Category Theory for Programmers Chapter 10: Natural Transformations

05 Oct 2018

A reminder of the naturality condition: G f ∘ α_a = α_b ∘ F f

1. Define a natural transformation from Maybe to the List functor. Prove naturality condition for it.

   asList :: Maybe a -> [a]
   asList Nothing = []
   asList (Just x) = [x]
   

   To prove the naturality condition, fmap f . asList = asList . fmap f, we have two cases, Nothing and Just x.

   fmap f (asList Nothing) = fmap f [] = []
   asList (fmap f Nothing) = asList Nothing = []
   

   fmap f (asList (Just x)) = fmap f [x] = [f x]
   asList (fmap f (Just x)) = asList (Just (f x)) = [f x]
   

2. Define at least 2 different natural transformations between Reader () and the list functor []. How many different lists of () are there?

   asEmptyList :: Reader () a -> [a]
   asEmptyList _ = []
   

   asOneList :: Reader () a -> [a]
   asOneList r = [r ()]
   

   asTwoList :: Reader () a -> [a]
   asTwoList r = [r (), r ()]
   

   And so on, there are infinitely many of these.

3. Define natural transformations between Reader Bool and Maybe.

   none :: Reader Bool -> Maybe
   none _ = Nothing
   

   true :: Reader Bool -> Maybe
   true r = Just (r True)
   

   false :: Reader Bool -> Maybe
   false r = Just (r False)
   

4. Show that horizontal composition of natural transformations satisfies the naturality condition.

   Let F, F' be functors from categories A to B and G, G' functors from categories B to C with natural transformations α from F to F' and β from G to G'. We must show the naturality condition holds for

   (G ∘ F) f ∘ (β * α) = (β * α) ∘ (G' ∘ F') f
   

   I don’t want to do it.

5. No essays.

6. Naturality conditions of transformations between differetn Op functors. Ie, Op A B is B -> A, and

   op :: Op Bool Int
   op = Op (\x -> x > 0)
   
   f :: String -> Int
   f x = read x
   

   fmap f op :: Op Bool String Don’t know where to go with this…