[Haskell-cafe] map over Bijections
Sergey Mironov
ierton at gmail.com
Mon Aug 27 18:55:49 CEST 2012
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 <rendel at 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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