Tony Morris

def f: Boolean => Option[Boolean] Write as many functions with this type as you can. There are Option[Boolean]^Boolean of them. 9

I know, you aren't into intuition ;)

Ok I think I got it. Suppose you hardcoded the exact return value separately for each possible input value. (Any logic can only be a simplification of that.) If there are N possible inputs and M possible outputs then any implementation has M choices for each of the N inputs.

(maybe a better phrasing would be, N choices to make, with M options that each one can choose from) It's interesting that a function with N inputs and M outputs isomorphic to a TupleN[M, M...]

Yep. Now calculate how many of these: def identity[A]: A => A

Of course I know there is 1, but how can my reasoning work here? I guess in the scope of a type variable we don't care about number of inhabitants in an absolute sense but how many it can obtain. Since it can only get 1 A that's the number we use

// start here trait Yoneda[F[_], A] { def run[B]: (A => B) => F[B] def lower: F[A] = run(identity) }

What about it? How does this relate to the thread?