Lists form a free monoid.

list-monoid.hs

import Data.Monoid

h x = [x]

{--

        h
    A ----->[A]
      \      |
       \     |
        \    |
      f  \   | f#
          \  |
           \ |
            \|
             B

For every monoid B and f :: A -> B, there is a unique homorphism f# from the monoid of list to B.

Which is the correct proof: ftofsharp or fold ?
--}

ftofsharp :: Monoid b       -- monoid (implictly determined by f)
             => (a -> b)    -- f
             -> ([a] -> b)  -- f#
ftofsharp f = f . head

fold :: Monoid b => (a -> b) -> ([a] -> b)
fold f [] = mempty
fold f (x:xs) = mappend (f x) (fold f xs)