Tony Morris

If Functor gives _map_ `(a -> b) -> F a -> F b`, then Applicative Functor gives us _functorial map_ `F (a -> b) -> F a -> F b`. #haskell

@ElAthanator Functor: a -> b Contravariant: b -> a Exponential: (a -> b,b -> a) Applicative: f (a -> b) Monad: a -> f b Comonad: f a -> b

@dibblego Don't forget the ubiquitous cofunctor https://hackage.haskell.org/package/acme-cofunctor …

@dibblego No it's not. Contravariant functors are `(a -> b) -> (f b -> f a)`. My package provides the true cofunctors.