4 ms·

What do you mean by "functions are exponentials"?

• Cu3PO42 6y ago
  A function with type A -> B is essentially equivalent to an object of type B^A. I can give you some relatively intuition for the case that A is finite. Imagine B^|A|, this is a list of B elements with one for every element in A. You can therefore interpret it as the return value for every possible input. Therefore it uniquely defines a function A -> B! The type of pairs is just the cartesian product. Therefore (A×B)->C ~ C^(A×B) = (C^B)^A ~ A->B->C. If you find this sort of thing, you might be interested in category theory :)
  • uryga 6y ago
    google keyword: "exponential object" ("exponential function" clashes pretty badly ;) )
    • epsilonclose 6y ago
      Thank you, that helped!
  • epsilonclose 6y ago
    Thank you for clarifying. I was familiar with the notation A^B for the set of all f: A -> B, but this seemed like something more (and it does appear to be). Putting some links I looked at here to save someone a few seconds of Googling: https://en.wikipedia.org/wiki/Exponential_object https://en.wikipedia.org/wiki/Exponential_object https://ncatlab.org/nlab/show/exponential+object https://ncatlab.org/nlab/show/exponential+object
    • dllthomas 6y ago
      I'm not sure if the notation you'd encountered differed, but in this case the equivalence is between A^B and B -> A (note the order). Interestingly, every identity you learned in grade school algebra which involved only addition, multiplication, and exponentiation holds here as an isomorphism. It can be a fun exercise to write the functions that take you back and forth. For instance: A^(B+C) = A^B * A^C to: (Either b c -> a) -> (b -> a, c -> a) to f = (f . Left, f . Right) fro: (b -> a, c -> a) -> (Either b c -> a) fro (f, g) = \case Left b -> f b Right c -> g c