Re: [Haskell-cafe] Applicative banana brackets

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 wrote:

On Mon, Dec 14, 2015 at 9:38 PM, martin 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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