Hi Brent,
Many thanks for taking the time to have a look! Hope you might be able to
cope with some more amateurish fumbling as I try and expand my
understanding...
On 14 November 2013 13:59, Brent Yorgey
In general one must consider what are called *di*natural transformations. (I ought to write a blog post about this.)...but I thought I would mention it in case you want to look it up later.
Thanks -- dinatural transformations are probably a bit beyond my comfort zone right now, but I'll add it to my list of things to look at. Interesting to discover that not all polymorphic functions are natural transformations. "fix" seems rather exotic -- are there any less unusual examples? Note that you can't always characterize a type class by a single
function like this. For example, consider class Foo ... There is no way to pick functors f and g such that a Foo dictionary is equivalent to f a -> g a.
I was wondering why the following wouldn't do the trick? sig :: Foo a => Either Int a -> Either a Int sig (Left n) = Left (bar n) sig (Right a) = Right (baz a)
This doesn't really make sense. If a natural transformation consists of a family of arrows which all live in Hask_T,
In this case, I think the family of arrows lives in Hask -- we can reinterpret our Hask endofunctors as instead being functors from Hask_T -> Hask. Then there's no requirement on our constrained polymorphic functions to commute with sig. In your example function f, the two functors would be the identify functor "Id" and a constant functor "K Foo", and naturality is just saying that, for any function g: A -> B that satisfies show == show . f, then: f = f . g Which is clearly true for f. Am I still barking up the wrong tree? Apologies if so, but it *feels* like there should be something here, with the intuition that: Functions that don't "inspect" their values ==> you can substitute values without fundamentally affecting the action of the function ==> natural transformations between Hask endofunctors. Functions that don't inspect their values other than through a particular interface T ==> you can substitute values without fundamentally affecting the action of the function, providing your substitution is invisible through the interface T ==> Natural transformations between functors Hask_T -> Hask. -- Matt