Is this function $\varphi:\mathbb R^2\to \mathbb R^2$ bijective? Let $f:\mathbb R\to \mathbb R$ be a function of class $C^1$ such that $|f'(x)|\le k\lt 1$ for every $x\in \mathbb R$. 
I would like to know whether this function $\varphi:\mathbb R^2\to \mathbb R^2$ defined as 

$$\varphi(x,y)=(x+f(y),y+f(x))$$

is bijective. I have already proved $|\varphi'(x,y)|>0$ but I don't know what to do with this information (if it were a real function of a real variable, it would be simpler).
 A: Write $\phi(x) = (x_1+f(x_2),x_2+f(x_1) = x + (f(x_2),f(x_1))$.
Look at the equation $y = \phi(x)$ which we can write as
$\xi_y(x) = x$, where $\xi_y(x) = y-(f(x_2),f(x_1))$.
It is straightforward to show that $\| {\partial \xi_y(x) \over \partial x} \|_2 = \sqrt{|f'(x_1) f'(x_2)|} \le k < 1$, hence $\xi_y$ is a contraction map, and hence for any $y$ there is a unique $x$ such that $y=\phi(x)$. Hence $\phi$ is a bijection.
A: Here is a solution which is obviously not optimal.
First, the Jacobian of $\phi$ has determinant $1-f'(x)f'(y) \geq 1-k^2>0$ so the map $\phi$ is orientation preserving and also a local homeomorphism. Thus, $\phi:\mathbb R^2 \to \mathbb R^2$ is an open map.
Next we show that if $(x_n,y_n)\to \infty$ then $\phi(x_n,y_n)\to \infty$. Wlog assume that $x_n\to \infty$. If $y_n$ is bounded then the first coordinate $x_n+f(y_n)$ of $\phi$ converges to $\infty$ so we are done. Thus, we assume that $y_n\to\infty$. 


*

*If $|x_n-y_n|$ is bounded, then by the mean value theorem we have
$$ |x_n+f(y_n)|= |x_n+f(x_n)+ f(y_n)-f(x_n)|\geq |x_n|-|f(x_n)| - k|x_n-y_n|$$
Since the term $|x_n|-|f(x_n)|$ converges to $\infty$ (by another application of the mean value theorem), it follows that $|x_n+f(y_n)| \to\infty$, thus $\phi(x_n,y_n)\to\infty$.

*If $|x_n-y_n|$ is unbounded, by passing to a subsequence we assume that it converges to $\infty$. Then, subtracting the coordinates of $\phi$ we have
$$|(x_n+f(y_n)- (y_n+f(x_n))|\geq |x_n-y_n|-|f(x_n)-f(y_n)|\geq |x_n-y_n|-k|x_n-y_n|$$
by the mean value theorem. This, again, converges to $\infty$. Since the difference of the coordinates of $\phi$ converges to $\infty$, at least one of them has to converge to $\infty$.


Therefore, $\phi(\infty)=\infty$ which shows that the map $\phi$ is a proper map, and thus closed, i.e. $\phi(\mathbb R^2)$ is a closed subset of $\mathbb R^2$. Since $\phi$ is also an open map, it follows that $\phi(\mathbb R^2)=\mathbb R^2$, i.e. $\phi$ is surjective. Furthermore, since it is a proper local homeomoephism it follows that it is a covering map.
In order to show that $\phi$ is injective, it remains to show that for any point $(a,b)\in \mathbb R^2$ the preimage $\phi^{-1}( (a,b))$ contains one point. Since $\phi$ is a covering map, it suffices to show this for a single point $(a,b)\in \mathbb R^2$.
Let $(x,y)\in \phi^{-1}((1,1))$, so $x+f(y)=1=y+f(x)$. Then 
$$|y-x|=|f(x)-f(y)|\leq k|x-y|$$
which can happen only if $x=y$. If $(x',x')$ is another preimage of $(1,1)$ then $x+f(x)=x'+f(x')$ and with the same argument this implies that $x=x'$. 
