Show that the function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ is bijective. Show that $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ with
$f(x,y):=(x^3+x+y,y^3+y)^\text{t}$
is bijective.
My idea:
So I know that a function needs to be injective and surjective to be bijective.
Injective means that every combination of x and y-value can only be allocated to one f(x,y) value, so:
$x_1^3+x_1+y_1=x_2^3+x_2+y_2 \Longrightarrow x_1=x_2 \ \text{and} \ y_1=y_2 $
as well as:
$y_1^3+y_1=y_2^3+y_2 \Longrightarrow x_1=x_2 \ \text{and} \ y_1=y_2$
The idea I had is that you could maybe take the derivative of a function to show that it's monotunes and consequently injective:
$g(x,y)=x^3+x+y$
$g'(x,y)=3x^2+1+1>0$
Doing the same for the second component of the vector yields
$h(x)=y^3+y$
$h'(x)=3y^2+1>0$
Surjectivity means that every element in $\mathbb{R}^2$ is hit.
You could show it by showing that they are linearly independent, because both vector components have an cubic polynom that can not be replicated by the other component. Does this make sense?
It's a question from an old Analysis exam so any help is appreciated. Thanks in advance!
 A: As you have shown, the function $h(y) = y^3 + y$ has derivative $h'(y) = 3y^2 +1$ and is therefore strictly increasing. Similarly, for each fixed $y$, the function $g_y(x) = x^3 + x + y$ has derivative $g_y'(x) = 3x^2 +1$ and is therefore strictly increasing.  Hence $h$ and $g_y$ map $\mathbb R$ bijectively onto $\mathbb R$.

*

*$f$ is surjective.
Let $(u,v) \in \mathbb R$. There exists a unique $y$ such that $h(y) = v$. For this $y$, there exists a unique $x$ such that $g_y(x) = u$. Hence $f(x,y) = (u,v)$.


*$f$ is injective.
Let $f(x,y) = f(x',y')$. Thus $h(y) = h(y')$ and therefore $y = y'$. Moreover, $g_y(x) = g_{y'}(x')$ and therefore $g_y(x) = g_y(x')$. We conclude $x = x'$.
A: Hint:
Your logic is a bit off; this is the correct formulation,
INJECTIVITY
$\text{IF}$
$x_1^3+x_1+y_1=x_2^3+x_2+y_2$
$\text{and}$
$y_1^3+y_1=y_2^3+y_2$
$\text{THEN}$
$\quad x_1=x_2 \ \text{and} \ y_1=y_2$
SURJECTIVITY
$\forall u,v \;  \exists x, y \text{ such that } x^3+x+y = u \; \land \; y^3+y = v$
By working on the surjectivity part first you'll discover at the same time that the function $f$ is also injective. You don't need Calculus for this problem; knowing how to factor $a^3 - b^3$ and the quadratic forumula is all that is needed for this problem.
