Consider the coordinate transformation $$ \varphi\colon\mathbb{R}^2\to\mathbb{R}^2, (x,y)\mapsto (y-\arctan(x),y+\arctan(x)). $$ To make it more easy, I set: $$ \xi:=y-\arctan(x),~~~~~\eta:=y+\arctan(x). $$ When I am not wrong, then this is no $C^2$-diffeomorphism, because the inverse transformation, which is concerning to my calculation $$ x=\tan(-\frac{1}{2}((\xi-\eta)+n\pi), n\in\mathbb{N},~~~~~y=\eta-\arctan(\tan(-\frac{1}{2}((\xi-\eta)+n\pi)), n\in\mathbb{N} $$ is not bijective.
So my question is which is an adequate domain of definition in order to get a bijective inverse transformation?
Or do I have to restrict the domain of dfefinition of $\varphi$?
The transformation $\varphi$ is injective, but not surjective. It is the composition of the diffeomorphism
$$\varphi_1 \colon \mathbb{R}^2 \to (-\pi/2,\pi/2)\times\mathbb{R};\quad \varphi_1(x,y) = (\arctan x, y),$$
the injection $\iota \colon (-\pi/2,\pi/2)\times\mathbb{R} \hookrightarrow \mathbb{R}^2$, and the linear isomporphism
$$\varphi_2 \colon (u,v) \mapsto (v-u, v+u)$$
of $\mathbb{R}^2$. The image of $(-\pi/2,\pi/2)\times\mathbb{R}$ under $\varphi_2$ is the strip
$$S = \left\lbrace (\xi,\eta) : \lvert \eta-\xi\rvert < \pi\right\rbrace,$$
so $\varphi \colon \mathbb{R}^2 \to S$ is a diffeomorphism, with inverse
$$\varphi^{-1} = \varphi_1^{-1} \circ \varphi_2^{-1}\lvert_S \colon (\xi,\eta) \mapsto \left(\tan \frac{\eta-\xi}{2},\, \frac{\eta+\xi}{2} \right).$$