# How can I prove the surjectivity of the following function

$F:\mathbb{R}^2 \to \mathbb{R}^2,\\ f(x,y)= ((x^3)-x),y)$

How can I check if this is surjective or not?

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Have you looked up the definition of surjective? – gebruiker May 4 '14 at 14:02
Try to balance the parentheses. – Marc van Leeuwen May 4 '14 at 17:32

This is surjective since $\forall\alpha\in\Bbb R:x^3-x=\alpha$ has at least one real root.

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$y$ has no influence to $x^3-x$ and we can choose $y$ any number.

It is enough to show that $r(x)=x^3-x$ is surjective. You can show it in many ways;

$lim_{x\to \infty}r(x)= \infty$ and $lim_{x\to -\infty}=-\infty$ and $r(x)$ continious so for any $a\in \mathbb R$ there exist $x$ such that $x^3-x=a$

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You can actually solve $x^3 + x = a$ for $x$.

Write it as $$x ^3 = a - x \tag1$$

Note that $(u-v)^3 = u^3 - 3u^2v + 3uv^2 - v^3$ can be rewritten as $(u-v)^3 = (u^3 - v^3) - 3uv(u - v)$

Now let $x = u - v$. You get $$x^3 = (u^3 - v^3) - 3uvx \tag2$$

Comparing (1) and (2), we see that we need to solve $u^3 - v^3 = a$ and $3uv = 1$.

From $3uv = 1$, we get $v = \dfrac{1}{3u}$. So

\begin{align} u^3 - v^3 &= a\\ u^3 - \dfrac{1}{27u^3} &= a\\ 27u^6 - 1 &= 27au^3 \\ 27u^6 - 27au^3 - 1 &= 0 \\ 27(u^3)^2 - 27a(u^3) - 1 &= 0 \\ u^3 &= \dfrac{27a \pm \sqrt{729a^2+108}}{54} & \{\text{Note that $729a^2 + 108 > 0$ for all $a$}\} \\ u^3 &= \dfrac{9a \pm \sqrt{81a^2+12}}{18} \\ u &= \left( {\dfrac{9a \pm \sqrt{81a^2+12}}{18}} \right)^{1/3}\\ \end{align}

If we let $u = \left( {\dfrac{9a + \sqrt{81a^2+12}}{18}} \right)^{1/3}$, then

\begin{align} u^3 - v^3 &= a\\ v^3 &= u^3 - a \\ v^3 &= \dfrac{9a + \sqrt{81a^2+12}}{18} - a\\ v^3 &= \dfrac{-9a + \sqrt{81a^2+12}}{18} \\ v &= -\left(\dfrac{9a - \sqrt{81a^2+12}}{18} \right)^{1/3} \\ \end{align}

So, a solution is

$$x = \left( {\dfrac{9a + \sqrt{81a^2+12}}{18}} \right)^{1/3} + \left( \dfrac{9a - \sqrt{81a^2+12}}{18} \right)^{1/3} \tag3$$

## To verify this solution

Let $x = u-v$, where

$u = \left( {\dfrac{9a + \sqrt{81a^2+12}}{18}} \right)^{1/3}$ and $v = -\left( {\dfrac{9a - \sqrt{81a^2+12}}{18}} \right)^{1/3}$

Note that \begin{align} uv &= -\left( {\dfrac{9a + \sqrt{81a^2+12}}{18}} \right)^{1/3} \left( {\dfrac{9a - \sqrt{81a^2+12}}{18}} \right)^{1/3} \\ &= -\left[ \left( {\dfrac{9a + \sqrt{81a^2+12}}{18}} \right) \left( {\dfrac{9a - \sqrt{81a^2+12}}{18}} \right) \right]^{1/3} \\ &= -\left( \dfrac{81a^2 - (81a^2+12)}{324} \right)^{1/3}\\ &= -\left(-\dfrac{12}{324} \right)^{1/3}\\ &= \dfrac 13 \end{align}

and $u^3 - v^3 = \left( {\dfrac{9a + \sqrt{81a^2+12}}{18}} \right) + \left( {\dfrac{9a - \sqrt{81a^2+12}}{18}} \right) = a$

So \begin{align} x^3 + x &= (u-v)^3 + (u-v)\\ &= (u^3 - 3u^2v + 3uv^2 - v^3) + (u-v)\\ &= (u^3 - v^3) -3uv(u-v) + (u-v) \\ &= a - (u-v) + (u-v) \\ &= a \end{align}

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That's the only complete solution among all the answers. +1 – Arghya Chakraborty Jun 16 at 14:43