[Haskell-cafe] map over Bijections

2012-08-27 Thread Sergey Mironov 
Hi. I need map equivalent for Bijection type which is defined in fclabels:

data Bijection (~) a b = Bij { fw :: a ~ b, bw :: b ~ a }

instance Category (~) = Category (Bijection (~)) where ...

I can define this function as follows:
mapBij :: Bijection (-) a c - Bijection (-) [a] [b] - Bijection (-) [a] [c]
mapBij b1 b = (map (fw b1)) `Bij` (map (bw b1))

but do I really need to do it explicitly? Can I obtain same result
using some Category combinators or other common stuff?

Sergey

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Re: [Haskell-cafe] map over Bijections

2012-08-27 Thread Erik Hesselink 
If you remove the second argument (which you don't use), you have the
function `liftBij` that is in fclabels.

Erik

On Mon, Aug 27, 2012 at 3:55 PM, Sergey Mironov ier...@gmail.com wrote:
 Hi. I need map equivalent for Bijection type which is defined in fclabels:

 data Bijection (~) a b = Bij { fw :: a ~ b, bw :: b ~ a }

 instance Category (~) = Category (Bijection (~)) where ...

 I can define this function as follows:
 mapBij :: Bijection (-) a c - Bijection (-) [a] [b] - Bijection (-) [a] 
 [c]
 mapBij b1 b = (map (fw b1)) `Bij` (map (bw b1))

 but do I really need to do it explicitly? Can I obtain same result
 using some Category combinators or other common stuff?

 Sergey

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 Haskell-Cafe@haskell.org
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Re: [Haskell-cafe] map over Bijections

2012-08-27 Thread Tillmann Rendel 

Hi,

Sergey Mironov wrote:

I need map equivalent for Bijection type which is defined in fclabels:

data Bijection (~) a b = Bij { fw :: a ~ b, bw :: b ~ a }

instance Category (~) = Category (Bijection (~)) where ...

I can define this function as follows:
mapBij :: Bijection (-) a c - Bijection (-) [a] [b] - Bijection (-) [a] [c]
mapBij b1 b = (map (fw b1)) `Bij` (map (bw b1))


Two observations.

First observation: The second argument seems unnecessary, so we have the 
following instead:



mapBij :: Bijection (-) a c - Bijection (-) [a] [c]
mapBij b = (map (fw b)) `Bij` (map (bw b))


Second observation: I guess this works for arbitrary functors, not just 
lists, so we get the following:



fmapBij :: Functor f = Bijection (-) a c - Bijection (-) (f a) (f c)
fmapBij b = (fmap (fw b)) `Bij` (fmap (bw b))


Lets check that fmapBij returns a bijection:

  fw (fmapBij b) . bw (fmapBij b)
  {- unfolding -}
= fmap (fw b) . fmap (bw b)
  {- functor -}
= fmap (fw b . bw b)
  {- bijection -}
= fmap id
  {- functor -}
= id


Looks good.


I guess we can generalize this to get: If f is a functor on a category 
c, it is also a functor on the category (Bijection c). But I am not sure 
how to express this with Haskell typeclasses. Maybe along the lines of:



import Control.Categorical.Functor -- package categories

instance Endofunctor f cat = Endofunctor f (Bijection cat) where
  fmap b = (fmap (fw b)) `Bij` (fmap (bw b))


So Bijection is a functor in the category of categories?

  Tillmann





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Re: [Haskell-cafe] map over Bijections

2012-08-27 Thread Sergey Mironov 
Yes, you are right, I don't really need the second argument. I am not
skilled enough to join the discussion, but I do understand your
solution. Thanks!

Sergey

2012/8/27 Tillmann Rendel ren...@informatik.uni-marburg.de:
 Hi,


 Sergey Mironov wrote:

 I need map equivalent for Bijection type which is defined in fclabels:

 data Bijection (~) a b = Bij { fw :: a ~ b, bw :: b ~ a }

 instance Category (~) = Category (Bijection (~)) where ...

 I can define this function as follows:
 mapBij :: Bijection (-) a c - Bijection (-) [a] [b] - Bijection (-)
 [a] [c]
 mapBij b1 b = (map (fw b1)) `Bij` (map (bw b1))


 Two observations.

 First observation: The second argument seems unnecessary, so we have the
 following instead:

 mapBij :: Bijection (-) a c - Bijection (-) [a] [c]
 mapBij b = (map (fw b)) `Bij` (map (bw b))


 Second observation: I guess this works for arbitrary functors, not just
 lists, so we get the following:

 fmapBij :: Functor f = Bijection (-) a c - Bijection (-) (f a) (f c)
 fmapBij b = (fmap (fw b)) `Bij` (fmap (bw b))


 Lets check that fmapBij returns a bijection:

   fw (fmapBij b) . bw (fmapBij b)
   {- unfolding -}
 = fmap (fw b) . fmap (bw b)
   {- functor -}
 = fmap (fw b . bw b)
   {- bijection -}
 = fmap id
   {- functor -}
 = id


 Looks good.


 I guess we can generalize this to get: If f is a functor on a category c, it
 is also a functor on the category (Bijection c). But I am not sure how to
 express this with Haskell typeclasses. Maybe along the lines of:

 import Control.Categorical.Functor -- package categories

 instance Endofunctor f cat = Endofunctor f (Bijection cat) where
   fmap b = (fmap (fw b)) `Bij` (fmap (bw b))


 So Bijection is a functor in the category of categories?

   Tillmann





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