**Puzzle 82** Again let \$$X\$$ be the monoidal preorder whose elements are words of finite length. \$$s \leq_X s'\$$ iff the length of word \$$s\$$ is smaller than the length of word \$$s'\$$. \$$\otimes_X\$$ is string concatenation and \$$1_X\$$ is the empty string.

Define a map \$$f: X \to X\$$ by word reversal. For example, \$f(\textrm{CAT}) = \textrm{TAC}.\$ First note that reversing a word doesn't change its length so \$s \leq_X s' \implies f(s) \leq_X f(s').\$ Also reversing the empty string is still the empty string, so \$$f(1_X) = 1_X\$$. The reflexive property then gives us \$f(1_X) \leq 1_X \textrm{ and } 1_X \leq f(1_X) .\$ Lastly, if I reverse two words and concatenate them, the resulting string has the same length as the string I get when I first concatenate them and then reverse them. So \$f(s) \otimes f(s') \leq f(s \otimes s') \textrm{ and } f(s \otimes s') \leq f(s) \otimes f(s').\$

Therefore \$$f\$$ is both lax and oplax! However, \$f(\textrm{CAT}) \otimes f(\textrm{DOG}) = \textrm{TAC} \otimes \textrm{GOD} = \textrm{TACGOD}\$ while \$f(\textrm{CAT} \otimes \textrm{DOG}) = f(\textrm{CATDOG}) = \textrm{GODTAC}.\$ Since \$f(\textrm{CAT}) \otimes f(\textrm{DOG}) \neq f(\textrm{CAT} \otimes \textrm{DOG})\$ \$$f\$$ is not strict monoidal monotone.