[Haskell-cafe] Functor and Haskell
dmehrtash at gmail.com
Wed Apr 22 18:14:03 EDT 2009
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
> 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,
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?
On Wed, Apr 22, 2009 at 12:18 AM, Kim-Ee Yeoh <a.biurvOir4 at asuhan.com>wrote:
> 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?
> 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:
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