**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.