[Haskell-cafe] Type of (>>= f) where f :: a -> m b

[Milind Patil]

For a function

f ::  a -> m b
f = undefined

I am having trouble understanding how the type of 

(>>= f) 

is

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

where, by definition, type of (>>=) is 

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

I do not see how (>>= f) even unifies.

I mean if I code a function with the same type as (>>=) ie. 

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

type of (tt f) does not infer to the same type as (>>= f), from ghc ...

(tt f) :: (Monad ((->) b)) => (m a -> b -> b1) -> b -> b1

There seems to something special about (>>=) apart from its type. And whats
(Monad ((->) b))? I am new to Haskell and I may have gaps in my understanding of
type inference in Haskell. 

regards,
Milind Patil