Problems with products, part one

Posing a difficulty with composition

Where we see a type parameter to the right of a function arrow (or with no arrow at all, since a is isomorphic to () -> a), we have a covariant functor (or simply, "functor").

newtype Decode k v a = Decode{ decode :: Map k v -> a }

Covariant functors quite often admit an applicative aspect that allows us to build up larger stuff within the type parameter.

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