Given a differentiable function $f:\mathbf{R} \to \mathbf{R},$ such that $|f'(x)| < c < 1$. Consider a function $g:\mathbf{R}^2 \to \mathbf{R}^2$, such that $g(x,y) = (x+f(y),y+f(x))$. Prove that g is surjective.
My attempt: determinant of derivative of g is always non-zero. Consider a point $(a,b)$ which we want to show will have a preimage, by implicit function theorem, there's a neighbourhood around the point and also a neighbourhood around the point $(a+f(b),b+f(a))$, such that function is a bijection. Now I wanted to make sure that the distance between $(a,b)$ and $(a+f(b),b+f(a))$ is small so that $(a,b)$ is attainable.
I think there is another possible approach. Let $(a,b)$ a point in the plane $\mathbb{R}^2$ and we want to find another point $(x,y)$ such that $a=x+f(y)$ and $b=y+f(x)$. So from the last equation we get $y=b-f(x)$. And inserting this into the first equation we get $a=x+f(b-f(x))$. So we reduce the problem to find such an $x$. Let us define $h:\mathbb{R}\to\mathbb{R}$ by the formula $$ h(x)=x+f(b-f(x)) $$ The derivative is $h'(x)=1-f'(b-f(x))f'(x)$ for every $x$. By hypothesis $0<1-c<h'(x) <1+c$. So integrating the inequality we get $$ h(0)+(1-c)x\leq h(x)\leq h(0) + (1+c)x $$ Thus checking the limits $x\to-\infty$ and $x\to +\infty$ we see that $h$ is surjective.