Can I show that this function is surjective? Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that 
$$f(x+f(x))+f(x-f(x))=x$$
for all $x \in \mathbb{R}$. Can I conclude that $f$ is surjective? If so, how can I prove it?
 A: Note that for all $x \in \Bbb R$, there exist real numbers $a = x + f(x)$ and $b = x - f(x)$ such that 
$$
x = f(a) + f(b)
$$
Let $S$ denote $f(\Bbb R)$, i.e. the image of $f$.  It's clear that $\{c + d : c,d \in S\} = \Bbb R$ (more concisely, $S + S = \Bbb R$).
If we know that $f$ is continuous, then the above allows us to conclude that $f$ is surjective.  In particular, it suffices to state that because $S + S$ is not bounded above and not bounded below, $S$ is also not bounded above and not bounded below.  Because $S= f(\Bbb R)$ is also connected, we can conclude that $f(\Bbb R) = \Bbb R$.
If we are to answer this without the assumption of continuity, things get tricky.  In order for $f$ to fail to be surjective, $S$ would need to be such that $S + S = \Bbb R$ with $S \neq \Bbb R$.  It is not obvious to me that this should be possible without invoking the axiom of choice.  Even if we could find such an $S$, it is unclear how one could get $f$ to actually satisfy the functional equation. 
