Presheaves in category theory

Patrick Reany

26 November 2023

In the abstract, the notion of a presheave in category theory doesn't seem too difficult, until one gets down to specific cases. Let C be a category, then a presheave based on C is a functor from C^op to the category Set. Generally speaking for any category C, C^op reverses all the arrows on all the objects. So, let's look specifically at the category of Set, whose objects are sets and whose morphisms are functions. Now, we all know that functions do not generally have inverses, and hence, cannot reverse direction -- unless the function is injective. But to reverse an arrow between sets, the function would also have to be surjective. Therefore, in the category of Set, C^op only makes sense if all the morphism are bijections. That's what commonsense would dictate, I suppose, but apparently it's more subtle than that.

What I would learn is that in C^op the objects don't change, but the morphisms do -- they are no longer functions in general. It seems intuitive that we could perhaps define the morphism in C^op in any manner we please so long as the following is an identity:

(C^op)^op = C.

However, this seems not to be the case. Most of this article is about the nature of f ^op when f is a morphism between sets, i.e., a regular function. On 22 November 2023, I had this discussion with ChatGPT in the hopes that the LLM would clarify the situation for me. Unfortunately, it did not.

Patrick:
I'm rather confused about the notion of presheaves in category theory. I have a question for you about it.

ChatGPT:
Of course! I'd be happy to help clarify the concept of presheaves in category theory. Please go ahead and ask your question.

Patrick:
If I'm not mistaken, presheaves is a functor from C^op to Set, where C is some category. But reversing the arrows in a category is not always possible, is it. One cannot reverse the arrows in the category of sets or groups, for examples. In the category of Set, reversing an arrow is reversing a function, which is not always possible.

ChatGPT:
You are correct that a presheaf is a functor from the opposite category C^op to Set, where C is some category. However, your concern about reversing arrows is understandable, and it's important to clarify how the opposite category works.

The opposite category C^op has the same objects as C, but the morphisms (arrows) are "reversed" in the sense that if f : A → B is a morphism in C, then in C^op it becomes f ^op : B → A. The composition of morphisms in C^op is also defined by reversing the order.

It's important to note that this notion of "reversing" arrows doesn't imply anything about the structure of the objects themselves; it's only about the morphisms. In Set, for example, the objects are sets, and reversing arrows doesn't mean reversing the elements of the sets or the functions between them. It's just a formal way to define a new category.

Editorial by Patrick: If category theory is to be of practical use, and not be mere abstract nonsense, then we need to know the exact nature of this morphism, which goes the other way in Set^op. Below, we'll get a better handle of what this thing is.

So, while reversing arrows might sound problematic, it's a well-defined process in category theory that allows us to work with the opposite category, and presheaves are functors defined on this opposite category.

If you have more specific questions or if there's a particular aspect you find confusing, feel free to ask for further clarification!

My conclusion on my response from ChatGPT:

I'm afraid that ChatGPT did not clarify things for me. Let's take a specific case. Let A be a set with three elements and let B be a set with two elements. If f is a function from A to B defined by sending all the elements of A to one of the elements of B, then what is f ^op, considering that there are many ways to define a function from B to A?

Follow up.

Instead of taking this situation to ChatGPT, I decided to take the following question to BingChat:

Patrick [23 November 2023]: Let C be a category of sets. Are the morphism in C^op multifunctions?

BingChat: [Bolding done by BingChat.]
In the category of Set, the morphisms are functions. However, in the opposite category of Set, denoted as Set^op, the morphisms are not necessarily functions [1,2,3].

Editorial by Patrick: Now we're getting somewhere. In Set^op the morphisms are not necessarily functions.

Set^op is the category of complete atomic Boolean algebras, or equivalently complete Boolean algebras of the form P(X) where X is a set 1. The morphisms in Set^op are morphisms of complete Boolean algebras that preserve arbitrary meets and negation, and hence preserve the Boolean operations as well [1]. It is important to note that the morphisms in Set^op are not necessarily functions with domain and codomain information [2,3]. I hope this helps! Let me know if you have any other questions. [The emphasis was provided by BingChat.]

[Ref. 1] https://math.stackexchange.com/questions/980933/what-is-the-opposite-category-of-set
[Ref. 2] https://math.stackexchange.com/questions/2092872/do-opposite-categories-always-exist
[Ref. 3] https://math.stackexchange.com/questions/4747736/how-to-understand-opposite-category

I then gave BingChat this specific example:

Suppose A and B are objects in Set. Suppose further that A has three elements and B has two elements. Lastly, suppose f is a morphism (function) from A to B such that all the elements of A are mapped to one element of B. Then what is f^op in Set^op?

