[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.

[crimsonalucard]

It's fine, no need to come from the top, thanks for the explanation. I mean if you want to take it from the top I'll read it really carefully, just saying I'm grateful that you both even take the time to explain what you guys already have.

I'm out of responses here on HN so I'll respond tomorrow.

I'm not a mathematician, so yeah I am a bit over my head.