# Concrete category with non-standard products

I am working on a paper in which I need to talk about what I call concrete categories with standard products. I write $U(X)$ for the underlying set of an object $X$ in a concrete category $\cal C$ and say $\cal C$ has standard products, if for any two objects $X$ and $Y$, there is an object $X \times Y$ of $\cal C$, such that (i) $U(X \times Y)$ is the set-theoretic product $U(X)\times U(Y)$, (ii) the set-theoretic projections of $U(X \times Y) = U(X) \times U(Y)$ onto its factors are $\cal C$-morphisms and (iii) if $f : Z \to X$ and $g : Z \to Y$ are $\cal C$-morphisms, then $\langle f, g\rangle : U(Z) \to U(X \times Y)$ is a $\cal C$-morphhism.

Many concrete categories have standard products, e.g., topological spaces, groups (or any other variety of algebras). Many concrete categories don't have products at all, e.g., fields, integral domains. I would like a natural example of a concrete category that has products but not standard products. The best I have come up with is the opposite of a concrete category like groups whose coproducts are not products. But that means I have to explain to my readers the device that makes the opposite of a concrete category into a concrete category. I would be very grateful for a nicer example, if anyone has one.

• Remark: Having standard products means that products exist and are preserved by the forgetful functor. And it seems that you want to restrict to binary products. Jun 19, 2015 at 17:30
• @egreg: this is incorrect. The forgetful functor from pointed sets to sets is representable (by the two-element set) so it preserves all limits. Jun 19, 2015 at 17:49
• @egreg: this is not the categorical product (check the universal property yourself). It is the left adjoint to the internal hom. But these don't agree. Jun 19, 2015 at 17:55
• This seems (to some extent) related: Why is the cartesian product so categorically robust? Jun 24, 2015 at 12:19
• @MartinSleziak: I am not sure exactly what you have in mind, but a possible answer is that it is because any class of structures that can be axiomatised by Horn clauses has standard products and that covers a lot of important examples. Jun 24, 2015 at 15:58

As Martin says in the comments, one way to rephrase the question is that you want an example of a concrete category $(C, U)$ where $C$ has binary products but the forgetful functor $U : C \to \text{Set}$ does not preserve them. In the most familiar examples $U$ not only preserves products but all limits because it has a left adjoint / is representable, so you need to avoid this.

A simple example, although it may be too simple, is the category of subsets of a set $S$ (with morphisms given by inclusions). Here the categorical product is given by intersection, so the forgetful functor fails pretty badly to preserve products.

An example that shows up in practice, although I'm not sure how to verify it, is the category of coalgebras. I think products in this category are quite complicated; in fact I don't know how to describe them explicitly.

• There are also artificial examples you can come up with by making strange choices of $U$, of course. Jun 19, 2015 at 17:58
• The product of two cocommutative coalgebras is just their tensor product with the evident coalgebra structure. I agree that the product of two arbitrary coalgebras is more complicted. Jun 20, 2015 at 9:00
• Thanks. You get the tick because, rather than being too simple, your example is ideal for the target audience of my paper. Jun 20, 2015 at 9:00

Here is a method to produce examples. Two specific examples will follow.

Let $$\mathcal{A} \to \mathsf{Set}$$ be a "standard" concrete category. For example, you may imagine an algebraic category. In particular, I want that $$\mathcal{A}$$ is complete. Let $$\mathcal{C} \subseteq \mathcal{A}$$ be a coreflective full subcategory. Then it is well-known that $$\mathcal{C}$$ is complete, too, and that the limits are computed via the coreflector applied to the limits in $$\mathcal{A}$$. In particular, usually the inclusion $$\mathcal{C} \to \mathcal{A}$$ is not continuous, and usually even the composition $$\mathcal{C} \to \mathcal{A} \to \mathsf{Set}$$ is not continuous.

1. Example. Let $$\mathsf{Grp}_*$$ be the category of pointed groups, i.e. pairs $$(G,g)$$ consisting of a group $$G$$ and an element $$g$$ of the underlying set of $$G$$. The morphisms should be clear. Let $$\mathsf{CycGrp}_*$$ be the full subcategory consisting of those $$(G,g)$$ such that $$G=\langle g \rangle$$. This is a coreflective subcategory, the reflector sends $$(G,g)$$ to $$(\langle g \rangle,g)$$. It follows that $$\mathsf{CycGrp}_*$$ is complete. For example, the product of $$(G,g)$$ with $$(H,h)$$ is $$(\langle (g,h) \rangle,(g,h))$$. You see that the forgetful functor $$\mathsf{CycGrp}_* \to \mathsf{Set}$$ doesn't preserve binary products.

2. Example. The category $$\mathsf{PrRing}$$ of prime rings (i.e. those rings $$R$$ for which the unique homomorphism $$\mathbb{Z} \to R$$ is surjective) is a coreflective subcategory of $$\mathsf{Ring}$$. The reflector sends $$R$$ to the image of $$\mathbb{Z} \to R$$, i.e. the prime subring of $$R$$. It follows that $$\mathsf{PrRing}$$ is complete. For example, $$\mathbb{Z} \times \mathbb{Z} = \mathbb{Z}$$ in that category. Hence, the forgetful functor $$\mathsf{PrRing} \to \mathsf{Set}$$ doesn't preserve binary products.

• @RobArthan: I've expaned my answer. Jun 24, 2015 at 21:54

Another way to get examples is to use non-standard forgetful functors. For example, $\mathsf{Set}$ is concrete over itself via the functor $X \mapsto X \amalg X$. But this functor doesn't preserve finite products.

• You can indeed get examples that way, as my definition of standard depends on the details of the forgetful functors. In the context of my paper I was after an example where the inutitive choice of forgetful functor fails to give standard products. Jun 24, 2015 at 15:55

A possibly interesting example is the category of torsion abelian groups.

In this example, finite products are standard, but infinite ones need not be: the product of an arbitrary (small) family of abelian groups is the torsion subgroup of its usual product as abelian groups.

Then, for example, the canonical map

$$\left| \prod_{n = 2}^{\infty} \mathbb{Z} / n \mathbb{Z} \right| \to \prod_{n = 2}^{\infty} \left| \mathbb{Z} / n \mathbb{Z} \right|$$

is not surjective: it's missing elements like $(1, 1, 1, \ldots)$ which must be in the codomain, but can't be in the image because its order in each place is unbounded.

• Thanks: my question was only about finite products, but your example is very nice to know. Jun 2, 2017 at 22:17