The list monad is given here . Also see Spivak's paper here . So list is a monad. Is it a comonad? How would you prove that?
The list type constructor a ↦ μ L. 1 + a * L
doesn't admit a comonad structure. Recall that if it were a comonad, we'd have (using the names from Haskell's Functor
and Comonad
typeclasses)
fmap :: ∀ a b. (a → b) → [a] → [b]
extract :: ∀ a. [a] → a
duplicate :: ∀ a. [a] → [[a]]
However, even without going into any required laws, extract
cannot be implemented, since its input could be the empty list, giving no way to come up with an a
.
The nonempty list type constructor a ↦ μ NE. a + a * NE
a ↦ μ NE. a + a * NE
does admit a comonad structure, with extract
returning the first element, and duplicate
mapping [x, y, ..., z]
to [[x], [x, y], ..., [x, y, ..., z]]
(note that each of them are non-empty by construction).
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