Is polymorphism lost by allowing empty
list handling? If so, how so, and why?
Are there particular cases which would
make this obvious?
The free theorem for head states that
f . head = head . $map f
Applying this theorem to [] implies that
f (head []) = head (map f []) = head []
This theorem must hold for every f, so in particular it must hold for const True and const False. This implies
True = const True (head []) = head [] = const False (head []) = False
Thus if head is properly polymorphic and head [] were a total value, then True would equal False.
PS. I have some other comments about the background to your question to the effect of if you have a precondition that your list is non-empty then you should enforce it by using a non-empty list type in your function signature instead of using a list.