[Haskell-cafe] Applicative banana brackets

[Erik Hesselink]

Every Arrow is a Functor through:

    fmapA :: Arrow arr => (a -> b) -> arr i a -> arr i b
    fmapA f a = arr f . a

Right?

Erik

On 14 December 2015 at 15:53, Kim-Ee Yeoh <ky3 at atamo.com> wrote:
> On Mon, Dec 14, 2015 at 9:38 PM, martin <monkleyon at googlemail.com> wrote:
>
>> 2) Every arrow a b c is an applicative (a b) c, because comma = (&&&).
>> Conversely, every applicative that is polymorphic over some "internal"
>> variable is automatically an arrow, through (&&&) = comma.
>
>
> Where is the proof that (a b) is a Functor?
>
> Recall that the class method arr -- the closest kin to fmap -- has the type:
>
> (b -> c) -> a b c.
>
> -- Kim-Ee
>
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