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How can I make (a, a) a Functor without resorting to a newtype ? Basically I want it to work like this: instance Functor (a, a) where fmap f (x, y) = (f x, f y) But of course that's not a legal way to express it: Kind mis-match The first argument of `Functor' should have kind `* -> *', but `(a, a)' has kind `*' In the instance declaration for `Functor (a, a)' What I really want is a type-level function like this: \a -> (a, a) (invalid syntax). So a type alias, perhaps? type V2 a = (a, a) instance Functor V2 where fmap f (x, y) = (f x, f y) I would think this would work, but it doesn't. First I get this complaint: Illegal instance declaration for `Functor V2' (All instance types must be of the form (T t1 ... tn) where T is not a synonym. Use -XTypeSynonymInstances if you want to disable this.) In the instance declaration for `Functor V2' If I follow the advice and add the TypeSynonymInstances extension, I get a new error: Type synonym `V2' should have 1 argument, but has been given 0 In the instance declaration for `Functor V2' Well, duh, that's the point! V2 has kind * -> * which is what is required of a Functor instance. Well, ok, I can use a newtype like this: newtype V2 a = V2 (a, a) instance Functor V2 where fmap f (V2 (x, y)) = V2 (f x, f y) But now I've got to sprinkle V2 s liberally throughout my code instead of just being able to deal with simple tuples, which kind of defeats the point of making it a Functor ; at that point I might as well make my own function vmap :: (a -> b) -> (a, a) -> (b, b) . So is there any way to do this nicely, i.e. without a newty

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