[evmar]

This post shows a Haskell-ish definition of Functor, then attempts to show the same thing in TypeScript.

    interface Functor<A> {
        map<B>(f: (a: A) => B): Functor<B>;
    }

But the TypeScript definition loses something important: the point of a Functor is that you get back the same data type -- there's one `f` in the Haskell defn both in the argument and return type, while this TS definition can give you back any random data type. E.g. the implementation of Array.map could give you back an Either.

I don't mean this comment to be a random nitpick of the post. Trying to think these things through is hard, and trying to use things you know to help understand the new idea is not unreasonable. But in particular with PL each thing has so much associated baggage (here in TS, subtyping) that reasoning "by metaphor" often means you end up missing the critical point.

[gcanti]

fp-ts [1] contains an implementation of Higher Kinded Types, which TypeScript doesn’t support natively (the idea for emulating higher kinded types in TypeScript is based on "Lightweight higher-kinded polymorphism" [2])

[1] https://github.com/gcanti/fp-ts [2] https://www.cl.cam.ac.uk/~jdy22/papers/lightweight-higher-ki...

[lacampbell]

Haven't had a coffee yet so go easy on me - doesn't this solve your issue?

    interface Functor<A> {
        map<B>(f: (a: A) => A): Functor<A>;
    }

You map over an option, you get an option. You map over an either, you get an either. etc etc.

[tel]

That gets closer to the problem, but the solution is further away. You don't want to return "Functor" (which is like a vtable, the interface itself) but instead the thing itself which is implementing the Functor interface.

So you end up with

    interface Functor<A> {
        map<B>(f: (a: A) => A): Self<B>
    }

where `Self` needs to recognize that the type being defined to implement this interface has a "slot". This tends to make things tough.

[rubyn00bie]

I'd like to +1 that "tends to make things tough." Not sure how it is today, I think it was remedied in Swift 4.1 or something... but holy-shit, that "slot" problem gave me endless grief trying to implement functors in swift for composing UI structs (I was ahead of the curve, heh).

[lacampbell]

You've lost me. Returning an interface isn't returning a concrete thing. It's returning the thing that implements the interface

I'm a very low level functional programmer. I'm big on immutability, big on not using loops and instead using map/flatMap/filter/fold, I tend to roll my own either and option implementations when they don't exist because it's the tidiest way of handling errors I've come across, etc etc. But when it comes to stuff like functors I don't get what it's buying me. What interesting stuff can I do if I know that both an option and a list can be mapped over?

I really need to look more deeply into it at some stage. I might be missing out on some powerful tools. Or it might be a bunch of stuff that's theoretically interesting but practically useless.

[rockostrich]

I think you may have misunderstood their point. They're just saying that specifying `Functor` in the return type of `map` isn't enough to resolve the issue because you could still have a case of `List#map` returning a `Maybe` or `Either` since they are all implementations of `Functor`.

This gets handle by the Cats library in Scala (https://typelevel.org/cats/typeclasses/functor.html) by defining the type class the functor is being defined for as an abstract type on the functor itself.

[tel]

Sorry, you're right. I wasn't thinking with subtyping. Rockostrich demonstrated the other weakness of this approach, but I'll say it outright: if you return a supertype then you've lost information.

It's very important for the type of `fmap f x` to be identical to `x` except in that its "inner" type has been modified. Without that, these kinds of interfaces lose all of their value.

[smegma2]

The functor it returns should be the same functor it was called on, i.e. calling map on a Maybe should return a Maybe. The parameterized type can change so map can bring you from a Maybe<Int> to a Maybe<String> which is why in the above description we have Functor<B> instead of A. But what typescript is lacking is a way to describe that the functor returned should be the same.

[crimsonalucard]

Actually both the Haskell and type script definitions don't communicate the concept of a functor well. Especially with the way it's used.

The interface only requires you to implement the mapping between morphisms, but a mathematical functor also includes mappings between the objects from one category to the next.

Additionally fmap is actually mapping

   (a -> b) to (f a -> f b) 

but the way it's typically used in haskell breaks this intuition. Most people see it as

   ((a -> b), f a) to (f b) 

as a result of how it's used in FP. Technically the usages are isomorphic but using it this way blocks your intuition from fully understanding the true nature of a functor.

