**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,

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