On the signature of >>= Monad operator

Question

This is the signature of the well know >>= operator in Haskell

>>= :: Monad m => m a -> (a -> m b) -> m b

The question is why

Answer 1

The reason is that (>>=) is more general. The function you're suggesting is called liftM and can be easily defined as

liftM :: (Monad m) => (a -> b) -> (m a -> m b)
liftM f k  =  k >>= return . f

This concept has its own type class called Functor with fmap :: (Functor m) => (a -> b) -> (m a -> m b). Every Monad is also a Functor with fmap = liftM, but for historical reasons this isn't (yet) captured in the type-class hierarchy.

And adapt you're suggesting can be defined as

adapt :: (Monad m) => (a -> b) -> (a -> m b)
adapt f = return . f

Notice that having adapt is equivalent to having return as return can be defined as adapt id.

So anything that has >>= can also have these two functions, but not vice versa. There are structures that are Functors but not Monads.

The intuition behind this difference is simple: A computation within a monad can depend on the results of the previous monads. The important piece is (a -> m b) which means that not just b, but also its "effect" m b can depend on a. For example, we can have

import Control.Monad

mIfThenElse :: (Monad m) => m Bool -> m a -> m a -> m a
mIfThenElse p t f = p >>= \x -> if x then t else f

but it's not possible to define this function with just Functor m constraint, using just fmap. Functors only allow us to change the value "inside", but we can't take it "out" to decide what action to take.