Ok, now I see a difference, why Kleisli can be used to relate
typeclasses (like Monad and ArrowApply) and Cokleisli can not:
"Kleisli m () _" = "() -> m _" is isomorphic to "m _"
whereas
"Cokleisli m () _" = "m _ -> ()" is not.
Can somebody point out the relevant category theoretical concepts,
that are at work here?
On Tue, May 28, 2013 at 5:43 PM, Tom Ellis
On Tue, May 28, 2013 at 05:21:58PM +0200, Johannes Gerer wrote:
That makes sense. But why does
instance Monad m => ArrowApply (Kleisli m)
show that a Monad can do anything an ArrowApply can (and the two are thus equivalent)?
I've tried to chase around the equivalence between these two before, and I didn't find the algebra simple. I'll give an outline.
In non-Haskell notation
1) instance Monad m => ArrowApply (Kleisli m)
means that if "m" is a Monad then "_ -> m _" is an ArrowApply.
2) instance ArrowApply a => Monad (a anyType)
means that if "_ ~> _" is an ArrowApply then "a ~> _" is a Monad.
One direction seems easy: for a Monad m, 1) gives that "_ -> m _" is an ArrowApply. By 2), "() -> m _" is a Monad. It is equivalent to the Monad m we started with.
Given an ArrowApply "_ ~> _", 2) shows that "() ~> _" is a Monad. Thus by 1) "_ -> (() ~> _)" is an ArrowApply. I believe this should be the same type as "_ ~> _" but I don't see how to demonstrate the isomorphsim here.
Tom
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