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I have to prove that the following is a bijection: $ T: \mathbb{R}^2 \rightarrow \mathbb{R}^2 $, with $T(x,y)= \left( \begin{array}{c} 5x + \sin(y)\\ 5y + \arctan(x) \end{array} \right)$

Now, to prove it is surjective I composed it with $x = \tan(z)$ obtaining:

$ T^1\colon \left]-\pi /2, \pi /2 \right[\times \mathbb{R} \rightarrow \mathbb{R}^2 $, with $T^1(z,y)= \left( \begin{array}{c} 5\tan(z) + \sin(y)\\ 5y + z \end{array} \right) $

Then, if we consider a generic $(z_0,y_0)$ we have $5y+z=y_0 \Rightarrow z = y_0 - 5y$ and $5 \tan(y_0 - 5y) + \sin(y)=g(y)$. Considering the limits for $g(y)$ at the inf and sup of its domain are $+ \infty$ and $-\infty$ and the continuity of $g(y)$ we prove the existence of $Y$ such that $g(Y) = z_0$ and then $Z = y_0 - 5Y$ and we've found a couple $(Y,Z)$ such that $T^1(Y,Z)=(z_0,y_0)$.

As for the injectivity I've noticed that, using the fact that sin and arctan are limited functions, we get $T(x,y)=T(x',y') \Rightarrow |x-x'|< \frac{2}{5}$ and $|y-y'|< \frac{2 \pi}{5}$ but I can't do any better.

My questions are:

1) Is there an easier or better proof of surjectivity? (given that mine is correct...if not please point out where I did wrong)

2) Can somebody give me hints to prove the injectivity?

Thank you very much in advance!

Arthur Jr.
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GivAlz
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2 Answers2

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One way to prove that $T \, : \, \mathbb{R}^{2} \, \rightarrow \, \mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $\mathrm{det} \, \mathrm{Jac}(T,(x,y))$ never vanishes and $\Vert T(x,y) \Vert \rightarrow +\infty$ as $\Vert (x,y) \Vert \rightarrow +\infty$.

However, there may be a better answer to the question.

pitchounet
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Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.

For the Jacobian part, compute it at $(x, y)$

$$ J = \begin{bmatrix}5 & \cos(y) \\ \frac{1}{1+x^2} & 5 \end{bmatrix}, $$

and check if the determinant of $J$ is zero for some pair $(x, y)$.

$$ \text{det}(J) = 25 - \frac{\cos(y)}{1+x^2} = \frac{25x^2 + 25 - \cos(y)}{1+x^2}\neq 0, \, \forall x, y \in \mathbb{R}, $$ since $|\cos(y)|\leq 1$.