Hi, Am Donnerstag, den 17.01.2019, 14:33 +0100 schrieb Olaf Klinke:
I have two type signatures where I conjecture the only functors satisfying these are the Identity functor and functors of the form (,)s. Can anyone give hints as how to tackle a proof?
Signature 1: What functors t admit a function forall f. Functor f => t (f a) -> f (t a)
What about the functor data Void1 a It seems I can write g :: forall f. Functor f => t (f a) -> f (t a) g x = case x of {} but Void1 is not the identity. (I guess it is `(,) Void`, if you want…) So if you allow the latter, let’s try a proof. Assume we have t, and g :: forall f. Functor f => t (f a) -> f (t a) We want to prove that there is an isomorphism from t to ((,) s) for some type s. Define s = t () (because what else could it be.) Now we need functions f1 :: forall a. t a -> (t (), a) f2 :: forall a. (t (), a) -> t a that are isomorphisms. Here is one implementation: f1 :: forall a. t a -> (t (), a) f1 = swap . g . fmap (λx → (x,())) {- note: x1 :: t a x2 :: t ((,) a ()) x2 = fmap (λx → (x,())) x1 x3 :: (,) a (t ()) x3 = g x2 x4 :: (t (), a) x4 = swap x3 -} f2 :: forall a. (t (), a) -> t a f2 (s, x) = x <$ s Now, are these isomorphisms? At this point I am not sure how to proceed… probably invoking the free theorem of g? But even that will not work nicely. Consider type T = (,) Integer g :: forall f. Functor => T (f a) -> f (T a) g (c, fx) = (,) (c + 1) <$> fx here we count the number of invocations of g. Surely with this g, f2 . f1 cannot be the identity. Cheers, Joachim -- Joachim Breitner mail@joachim-breitner.de http://www.joachim-breitner.de/