Suppose you have a continuously differentiable function f: R -> R with |f'(x)| <= c < 1 for all x in R. Define a second function [tex]F: R^2 \rightarrow R^2[/tex] by [tex]F(x,y) = (x + f(y), y + f(x)).[/tex] F is supposed to be onto. Why is that so? Intuitively, I would say that F is locally onto by the Inverse Function Theorem (since det(DF) = 1 - c^2 > 0). And globally the identity portion of F dominates over the little perturbation from f, so I would expect the function to be onto. But that's far away from a proof. Any suggestions?