Having a crack at these:

> **Puzzle 213.** Show that for any preorder \\(A\\), the preorder
\\( A \times \textbf{1}\\) is **isomorphic** to \\(A\\):: in other words, there is a monotone function from \\( A \times \textbf{1}\\) to \\(A\\) with a monotone inverse. For short we write \\(A \times \textbf{1} \cong A\\). (This is one way in which \\(\textbf{1}\\) acts like 'nothing'.)

> **Puzzle 214.** Show that for any preorders \\(A\\) and \\(B\\) we have \\( (A \times B)^{\text{op}} \cong A^{\text{op}} \times B^{\text{op}}\\).

> **Puzzle 215.** Show that for any preorder \\(A\\) we have \\( (A^{\text{op}})^{\text{op}} \cong A\\).

Let's write \\(\bullet\\) for the sole element in \\(\textbf{1}\\). Define \\(f : A \times \textbf{1} \to A\\) and \\(g : A \to A \times \textbf{1}\\) by \\(f(a, \bullet) = a, g(a) = (a, \bullet)\\).

\\(f\\) and \\(g\\) are clearly both monotone and inverse to each other. So \\(A \times \textbf{1} \cong A\\).

Now define \\(h : (A \times B)^\text{op} \to A^\text{op} \times B^\text{op}\\) and \\(k : A^\text{op} \times B^\text{op} \to (A \times B)^\text{op} \\) by \\(h(a, b) = (a, b), k(a, b) = (a, b)\\).

\\(h\\) and \\(k\\) are clearly inverse to each other, but are they monotone? Yes, because

\\(\qquad(a, b) \leq_{(A \times B)^\text{op}} (a', b')\\)

\\(\qquad\iff (a', b') \leq_{A \times B} (a, b)\\)

\\(\qquad\iff a' \leq_A a \text{ and } b' \leq_B b\\)

\\(\qquad\iff a \leq_{A^\text{op}} a' \text{ and } b \leq_{B^\text{op}} b'\\)

\\(\qquad\iff (a, b) \leq_{A^\text{op} \times B^\text{op}} (a', b')\\)

So \\((A \times B)^\text{op} \cong A^\text{op} \times B^\text{op}\\).

Finally, define \\(m : (A^\text{op})^\text{op} \to A\\) and \\(n : A \to (A^\text{op})^\text{op}\\) by \\(m(a) = a, n(a) = a\\).

\\(m\\) and \\(n\\) are clearly inverse to each other, but are they monotone? Yes, because

\\(\qquad a \leq_{(A^\text{op})^\text{op}} a'\\)

\\(\qquad\iff a' \leq_{A^\text{op}} a\\)

\\(\qquad\iff a \leq_A a'\\)

So \\((A^\text{op})^\text{op} \cong A\\).