The reply was rather long, and I want to report just the important snipit of it:

Since f maps all the elements of A to one element of B, f^op maps that one element of B^op to all the elements of A^op.

So, this is why I asked if the morphisms in Set^op are generally multifunctions. Mapping one element to multiple elements is a multifunction. This is not okay in Set, but, apparently, it is okay in Set^op. In other words, the morphism in general category C^op are not necessarily that same nature as those in general category C. Furthermore, f^op ignores some elements in its mapping, and this makes it a 'partial function'.

Now, it may be canonical or natural to just reverse the mapping in Set^op between each pair of sets, but that may not be the only way to define the morphism in Set^op. Generally speaking, it may be that the only requirement we need to fulfill in defining the morphisms in C^op, given the arbitrary category C, is that

(C^op)^op ≅ C,

which is a theorem in category theory (?) for some but not all categories.

My last followup is with ChatGPT (23 November 2023):

Please explain the double dualization property of category theory if such a thing exists.

ChatGPT:
The term "double dualization" in category theory typically refers to the process of taking the double dual of an object in a category. To understand this, let's break down the key concepts.

Dual Category:
In category theory, given a category C, you can form its dual category, denoted as C^op. The dual category has the same objects as C, but the morphisms (arrows) are reversed. If f: A → B is a morphism in C, then in C^op, the morphism goes in the opposite direction: f^op: B → A.

Double Dual of an Object:
Now, the double dual of an object A in a category C refers to taking the dual of the dual of A. Symbolically, if A is an object in C, then its double dual, denoted as A^op ^op, is the object in C^op ^op.

Start with A in C.
Take the dual to get A^op in C^op. Take the dual again to get A^op ^op in C^op ^op. The double dualization process is interesting because in many categories, the double dual of an object is naturally isomorphic (essentially, structurally identical) to the original object. The canonical isomorphism is given by the evaluation map, which provides a natural transformation from an object to its double dual.

This property is often encountered in categories where morphisms have some notion of "duality," such as vector spaces and linear maps. In such categories, the double dualization process essentially brings you back to where you started.

It's important to note that not all categories exhibit the double dualization property. The existence and properties of the double dual depend on the specific structure and properties of the category in question.

Patrick:
Is there some sufficiency condition by which we can claim that for category C, C^op ^op is isomorphic to C?

ChatGPT:
Yes, there is a condition that ensures that the double dual of a category is isomorphic to the original category. This condition is known as the self-duality of a category.

A category C is said to be self-dual if it is isomorphic to its own dual, i.e., if C ≅ C^op. In this case, taking the dual again results in C^op ^op ≅ C. Therefore, the self-duality condition is a sufficient condition for a category to have the property that its double dual is isomorphic to itself.

The self-duality condition is not always satisfied, and it depends on the specific properties of the category. Categories with self-duality are often found in contexts where morphisms have some notion of "reversibility" or "duality," such as in certain algebraic structures or in categories related to linear algebra.

It's worth noting that not all categories are self-dual. Some categories are self-dual, some are not, and for others, it might be an open question. The study of self-duality and related concepts is part of the broader investigation into the structure and properties of categories in category theory.

Patrick:
Clearly then the category of Set is not self-dual, right?

ChatGPT:

That's correct. The category of sets, often denoted as Set, is not self-dual. In Set, the objects are sets, and the morphisms (arrows) are functions between sets. Taking the dual of Set involves reversing the direction of morphisms, which means considering functions in the opposite direction. This doesn't lead to an isomorphic category to Set.

To be more specific:

In Set, if you have a function f: A → B, its dual (in Set^op) would be f^op: B → A, where functions are reversed. If you take the dual again (in (Set^op)^op), you'd get f^op ^op: A → B. However, this is not isomorphic to the original function f: A → B. So, the category of sets is not self-dual. Self-duality is a specific property that not all categories possess, and it depends on the nature of the morphisms and objects in the category.

Afterthoughts.

Okay, the category of Set is not self-dual, but that doesn't by itself preclude Set from having the double dualization property, anyway. In fact, if we adopt that algorithm given to us by BingChat, not only is Set^op ^op ≅ Set, they're equal: Set^op ^op = Set. This isn't hard to understand. The functions between the sets in category Set get mapped to multifunctions between the sets (Objects) of Set^op

I now offer a notation for a map which is in Set^op, which I'll refer to as f °, pronounced as 'f circ', and it can be at once a partial function and a multifunction. The advantage of defining f ° is that one does not have to be inside a categorical framework to use it, but if one is, then if f is a morphism in Set, then f ^op = f °.