Remark 1. Before reading the formal definition of a category, keep in mind the big picture and some examples.

A category is just a collection of objects and arrows (called morphisms) between them. There’s the category \(\mathsf{Top}\) which has objects being topological spaces, and morphisms being continuous maps (i.e., an arrow for each continuous map). There’s the category \(\mathsf{Group}\) which has objects being groups, and morphisms being group homomorphisms.

At the same time, a category is a very general definition, and some unexpected things can actually fit the axioms. In general we think of morphisms as being functions, but there are examples where morphisms aren’t literally functions, which we’ll get into later.

Remark 2. We outline some notation.

• Instead of writing \(A\in \mathop{\mathrm{Ob}}(\mathcal{C})\), we often just write \(A\in \mathcal{C}\).

• Instead of writing \(f\in \mathop{\mathrm{Hom}}_\mathcal{C}(A,B)\), we often just write \(f\in \mathop{\mathrm{Hom}}(A,B)\) when \(\mathcal{C}\) is clear, or just \(f: A\to B\).

• For \(f:A\to B\), we call \(A\) the domain of \(f\) and \(B\) the codomain of \(f\).

• Instead of writing \(g\circ f\) for \(f:A\to B\), \(g:B\to C\), we often just write \(gf\) (unless this could be confused for a coordinate-wise product in certain categories).

Remark 3. Our goal is to understand what it means for something to be a product/coproduct in the category theory sense, and then see how this perspective can help us prove some things in topology.

Remark 4 (Intuition). You’ve definitely seen some products before. The cartesian product of two sets \(A,B\) is a new set \[A\times B = \{(a,b) \mid a\in A, b\in B\}.\]

The product of topological spaces \(X,Y\) is the cartersian of the underlying sets \(X\times Y\), with the topology generated (i.e., having subbasis) sets \(U\times Y\), \(X\times V\) for \(U\subseteq X\) open, \(V\subseteq Y\) open. This is kind of a weird way to phrase it, but it turns out to be useful later. This is exactly the same as the topology generated by \(U\times V\) for \(U,V\) open, since we can get \(U\times V\) by doing \[U\times Y \cap X\times V.\]

We can also talk about products of groups, rings, etc. All of these constructions feel kinda natural in some sense. For starters, these constructions make it so the projection maps are morphisms in the appropriate category (i.e., functions, continuous maps, group/ring homomorphisms). However, this is clearly not a sufficient requirement. For example, \(\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\) has two projection maps to \(\mathbb{Z}\), but this thing is definitely not the product of \(\mathbb{Z}\times \mathbb{Z}\).

In category theory we like to zoom out and talk about objects and morphisms, usually trying to somewhat avoid talking about elements of objects. Can we come up with a category theoretic definition of product that somehow encapsulates all of these examples? It probably should say something about these projection maps, but what should it say?

Remark 5. So that’s a lot of text, and we should make some comments.

First, this universal property holds for the product constructions that we’re familiar with. For example, if \(A,B,C\) are sets, and we have a functions \(f_A: C\to A\) and \(f_B: C\to B\), we can assemble these two coordinate functions into a single function \(f:C\to A\times B\) defined by \(f(c) = (f_A(c),f_A(c))\).

Second, we said “a” product, not “the” product. For starters, there are some categories where certain objects do not have products. Still, we’ll eventually see that the product (if it exists) is unique up to unique isomorphism that makes the projections commute. So this makes it so products are pretty much unique, but not quite. We’ll usually say “the” anyways.

For example, if \(\varphi: C\to A\times B\) is an isomorphism between an object \(C\) and the product \(A\times B\), then we could also \(C\) is a product with projections \(\pi_A\circ \varphi, \pi_B \circ \varphi\).

Remark 6. I’m going to be a little bit intentionally vague in this remark. It’ll be more clear in the actual proof of uniqueness. I just wanted to get some motivation down first.

Let’s turn our attention to uniqueness of the product. Our definition of a categorical product is an example of a universal property. There’s a precise definition of universal property somewhere out there, but it’s not worth getting at for now (I’m not even sure what it is either). Essentially, something has a universal property when we have a condition that says “whenever \(-\) morphisms exist, then there exists a unique morphism \(-\) such that \(-\) commutes”.

Then, there’s a common strategy to prove that an object with a universal property is unique up to unique isomorphism (that makes some diagram commute). Given two objects \(C,D\) with the universal property, we use the universal property to come up with a morphism \(C\to D\) and a morphism \(D\to C\), and then we use uniqueness to argue that the composition \(C\to D\to C\) is just the identity morphism.

Remark 7 (Intuition). We saw before that products generalize cartesian products. In the same way, coproducts generalize disjoint unions. Roughly speaking, a coproduct is the free-est way to put together two objects in a category. In particular, we’ll see later that direct sums are an example of coproducts.

First things first, we should clarify what we mean by a disjoint union. There’s a slightly annoying abuse of notation that goes on here. You might have heard of internal/external direct sum—we’re going to have the same problem.

Remark 8. In case of external disjoint unions, we put together two sets \(A,B\) that might overlap, but we assert that their elements be different in the disjoint union by “tagging” them with either a \(0\) or a \(1\). These tagged versions of \(A,B\) are still in bijection with \(A,B\).

Remark 9. As before, coproducts need not exist in every category, and they will only be unique up to unique isomorphism. We will see a proof of this later.

Remark 10. We recall that the direct product and direct sum of finitely many \(R\)-modules are isomorphic, i.e., \(A\times B \cong A\oplus B\). In some category, for objects \(A,B\), when the product and coproduct are isomorphic, we call it the biproduct. The category of \(R\)-modules is a particularly nice category because it has all finite biproducts. It has lots of other nice properties too—it’s what’s called an abelian category.

Remark 11. Before proving the uniqueness of coproducts, we’re going to go on a tangent about duality in category theory, which in the process will give us a nice proof of uniqueness of coproducts for free.

Remark 12. Observe that the diagram giving the universal property of a coproduct has the same shape as that of products, except all of the arrows are reversed. We can actually take advantage of this symmetry to translate proofs about products into proofs of dual propositions about coproducts.

For starters, we could just say, “take the proof of uniqueness of products, except flip all arrows, and then we have a proof of uniqueness of coproducts” (using the fact that “reversed” isomorphisms are still isomorphisms). So you could try to write down a proof of uniqueness of coproducts already.

Instead, we’re going to do essentially the same thing with more formalism. If we have a category \(\mathcal{C}\) and we want to prove uniqueness of coproducts in \(\mathcal{C}\), we’re going to cook up another category called \({\mathcal{C}}^{\mathrm{op}}\). Then, in \({\mathcal{C}}^{\mathrm{op}}\), coproducts of \(\mathcal{C}\) are literally translated into products.

Remark 13. We make a few remarks.

• The opposite category is really is just taking a category, and flipping all of the arrows around. The thing to note is that reversing all the arrows still gives you a category—you should check the axioms and see that identities and associativity still works out.

• For \(f\in \mathop{\mathrm{Hom}}_\mathcal{C}(B,A)\), sometimes (for clarification) we write the corresponding morphism in \(\mathop{\mathrm{Hom}}_{{\mathcal{C}}^{\mathrm{op}}}(A,B)\) by \({f}^{\mathrm{op}}\).

• The opposite category is an example of a category where the morphisms aren’t literally functions—they are just arrows that are “labeled” by functions, but the arrows are pointing opposite of the functions.