Natural Transformations From Reader Bool To Maybe

Question

I'm reading Bartosz's awesome blog, and following along on challenge question 3, I'm a bit stumped:

https://bartoszmilewski.com/2015/04/07/natural-transformations/

Q3: define some natural transformations from Reader Bool to Maybe

I have defined the Reader functor as:

newtype Reader e a = Reader (e->a) runReader :: Reader e a -> e -> a runReader (Reader f) env = f env instance Functor (Reader e) where fmap f (Reader g) = Reader (\x -> f (g x))

and I want to find a natural transformation

n :: Reader Bool a -> Maybe a

My intuition is that if Reader's environment is True, I can have a Just a, and if False, then the natural transformation projects to Nothing. But that feels slightly monadic, or like a Maybe nested inside a Reader, like Reader Bool (Maybe Int), so I'm not sure.

I can't concieve of how to do that. The best I have is:

n :: Reader Bool a -> Maybe a n (Reader f) = Just (f True)

But this cannot take into account the environment, nor return Nothing.

I want to build a structure that commutes, building a square out of fmap, and natural transformations.

For ex:

 nat Reader Bool Int +---------------------------> Maybe Int + + | | | | | | | fmap show | fmap show | | | | | | | | v nat v Reader Bool String +--------------------------> Maybe String

can you help me fill in the gaps?

You want n to use the environment, but, since you have only the Reader, there is no environment and you have to provide it yourself. I feel that Just (f True), Just (f False) and Nothing are all natural transformations between these functors. — lisyarus
– lisyarus, Commented Aug 17, 2016 at 22:21

lisyarus · Accepted Answer · 2016-08-17 22:27:16Z

As I've already said in a comment, you want n to use the environment, but, since you have only the Reader, there is no environment and you have to provide it yourself.

Evidently, these are in fact three distinct natural transformations Reader Bool a -> Maybe a:

n (Reader f) = Just (f True) n (Reader f) = Just (f False) n (Reader f) = Nothing

Let's prove there are no more. In category-theoretic terms, the functor Reader Bool a is just Hom(Bool,_). Now, Yoneda lemma tells us that

Nat(Hom(X,_), F) = F(X)

That is, natural transformations from the functor Hom(X,_) to a functor F are in one-to-one correspondence with elements of the set F(X).

In our case, X = Bool and F = Maybe, so we obtain

Nat(Hom(Bool,_),Maybe) = Maybe Bool

Maybe Bool contains exactly three elements: Just True, Just False and Nothing, which correspond to the implementations in the beginning of the answer.

That makes sense! So, there are only 3 possible nat. transformations between Reader Bool and Maybe, right? Then on question #2 he asks for natural transformations from Reader () to List. I came up with: n (Reader f) = [f ()] and infinite of n (Reader f) = [() ... ()]. Is this accurate?
@Josh.F, Reader () is morally isomorphic to Identity. So you get one element and need to make a list. There is thus one such natural transformation for each natural number (don't forget 0), plus the infinite one. I think.
@dfeuer Exactly! And Yoneda gives the same: natural transformations Reader () -> List correspond to List ().

Collectives™ on Stack Overflow

Natural Transformations From Reader Bool To Maybe

1 Answer 1

3 Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Related