Category Theory Fundamentals Sachi Hashimoto 1 Category Theory Basics Category theory may seem abstract but I hope to convince you that it is highly understandable and useful! The goal is to get at some unifying concepts in mathematics by just studying maps. For example, one can study sets simply by studying the maps between them instead of by studying the elements inside of the sets. More precisely, here are some basic definitions. Definition. A category C is comprised of objects Ob(C) together with for any two objects X; Y 2 Ob(C) a collection of morphisms f : X ! Y and a composition operation ◦ which allows us to compose morphisms f : X ! Y and g : Y ! Z for X; Y; Z 2 Ob(C) and f; g morphisms. We require that composition is associative and each object X has the identity map idX : X ! X such that idX ◦ f = f and f ◦ idX = f for all f such that those compositions are defined. Example 1. • For example, there is a category of sets with objects which are sets and morphisms which are functions between sets. • There is a category of groups with morphisms which are group homomorphisms. • There is a category of vector spaces over a fixed field k and morphisms which are linear transformations. If C and D are categories, we can construct maps of categories. Definition. A (covariant) functor F : C ! D is a map which • for each object X 2 C assigns an object F (X) in D, and • for each morphism g : X ! Y in C assigns a morphism F (g): F (X) ! F (Y ) in D, such that F (g ◦ f) = F (g) ◦ F (f), and F (IdX ) = IdF (X). 2 The Yoneda Embedding A key example of a functor is the functor X 7! Hom( ;X), called the \functor of points". This is a op functor from a category C to Set. Let's call the functor HX and we will look into where it sends objects and morphisms in C. Fix Y 2 C. Then HX (Y ) = Hom(Y; X) takes the object Y 2 C and sends it to the set Hom(Y; X) in Set. Fix f : Y ! Z. Then HX (f) needs to take f to HX (f): HX (Y ) ! HX (Z). Let's unpack this! This is equivalent to HX (f) : Hom(Y; X) ! Hom(Z; X), so we need a morphism in Set from taking 1 a map s : Y ! X and turning it into a map Z ! X. But, we should use the fact that f : Y ! Z. So, the answer is to take s and pre-compose it with f. Thus HX (f) is pre-composition with f, HX (f) = ◦ f. Now, we can turn this on its head a bit: for each object X in C we get a functor HX . That sounds like a functor! In particular, this is now a function from C to functors from Cop ! Set. Let's op write the latter set as [C ; Set]. Then we want to prove that the assignment X 7! HX is a functor op H∗ : C! [C ; Set]. This is called the Yoneda Embedding. First, we need to understand what H∗ does to morphisms a little better. A morphism between functors is a natural transformation, so in order to understand this, we have to understand natural transformations. If this helps, natural transformations are like homotopies. If not, don't worry about it. In the abstract, suppose that I have two functors F and G taking a category C to a category D. When would we say that these functors are equivalent? I want to assign every object F (X) to the object G(X) and every morphism F (f) to the morphism G(f) in a compatible way. If I fix an object X 2 C and assign F (X) ! G(X) by ηX (and similarly assign for every object in C a pairing η), then for each morphism f : X ! Y in C, I will get a constraint, depicted by the following diagram: F (f) F (X) / F (Y ) ηX ηY G(X) / G(Y ) G(f) F (f) sends F (X) to F (Y ), but G(f) sends G(X) to G(Y ) and we have assigned F (Y ) to G(Y ) by ηY . We need the square to commute. In our example above, we will let F = HA and G = HB for two objects A and B. If f : A ! B is any morphism, it gives rise to a natural transformation given by composition with f. 3 Representability and Pro-Representability Representability is sort of the converse of what we just discussed. We showed how to take an object X and turn it into a functor HX : C ! Set by setting HX = Hom(X; ). Suppose instead we were handed a (now covariant) functor F : C ! Set. Could we find some X 2 C such that F = Hom(X; ) (the covariant version)? In other words, can we find X and a natural isomorphism φ : F ∼= Hom(X; )? If so, we say that F is representable or F is represented by (X; φ). (We can play the same game with contravariant too but all my examples later are covariant). Yoneda's lemma says the following thing about the functor HX : Lemma 1 (Yoneda). Let F be an arbitrary functor C ! Set. Then natural transformations from F to HX are in bijection with the set F (X). This is quite easy to prove from what we have done so far, the idea is to consider the distinguished element which is the identity element, but I will leave it as an exercise for the listener. Example 2. There is a forgetful functor from groups to sets, which takes a group and forgets about the group structure. This is representable by Z. 2 Similarly, the forgetful functor from rings to sets is representable by Z[x]. You can play this game with other forgetful functors: try it out for yourself! Sometimes the representing object that you want doesn't quite live in the category you are in. I'm not going to belabor this point, mostly because it is a relatively straightforward leap of intuition and we don't have much time, but it's possible that instead of being representable, F is of the form limi Hom(Xi;Y ) where we are taking a direct limit, for example if we are interested in a n limit of Z=p Z as n increases, but Zp is not an object in our category because are are only looking at Artinian rings. 4 Products and Fibre Products Before we talk about fibre products, let's review the definition of product, placing it in the setting of a category. This generalizes the notion of a cartesian product. Definition. Let A and B be objects in a category. Then the product of A and B is an object A×B together with projection maps π0 : A × B ! A and π1 : A × B ! B such that whenever we have an object C with maps f0 : C ! A and f1 : C ! B there is a unique map f : C ! A × B such that the following diagram commutes: C f0 f1 f { # AAo × B / B π0 π1 We can prove the product is unique up to unique isomorphism. Example 3. In the category of sets the product is simply the cartesian product. In the category of groups the product is the direct product of groups. A fibre product can be see as a special kind of product. Often, we like to think of categories where something lives over a base field or base ring. These are categories of objects that come with a morphism to a certain object. For example, we might think of the category of finite dimensional k-algebras: these are algebras A that come with a structure map A ! k. Similarly, we could be in a category of k-schemes, or S-schemes, and this will be a category of schemes which all have a map to Spec k or Spec S. Morphisms between these objects will have to preserve these structure maps. We will see that one way to define a fibre product is simply to say that it is a product in such a category. That is what makes fibre proudcts so important. The diagram for the fibre product of A and B over s and t looks like this: C f1 f " π1 $ A ×Z B / B f0 π0 t A s / Z 3 The fibre product does not need to exist, but if it does it exists up to isomorphism. What is the fibre product? Let's do an example in the category of sets. The product A × B is just the cartesian product, but if we have maps s : A ! Z and t : B ! Z then A ×Z B = f(a; b) 2 A × Bjs(a) = t(b)g: This we can compute by tracing a pair (a; b) around the diagram: going around the right side, we get (a; b) 7! b 7! t(b). Going around the left side we get (a; b) 7! a 7! s(a). So for that pair to be in the fibre product we must have s(a) = t(b), and we can see this is an if and only if statement. It is not hard to show that this satisfies the universal property above. 5 Mayer-Vietoris The Mayer-Veitoris principle will give another criterion for representability. We are motivated by the following situation from topology.