>**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'.)

If we note that \$$A \cong A^{\mathbf{1}}\$$ for all categories (and therefore also all preorders), then under currying,

\$A \times \mathbf{1} \cong A^{\mathbf{1}} \$

and since,

\$A^{\mathbf{1}} \cong A \$

then by transitivity of \$$\cong\$$,

\$A \times \mathbf{1} \cong A \$

\$$\blacksquare\$$.