Showing that `newtype T a = T (a -> Int)` is a Type Constructor that is Not a Functor

Question

It's been claimed that newtype T a = T (a -> Int) is a type constructor that is not a functor (but is a contravariant functor). How so? Or what is the contravariant functor (whence I assume it will be obvious why this can't be made a normal functor)?

Answer 1

Given

newtype T a = T (a -> Int)

let's try to build the Contravariant instance for this datatype.

Here's the typeclass in question:

class Contravariant f where
  contramap :: (a -> b) -> f b -> f a

Basically, contramap is akin to fmap, but instead of lifting a function a -> b to f a -> f b, it lifts it to f b -> f a.

Let's begin writing the instance...

instance Contravariant T where
  contramap g (T f) = ?

Before we fill in the ?, let's think about what the types of g and f are:

g :: a -> b
f :: b -> Int

And for clarity, we might as well mention that

f a ~ T (a -> Int)
f b ~ T (b -> Int)

So we can fill in the ? as follows:

instance Contravariant T where
  contramap g (T f) = T (f . g)

To be super pedantic, you might rename g as aToB, and f as bToInt.

instance Contravariant T where
  contramap aToB (T bToInt) = T (bToInt . aToB)

The reason you can write a Contravariant instance for T a boils down to the fact that a is in contravariant position in T (a -> Int). The best way to convince yourself that T a isn't a Functor is to try (and fail) to write the Functor instance yourself.