Now that I think about it, the answer in puzzle 170 implies the answer in puzzle 169.

But to answer 170, a morphism \\(h\\) in \\(\mathcal{C} \times \mathcal{D}\\) is equal to a pair of morphisms, one morphism \\(f\\) in \\(\mathcal{C}\\) and a morphism \\(g\\) in \\(\mathcal{D}\\),

\\[

h: \mathcal{C} \times \mathcal{D} \to \mathcal{C} \times \mathcal{D} \\\\

h = \langle f, g \rangle

\\]

Setting, \\(\mathcal{C}\\) to \\(X^{op}\\), \\(\mathcal{D}\\) to \\(Y\\), \\(f\\) to \\(\leq\_{X^{op}}\\) and \\(g\\) to \\(\leq_Y\\) answers question 169.

But to answer 170, a morphism \\(h\\) in \\(\mathcal{C} \times \mathcal{D}\\) is equal to a pair of morphisms, one morphism \\(f\\) in \\(\mathcal{C}\\) and a morphism \\(g\\) in \\(\mathcal{D}\\),

\\[

h: \mathcal{C} \times \mathcal{D} \to \mathcal{C} \times \mathcal{D} \\\\

h = \langle f, g \rangle

\\]

Setting, \\(\mathcal{C}\\) to \\(X^{op}\\), \\(\mathcal{D}\\) to \\(Y\\), \\(f\\) to \\(\leq\_{X^{op}}\\) and \\(g\\) to \\(\leq_Y\\) answers question 169.