The following would be more a more accurate type class for a functor:

   class Functor f where 
      fmap :: (a -> b) -> (f a -> f b) 
      tmap :: a -> f a

See here:

https://en.wikipedia.org/wiki/Functor

Note that there are TWO axioms for functor, I added the second axiom to complete the definition.

I think the reason why tmap doesn't exist is because it's trivial? Not sure maybe some expert can pitch in as I'm certainly not an expert in haskell.

[kmill]

Functors don't include tmap. Yes, a functor maps objects from one category to objects of the second, but it does so on the sets of objects. There might not be a corresponding morphism between "a" and "f a".

The typeclass you created would be a functor along with a natural transformation from the identity functor to f. I don't think these always exist, but they do for applicative functors (pure/return).

The object mapping is reflected at the type level. "f a" is the object that "f" sends "a" to.

In Haskell, functors are all endofunctors. In math, functors can be between categories, and in such a case a -> f a might not make any sense because there are no morphisms between categories.

Using some quasi-Agda syntax, here would be a full definition of a functor between two categories. Maybe you could overload "->" so that "X -> Y" could mean "Mor X Y" for the set of morphisms between X and Y in C or D (depending on context). Curly brackets mean optional parameters, which Haskell approximates via typeclasses.

  record Functor where
    field
      C, D : Cat
      F : Ob C -> Ob D
      fmap : {X Y : Ob C} -> Mor{C} X Y -> Mor{D} (F X) (F Y)

      fmapId : {X : Ob C} -> fmap id == id
      fmapComp : {X Y Z : Ob C} -> {f : Mor X Y} -> {g : Mor Y Z}
                 -> fmap g . fmap f == fmap (g . f)

[crimsonalucard]

>and in such a case a -> f a might not make any sense because there are no morphisms between categories

In CAT the category of small categories objects are categories and morphisms are functors. Haskell types can be thought of as a small categories so I don't think this statement is correct? I'm no expert but can you clarify, the definition on wikipedia says that functors are morphisms between categories in CAT.

>There might not be a corresponding morphism between "a" and "f a".

Hmm. if haskell types make up a small category that means 'a' and 'f a' are sets. I can't imagine a case where two sets cannot have a mapping between them. Can you give a specific example?

>The typeclass you created would be a functor along with a natural transformation from the identity functor to f.

I'm confused about this. Isn't the natural transformation from identity to another category (at least for small categories) equivalent to a functor? In my head they look identical.

https://en.wikipedia.org/wiki/Natural_transformation

If you look at the commutative diagram on this page and replace F(X) and G(X) with identity X and G the natural transformation diagram looks just like a diagram for a functor.

btw I'm not an expert in haskell (let alone agda) or category theory so forgive me if the stuff I talk about is totally off.

[kmill]

Are you familiar with how the word "vector" changes meaning depending on which vector space your talking about? The same goes for "morphism," which depends on the category. In your example of "tmap : a -> f a", the only way this would make sense is as a morphism from object 'a' of the first category to an object 'f a' of the second, but this is a semantic error (a categorical error, if you will).

It is true that functors are morphisms between categories, but that's a morphism in CAT.

> I'm confused about this.

That's how these things go :-) Anyway, natural transformations are between two functors C -> D. This is an example of a 2-morphism in a 2-category. It's hard to come up with something to say other than they're just different, but think about this: a natural transformation is a consistent choice of morphisms F X -> G X (in D) for each object X (in C), but a functor needs to know where the morphisms of C go, too.

(An intuition is a natural transformations is kind of like a continuous path transforming one functor into another. This might just make things confusing, though.)

[Sniffnoy]

> Hmm. if haskell types make up a small category that means 'a' and 'f a' are sets. I can't imagine a case where two sets cannot have a mapping between them.

This is irrelevant. The question isn't, does there exist a function, it's does the functor determine one. (Given any sets A and B there is always a function from A to B unless B is empty and A is nonempty. But that's not very helpful, is it?)

> I'm confused about this. Isn't the natural transformation from identity to another category (at least for small categories) equivalent to a functor?

This doesn't mean anything. A natural transformation is from one functor F:C->D to another functor G:C->D. Not from a functor to... a category? Huh?

