Relationship between fmap and bind

Question

After looking up the Control.Monad documentation, I'm confused about this passage:

The above laws imply:

fmap f xs = xs >>= return . f

How do they imply that?

Answer 1

Control.Applicative says

As a consequence of these laws, the Functor instance for f will satisfy

fmap f x = pure f <*> x

The relationship between Applicative and Monad says

pure = return

(<*>) = ap

ap says

return f `ap` x1 `ap` ... `ap` xn

is equivalent to

liftMn f x1 x2 ... xn

Therefore

fmap f x = pure f <*> x
         = return f `ap` x
         = liftM f x
         = do { v <- x; return (f v) }
         = x >>= return . f