Also here's a [diagram of the commutative square]( in puzzle 170 (similar to the one drawn by John above).

Because the diagram commutes, we have \\( (f,g) = (f,1_{y'}) \circ (1_x,g) = (1_{x'},g) \circ (f,1_y) \\). If \\( (f,g) \\) is a preorder on \\( X^\mathrm{op} \times Y\\), then the diagram is the solution to puzzle 169. You can translate the category theoretic concepts back into preorder ones for a proof: If \\(x' \le x\\) in \\(X\\), then \\(x \le x'\\) in \\(X^\mathrm{op}\\) and if \\(y \le y'\\) in \\(Y\\), then by reflexitivity, we have \\((x,y) \le (x',y)\\) and \\((x',y) \le (x',y')\\). By transitivity, we have \\((x,y) \le (x',y)\\) and \\((x',y) \le (x',y')\\) implies \\((x,y)\le (x',y')\\).