Also smallness is irrelevant to all of this.

[crimsonalucard]

>(Given any sets A and B there is always a function from A to B unless B is empty and A is nonempty. But that's not very helpful, is it?)

Can you clarify this? I can't define a -> f a because the functor can't determine one?

What about like for list functor?

   Int -> [Int]

Doesn't this work?

>This doesn't mean anything. A natural transformation is from one functor F:C->D to another functor G:C->D. Not from a functor to... a category? Huh?

I'm still kind of confused. Yeah you're right it doesn't make sense. But then the parent poster is saying that

    a -> f a

is the natural transformation from identity functor to f it seems off to me. It looks to me like it's just a functor from a to f a. So I'm confused with the nomenclature here. a is a category and f a is another category how is a mapping between categories a natural transformation?

Isn't this type signature the natural transformation from identity to f?:

   (a -> a) -> f a

--edit:

I think I see where I'm confused. Your last comment on the parent helped. Thanks.

[kmill]

> is the natural transformation from identity functor

Sorry I confused you here. I meant, if you had a function of that type (a -> f a), you would want a naturality relation to hold for it, like how 'return' for monads is meant to relate to fmap.

As it is, just having a map "a -> f a" is not a natural transformation.

(I think you might be confusing objects of Hask with the category Hask itself. Remember that "a -> f a" means "for each object 'a' of Hask, a function that takes an element of 'a' to an element of 'f a'. The functor is f : Hask -> Hask, an endofunctor. (In haskell, Hask is replaced by stars.) "f a" is the object of Hask that F sends "a" to.)

[Sniffnoy]

> Can you clarify this? I can't define a -> f a because the functor can't determine one?

In mathematics, it's important to distinguish between "an X such that there exists a Y" and "an X together with a Y".

We are discussing, what data determines a functor? You seem to be asserting that part of this is something you're calling tmap, which for a specific functor F would have type a -> F a.

This is mistaken; no such thing is part of the definition of a functor.

But you are confusing the issue of whether such a map is part of the definition of a functor (or is determined by the functor and so could be included in the definition with no change in the actual meaning; let's just group these two cases together as "part of the functor"), with the issue of whether such a map exists.

You pointed out that such a map exists -- as if that made it part of the functor. I pointed out that such a map does exist, but that's not helpful, because this has no relation to the functor, and that it's not part of the functor.

Your reply confuses these issues again, suggesting that because it's not part of the functor like I said (true), therefore it doesn't exist (false). No. It does exist, but it's not part of the functor.

Because it's not part of the functor -- not determined by the functor -- that means that, if you were to tack it on to the definition of the functor -- let's call this new object you're defining a functor' -- then, for any one functor, there would be multiple possible functor' that result, because of the different choices that can be made. This demonstrates that a functor and a functor' are not in fact the same thing; the functor' contains extra information that the functor does not, information that cannot be determined from the functor alone.

> Isn't this type signature the natural transformation from identity to f?: > (a -> a) -> f a

No. You are mixing things in ways that do not make sense.

Look, apologies, but I think you're in over your head here. And I'm afraid your notation is confusing the issue by mixing things inappropriately. There are too many errors here to fix individually; if you want to get this right, you're going to have to take it from the top. I could try to write an explanation that I'd hope you might understand, but my point is that that's what I'd have to do, write a whole explanation from the top. I don't see this conversation going anywhere otherwise.

[tel]

> Hmm. if haskell types make up a small category that means 'a' and 'f a' are sets. I can't imagine a case where two sets cannot have a mapping between them. Can you give a specific example?

This is incorrect in a bunch of ways. First, Haskell types aren't truly just sets. As a simple, practical example you might have

    module Container (makeInt, Container, getValue) where
  
    data Container a 
      = Container { getValue :: a }
      deriving Functor

    makeInt :: Int -> Container
    makeInt = Container

And now we've got (externally) a type which is equivalent to the identity functor in structure, but has a weird interface which disallows tmap.

More powerfully, we're only actually interested in discussing an interface in and of itself. A type may instantiate many interfaces of varying levels of power, but each interface needs to be well-defined and well-behaved on its own. Then their compositions need to be "glued together" properly.

