Is a continuous function one-to-one if it is one-to-one almost everywhere on its range?

Question

Let $A\subset \mathbb{R}^n$ be open, $B\subset A$ be Lebesgue measurable, $f:A\to\mathbb{R}^n$ be continuous on $A$ and one-to-one on $B$, and $m(f(A-B)) = 0$ where $m$ is the Lebesgue measure. Is $f$ one-to-one on $A$?

Answer 1

No, not even if we assume that $f$ is algebraic: Let $A=\Bbb R^2$ and $B=\Bbb R^2\setminus(\Bbb R\times\{0\})$, and $f(x,y)=(xy,y)$.