Show $g$ is bijective onto $\mathbb{R}^2$ Let $f\in C^1(\mathbb{R})$ and $|f'(x)|\leq\frac{1}{2}$ for all $x\in\mathbb{R}$. Show that $g:\mathbb{R}^2\to\mathbb{R}^2$ where $g(x,y)=(x+f(y),y+f(x))$is bijective onto $\mathbb{R}^2$.
I'm having trouble with both injectivity and surjectivity. For injectivity: I know $f$ is injective since it's a contraction. Suppose $g(x_1,y_1)=g(x_2,y_2)$, then
$(1) \ \ x_1+f(y_1)=x_2+f(y_2)$ and 
$(2) \ \ y_1+f(x_1)=y_2+f(x_2)$.
Assume $(x_1,y_1)\neq (x_2,y_2)$ and derive a contradiction: Assume $x_1\neq x_2$. Then from $(1)$ above, we must have $f(y_1)\neq f(y_2)$, so $y_1\neq y_2$. Since $f$ is injective it follows from $(2)$ that $f(x_1)\neq f(x_2)$ and so $y_1\neq y_2$. This isn't a contradiction though, I don't see how to proceed to derive a contradiction or if this is the wrong approach.
For surjectivity, I'm not sure where to start. Can someone provide a hint with how to proceed for each parth?
Thank you.
 A: Injectivity
Assume $(x_1,y_1)\neq (x_2,y_2).$ Then
$$ x_1+f(y_1)=x_2+f(y_2)\implies |x_1-x_2|=|f(y_2)-f(y_1)|\le \frac12 |y_2-y_1|$$ and 
$$ y_1+f(x_1)=y_2+f(x_2)\implies |y_1-y_2|=|f(x_2)-f(x_1)|\le \frac12 |x_2-x_1|.$$ Is this possible if $(x_1,y_1)\neq (x_2,y_2)?$
A: You can prove injectivity as outlined in the answer by mfl.
You can prove surjectivity as follows. We can write $g=I+h$ where $I$ is the identity mapping on $\Bbb R^2$, and $h(x,y) = (f(y),f(x))$. Note that $h$ is a contraction since $|f'(x)| \leq 1/2$ for all $x$.
Now fix a point $u_0 \in \Bbb R^2$. Define a map $q: \Bbb R^2 \to \Bbb R^2$ as $q(u)=-h(u)+u_0$. Since $h$ is a contraction it follows that $q$ is a contraction. Thus by the contraction mapping principle it follows that $q$ has a fixed point $v_0$. Then $q(v_0)=v_0$ which implies that $g(v_0)=u_0$. This proves surjectivity.
This method can be used to prove (more generally) that in any Banach space, the identity plus a contraction is always a homeomorphism.
A: I disagree with your (apparent) belief that contractions are injective.  (Others have given counterexamples.)
I think, a key use of the bound on $f'$ is this:
$$
 \left|f(x_{2}) - f(x_{1})\right| = \left| \int_{x_{1}}^{x_{2}} f'(t) dt \right| \leq {1 \over 2}|x_{2}  - x_{1}|
$$
This allows you, for $x_{2} > x_{1}$ (for definiteness), to estimate from below the absolute difference between $x_{1} + f(x_{1})$ and $x_{2} + f(x_{2})$.
