>**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\\).