How to understand functor in \g x -> fmap ($) g x

Question

I am a beginner to Haskell. I know that the function fmap takes two arguments ((a->b) -> f a ->) and returns a functor (f b), but I cannot understand the following lambda expression

*Main> :t \g x -> fmap ($) g x
\g x -> fmap ($) g x :: (t -> a -> b) -> t -> a -> b

where is the functor?

Btw, I have tried several similar expression with different brackets, which give different results:

*Main> :t \g x -> fmap ($) (g x)
\g x -> fmap ($) (g x) :: Functor f => (t -> f (a -> b)) -> t -> f (a -> b)

*Main> :t \g x -> fmap $ g x
\g x -> fmap $ g x :: Functor f => (t -> a -> b) -> t -> f a -> f b

I don't understand this.

Answer 1

From

\g x -> fmap ($) g x :: (t -> a -> b) -> t -> a -> b

we have

g            :: t -> a -> b
x            :: t
fmap ($) g   :: t -> a -> b
fmap ($) g x :: a -> b

To find the functor, it suffices to write the type of g as the type-level application F X for some F and X.

We have

t -> a -> b    =    (->) t (a -> b)

hence, F = (->) t and X = (a -> b).

Therefore, the fmap call works in the (->) t functor. We need to think to t -> a -> b as a value of type a -> b "wrapped" under the functor (->) t.

Now, we have

($) :: (a->b) -> a -> b
-- which means
($) :: (a->b) -> (a -> b)

g   :: F (a->b)
fmap ($) :: F (a->b) -> F (a->b)
fmap ($) g :: F (a->b)
-- i.e.
fmap ($) g :: t -> (a->b)