Got a bijection?

This is an informal solution (informal for me, at least).

Consider a lattice grid, and look at the grid points. Draw the line $y = \frac{A}{B} x$ , and observe the rectangle bounded by $y = 0, y = \frac{A}{2}, x = 0, x = \frac{B}{2}$ . Note that the line $y = \frac{A}{B} x$ passes through two of the corners ( $(0,0)$ and $\left( \frac{A}{2}, \frac{B}{2} \right)$ ).

Now note that $\left\lfloor \frac{A}{B} x_0 \right\rfloor$ counts the number of grid points below the line (but above the x-axis) in column $x = x_0$ , so the first summation gives the number of grid points within the rectangle, below the line (green region). Likewise, $\left\lfloor \frac{B}{A} y_0 \right\rfloor$ counts the number of grid points to the left of the line (but to the right of the y-axis) in row $y = y_0$ , so the second summation gives the number of grid points within the rectangle, to the left of the line (purple region). But to the left of the line is equal to above the line. Also, no grid points other than those in the axes are passed by the sides or the line (because $A,B$ are relatively prime and odd). This means all grid points are accounted for. But the number of grid points is trivially $mn$ . Thus $f(m)f(n) = mn$ , so $f(n) = n$ .