Generalization of Exponential Type

Question

How (if at all) does the exponential interpretation of (->) (a -> b as b^a) generalize to categories other than Hask/Set? For example it would appear that the interpretation for the category of non-deterministic functions is roughly Kliesli [] a b as 2^{a * b} (a -> b -> Bool).

I'm not sure I understand what you are asking. Exponentials can be found in any cartesian closed category. http://en.wikipedia.org/wiki/Cartesian_closed_category — chi, Dec 26 '14 at 17:59
The exponential interpretation of `a -> b` is `b^a`. Consider `a -> ()` which has only `1^a=1` inhabitant, namely `const ()`. On the other hand, `() -> b` has `b^1=b` inhabitants, one for each inhabitant of `b`. — Cirdec, Dec 26 '14 at 18:51
@chi Expand that a tiny bit -- basically just copy some definitions from Wikipedia -- and I'll upvote your answer. — Daniel Wagner, Dec 26 '14 at 20:06
@Cirdec - Do you know of an existing SO answer to that effect? I want to add a cross-reference to an unrelated answer of mine that touches on the subject. (This off-topic comment will self-destruct whenever I remember to destruct it.) — Christian Conkle, Dec 27 '14 at 03:53
@ChristianConkle This answer contains a similar explanation http://stackoverflow.com/a/9197803/414413 — Cirdec, Dec 27 '14 at 04:07

chi · Accepted Answer · 2014-12-27T09:53:35.413

The notion of exponential can be defined in general terms, beyond Hask/Set. A category with exponentials and products is called a cartesian closed category. This is a key notion in theoretical computer science since each c.c. category is essentially a model of the typed lambda calculus.

Roughly, in a cartesian closed category for any pair of objects a,b there exist:

a product object (a * b), and
an exponential object (b^ab)

with morphisms

eval : (b^a)*a -> b (in Haskell: \(f,x) -> f x, AKA apply)
for any f : (a*b)->c, there exists Lf : a -> (c^b) (in Haskell: curry f)

satisfying the equation "they enjoy in the lambda calculus", i.e., if f : (a*b)->c, then:

f = (Lf * id_a) ; eval

In Haskell, the last equation is:

f = \(x :: (a,b), y :: a) -> apply (curry f x, id y) where apply (g,z) = g z

or, using arrows,

f = (curry f *** id) >>> apply where apply (g,z) = g z

Generalization of Exponential Type

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