$f:A→B$ is surjective but not injective, so it has a left-inverse, but no right-inverse? If $f:A→B$ is surjective but not injective, it has a left-inverse, but no right-inverse? It would make sense to me if it were the opposite.
This answer says:

If $f:A→B$ is surjective but not injective, it has a left-inverse, but
  no right-inverse.

But this answer says:

For example let $f: R → [0, ∞)$ denote the squaring map, such that
  $f(x) = x^2$ for all x in R, and let $g: [0, ∞) → R$ denote the square
  root map, such that g(x) = √x for all x ≥ 0. Then f(g(x)) = x for all
  x in $[0, ∞)$; that is, g is a right inverse to f. However, g is not a
  left inverse to f, since, e.g., $g(f(−1)) = 1 ≠ −1$.

The second answer makes sense to me. But aren't the two statements are saying the negations of each other? So that would make the first answer false?
 A: I think there is some confusion on what is left and what is right, stemming from our culture: We like to depict functions as sending things towards the right, and we like to write functions to the left of their argument.
If $f:A\to B$ is surjective, then we can compose it with a $g:B\to A$ to get $B\overset g{\to} A\overset f\to B$ which becomes the identity on $B$. In this case, $g$ is put on the left of $f$ (technically, the existence of $g$ does require the axiom of choice in the general case, but let's not get into that).
However, if we write things algebraically, we denote this function as $f\circ g$ and its value at $b\in B$ as $f(g(b))$, which puts $g$ to the right of $f$.
I believe this is the confusion that gives the two seemingly conflicting answers. For completeness, it is the latter which is the conventional way of writing it, which means that if $f$ is surjective, but not injective, then it does have a right inverse, but not a left inverse.
After reading the first answer, I can also see that he puts the composition in the wrong order (i.e. he writes $f\circ g$ for the function that first uses $f$, then applies $g$ to the result), so this is clearly what has happened.
