Question

(=<<) :: (a -> m b) -> m a -> m b
id :: a -> a
join :: m (m a) -> m a

So shouldn’t,

(=<<) id

give an error because,

id :: a -> a

and not,

id :: a -> m a

Doesn’t (=<<) expect,

(something -> m anything)

as its first argument?

Answer 1

m a' -> m a' is also a kind of a -> a, so we can have

      id ::  m a' -> m a'                        -- a = m a'
   (=<<) :: (m a' -> m a') -> m (m a') -> m a'   -- a = m a', b = a'

(=<<) id ::                   m (m a') -> m a'