So even if all Haskell Functors were actually TMapFunctors, it's still important to note that the interface TMapFunctor is a sub interface to Functor which allows more operations and has more laws.

An even stronger example of this is the fact that due to the way Haskell arrows work, all Haskell Functors are actually "strong" functors in the sense that we cannot even truly specify a non-string functor.

Strength is a way that fmap and products (or, really, category tensors) interact.

    strength :: Functor f => (a, f b) -> f (a, b)
    strength (a, fb) = fmap (\b -> (a, b)) fb

This seems completely obviously possible, but it's only because we can crack open products in building general anonymous arrows. That's not supported in every category and a generic functor should not be expected to satisfy it.

[crimsonalucard]

>An even stronger example of this is the fact that due to the way Haskell arrows work, all Haskell Functors are actually "strong" functors in the sense that we cannot even truly specify a non-string functor.

Can you explain this? What is a strong functor and how does it have to do with arrows? Also what do you mean by non-string functor?

[tel]

A strong functor is one which supports the strength operation above. Haskell -> arrows being based on any lambda term are rich enough to make all Haskell functors strong. But not all functors are!

And “non-string” was just a typo.

[a1369209993]

FWIW,

  Const q a = Const q
  fmap _ (Const q) = Const q

is a perfectly viable Functor on a. Good luck converting a a into one of those.

[tel]

What's hard about that? C and D are Hask. F_Q maps objects A to objects Const Q A, and fmap takes functions in Hom(a, b) to Hom(Const Q a, Const Q b). We identify that Const Q a is just a little subset of Hask, so Hom(Const Q a, Const Q b) is Hom(Q, Q) and fmap works by mapping all functions to id_Q. Now fmapId and fmapComp are satisfied trivially.

[kmill]

a1369209993 was answering my "I don't think [natural transformations from the identity functor] always exist" with "no, they don't."

The natural transformation would be a function "a -> Const q a" for all a. There can't be a way to do this for all q that is somehow natural in q, since this would have to construct a value of q to put into the Const constructor. Worse, (even though vanilla Haskell doesn't allow it), q might be the empty type, so there might be no functions "a -> Const q a" whatsoever. I think the point of the example is that keeping q polymorphic is a simulation of the empty type.

Allowing empty types, I'm happy with the example. However, this is technically a natural transformation in Haskell:

  f :: a -> Const q a
  f x = Const undefined x

since every type contains bottom (undefined).

[tel]

Ah, I misunderstood! And yes, absolutely. An even easier example

    data Nullity a

This is a perfectly valid functor, too. Ignoring undefined, there are no values of it, but fmap cannot be used to create values of the mapped type, just manipulate them.

[crimsonalucard]

I wrote something to refute your example, but then I realized that you're right.

[Sniffnoy]

Other commenters have already pointed out some of the ways you're confused about categories, but let me explicitly answer your question of, why doesn't the functor include the mapping on the objects and not just on the morphisms?

The answer is it does. In this context, the types are the objects. So if F is your functor, then F itself -- which does not have a type, but has kind * -> * -- is the mapping on the objects.

[crimsonalucard]

So you're saying the constructor of the type itself is a -> f a. Right?

I think I get it. Thanks.

[kmill]

I put this in another comment, but I think you're thinking 'a' is a category, but it's an object in a category. The type for a functor (in Haskell) is Hask -> Hask, which is represented by the star notation Sniffnoy uses. That is, it sends objects of Hask to objects of Hask. The object mapping for a functor in Haskell is defined by its "data" or "type" definition.

The notation "a -> f a" means a function that sends elements of 'a' to elements of 'f a', where 'a' is an object of the category Hask. (Functors don't look into the objects at the element level.)

[crimsonalucard]

Thanks for the clarification. I think I get it despite the incorrect notation. You guys are talking about some sort of type of types?

[kmill]

Yeah, the "type of types" is called a kind in Haskell https://wiki.haskell.org/Kind https://en.wikipedia.org/wiki/Kind_(type_theory)

Caveat: there seems to be some reason not to think of kinds as actually being the types of datatypes, but I don't understand why that is.

[Sniffnoy]

No. It's not a -> f a. It's * -> *. There is no a -> f a here. You are simply mistaken in thinking that any such thing is part of the definition of a functor.