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.
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?