**Puzzle 213:** To show \\(A \times 1 \simeq A \\), we need to show that there's a monotone function \\(f\\) with an inverse \\(f^{-1}\\) between these preorders.

Define \\(f : (a, 1) \mapsto a\\). This is a monotone function, whenever \\((a,1) \le (b,1)\\) we have \\(a \le b\\) because by the definition of a product preorder, \\((a,1) \le (b,1)\\) iff \\(a \le b\\) and \\(1 \le 1\\).

The inverse \\(f^{-1} : a \mapsto (a,1)\\) is also monotone since whenever \\(a \le b\\) we have \\((a,1) \le (b,1)\\) as \\(1 \le 1\\) holds for all pairs in \\(A \times 1\\). Composing both functions (\\(f f^{-1} = 1_A\\) and \\(f^{-1}f = 1_{A \times 1} \\)) proves \\(A \times 1 \simeq A \\).

Note \\(f\\) is the projection map that appears the categorical definition of a product.

**Puzzle 214:** Since the opposite preorder \\(X^\mathrm{op}\\) has the same objects as \\(X\\), it holds that there's a bijection from \\( (A \times B)^\mathrm{op} \\) to \\(A^\mathrm{op} \times B^\mathrm{op}\\). We just have to prove that both product preorders have the same order relation.

For \\((a,b) \le (a',b')\\) in \\(A \times B\\), we have \\((a,b) \ge (a',b')\\) in \\((A \times B)^\mathrm{op} \\). However, we have \\(a \ge a'\\) in \\(A^\mathrm{op}\\) if \\(a \le a'\\) in \\(A\\) (the same is true for \\(B\\)), and thus \\((a,b) \ge (a',b')\\) in \\(A^\mathrm{op} \times B^\mathrm{op}\\) if \\((a,b) \le (a',b')\\) in \\(A \times B\\). So any inequality in \\((A \times B)^\mathrm{op} \\) is also in \\(A^\mathrm{op} \times B^\mathrm{op}\\) and vice versa, therefore \\((A \times B)^\mathrm{op} \simeq A^\mathrm{op} \times B^\mathrm{op}\\).

Alternatively, here's an attempt to do a more category-theoretic version of the previous paragraph: Consider the diagram \\( A^\mathrm{op} \overset{\pi}{\leftarrow} (A \times B)^\mathrm{op} \overset{\varphi}{\rightarrow} B^\mathrm{op} \\) and the diagram \\( A^\mathrm{op} \overset{\pi'}{\leftarrow} A^\mathrm{op} \times B^\mathrm{op}\overset{\varphi'}{\rightarrow} B^\mathrm{op} \\). It's not hard to see that the projections are equal \\(\pi =\pi', \varphi =\varphi'\\). By the universal property of products, there must be an unique monotone function \\(h: (A \times B)^\mathrm{op} \to A^\mathrm{op} \times B^\mathrm{op} \\) such that \\(\pi = \pi'h \\) and \\(\varphi = \varphi' h\\), but since both projections are equal, \\(h\\) must be the identity map. Therefore \\((A \times B)^\mathrm{op} \simeq A^\mathrm{op} \times B^\mathrm{op}\\).

**Puzzle 215:** If \\(a \le a'\\) in \\(A\\), then \\(a \ge a'\\) in \\(A^\mathrm{op}\\). If \\(a \ge a'\\) in \\(A^\mathrm{op}\\), then \\(a \le a'\\) in \\(\left(A^\mathrm{op}\right)^\mathrm{op}\\). By transitivity, whenever \\(a \le a'\\) in \\(A\\), \\(a \le a'\\) in \\(\left(A^\mathrm{op}\right)^\mathrm{op}\\) so \\(A \simeq \left(A^\mathrm{op}\right)^\mathrm{op}\\).

**Puzzle 216:** Rewritten as a monotone function, we have \\(\cap_X : 1^\mathrm{op} \times X^\mathrm{op} \times X \to \mathbf{Bool}\\). By the above theorems (puzzle answers), this function is isomorphic to a function \\(X^\mathrm{op} \times X \to \mathbf{Bool}\\). But this is just the hom-functor! So we have

\[ \cap_X (x,x') = \hom(x,x')\]

In other words,

\[ \cap_X(x,x') =\begin{cases}\mathrm{True} & \mathrm{if} \ x \le x' \\\ \mathrm{False} & \mathrm{otherwise} \end{cases}\]