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\$$.