Here F is the identity functor, and G is the list functor. And yes, C=D= category of (a subset of) Haskell types. Are you saying the function that goes from list functor to singleton funtor is a natural transformation? But aren't they functors to different subset of Haskell Types? The Haskell Wikibooks also sayshttp://en.wikibooks.org/wiki/Haskell/Category_theory#Functors_on_Haskthe same thing:
Functors in Haskell are from *Hask* to *func*, where *func* is the subcategory of *Hask* defined on just that functor's types. E.g. the list functor goes from *Hask* to *Lst*, where *Lst* is the category containing only *list types*, that is, [T] for any type T. The morphisms in *Lst* are functions defined on list types, that is, functions [T] -> [U] for types T, U.
So in your example there is C that is Hask. But there are two D's, D1 that
is all List types, and D2 all singleton types. In this example I guess,
the Singleton types are subset of List types which are subset of Hask.
Is that related to natural transformation or unrelated?
Daryoush
On Wed, Apr 22, 2009 at 12:18 AM, Kim-Ee Yeoh
Daryoush Mehrtash-2 wrote:
I am not sure I follow how the endofunctor gave me the 2nd functor.
As I read the transformation there are two catagories C and D and two functors F and G between the same two catagories. My problem is that I only have one functor between the Hask and List catagories. So where does the 2nd functor come into picture that also maps between the same C and D catagories?
Consider singleton :: a -> [a] singleton x = [x]
Here F is the identity functor, and G is the list functor. And yes, C=D= category of (a subset of) Haskell types.
-- View this message in context: http://www.nabble.com/Functor-and-Haskell-tp23166441p23170956.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
-- Daryoush Weblog: http://perlustration.blogspot.com/