Also with your edit it's worth noting that your given fmap cannot be: it violates both functor laws. As it turns out, for Haskell and most datatype-like functor

Also with your edit it's worth noting that your given fmap cannot be: it violates both functor laws. As it turns out, for Haskell and most datatype-like functors there is exactly one law abiding implementation of fmap implicit in the structure of the type.