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
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@dibblego No it's not. Contravariant functors are `(a -> b) -> (f b -> f a)`. My package provides the true cofunctors.
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@dibblego Yes, is what I said. This is different from cofunctors, though. Also note which category my package is in.#typetheoryjokes -
@jaspervdj Sorry mate :)
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