to be $0$ and the other one would be $\sqrt{|a|}$)

If $b\neq 0$ the solution to system of equations is decribed by the intersection of the graphs of $y=\frac{b}{2x}$ and $x^2-y^2=a$; But it is quite immediate that the graphs must have an intersection.

Also note that the intersection points of the graphs are symmetric about the origin, and that's another way of seeing that $f$ is 2 to 1 function except at origin.