Proving $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ function is bijective. Given $f: \mathbb{R}\rightarrow\mathbb{R}$ is a $C^1$ function and $\forall t.|f'(t)| \leq c &lt; 1$,$g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is defined as $g(x,y) = (x + f(y), y+f(x)).$ I am trying to show $g$ is bijective.

It is sufficient to show $g$ is 1-to-1 and onto, but how exactly is this approached with such an $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ function?
You can construct the inverse to show the function is bijective: Assuming f bijective:

take $h:= (x-f(y), y-f(x) )$ gives you the inverse, so $g$ is bijective.
You can construct the inverse to show the function is bijective: Assuming f bijective:

take $h:= (x-f(y), y-f(x) )$ gives you the inverse, so $g$ is bijective.