Categories that admit all products but not all coproducts

Question

What are examples for categories that admit all products but not all coproducts.

Answer 1

Now that the question was changed to existence of finite (co)products, here is a well-known example: the category $\mathbf{Gp}_{\text{fin}}$ of finite groups has finite limits, but not finite coproducts. The problem is that the coproduct in $\mathbf{Gp}$ of two finite groups need not be finite; e.g. $C_2 \ast C_3 \cong \operatorname{PSL}_2(\mathbf Z)$. The coproduct of $C_2$ and $C_3$ in finite groups cannot exist, as $\operatorname{PSL}_2(\mathbf Z)$ does not have a largest finite quotient.

Answer 2

Here is an example appearing 'in nature': the category $\mathbf{Sch}$ of schemes has arbitrary (small) coproducts (disjoint unions), but existence of (small) products is a subtle question (on the other hand, finite limits do exist, whereas finite colimits are subtle). For instance, the product $(\mathbf P^1_{\mathbf Z})^{\mathbf N}$ does not exist in $\mathbf{Sch}$; see [Stacks, Tag 078E].

As noted before, to get an example where products exist but coproducts do not, take the opposite category $\mathbf{Sch}^{\text{op}}$.

Answer 3

• Jeremy Rickard gives a nice example of a (locally small) category with all small colimits but not all small limits in an answer to Cocomplete but not complete abelian category. His category is even abelian (and in particular has finite limits), so the failure to have all small limits must come down to a failure to have all small products. You can dualize his example.

To underscore Max New’s point, there is some pressure in category theory for a category with all small limits to have all small colimits and vice versa. For instance, as Max New points out, this is the case when the category is small (and hence is a poset). It’s more generally true that if $\mathcal C$ is accessible (which includes most “categories of structures”), then it has all small limits if it has all small colimits (in which case the category is called locally presentable). So in some sense, you should expect the examples here to all be in some sense “artificial”.

On the other hand, I just remembered a good source of accessible categories with all products — if you take a locally presentable category $\mathcal K$ and then pass to a full subcategory $\mathcal L$ defined by some injectivity conditions (i.e. you have a set $I$ of morphisms $(A_i \to B_i)_{i \in I}$ such that $X \in \mathcal L$ iff for all $i$ and all maps $A_i \to X$, there exists a map $B_i \to X$ making the obvious triangle commute), then $\mathcal K$ has all products, but not all coproducts in general.

• For example, there are rings $R$ such that the category of injective $R$-modules (which always has all products) is not closed under coproducts. I don’t remember an example of this example offhand though….

• Even better — the category of free abelian groups has all small coproducts but not all small products. Similarly the category of free (nonabelian) groups has all small coproducts but not all small products. Dualize this.

• Actually — that’s true for almost any “category of algebraic structures” — the full subcategory of free algebras has all small coproducts, but almost never has all small products. Dualize this. (I suppose this is an instance of Alex Kruckman’s example where you take all coproducts of copies of some fixed object — here the free algebra on one generator.)

• For example, the category of extremeally disconnected spaces has all coproducts but not all products.

Answer 4

If $X$ is a large set and $\mathsf{Set}$ is the category of small sets, then $\mathsf{Set}/X$ is small-cocomplete, but it fails to have a terminal object. Now dualize this.

Answer 5

For the version of the question about finite products and coproducts:

• Any meet-semilattice that does not have joins. For example, a linear order with no least element (which has no empty join).

• Take any object $A$ in a category with all finite products and consider the full subcategory whose objects are all finite powers of $A$. This will usually fail to have coproducts. For example, the category of all finite sets whose cardinalities are powers of $2$.

For the version of the question about small products and coproducts:

• The second example generalizes easily to arbitrary powers. For example, the category of all sets of cardinality $2^\kappa$ for arbitrary cardinals $\kappa$.

• I originally claimed that the first example above generalizes easily as well, to complete meet-semilattices, but Max New correctly points out in the comments that every complete meet-semilattice is a lattice. However, as Tim Campion also correctly points out in the comments, a large semilattice may have all small meets without having all small joins. For example, the dual of the join-semilattice of all small subsets of a large set (which has no bottom element, hence no empty join).

Yet another naturalish example is the category of non-empty sets, which has all products but no initial object, hence no empty coproduct.

Answer 6

As the question is stated, there are actually no examples, at least assuming classical logic.

1. Classically, every small complete category is thin, i.e., a preorder with arbitrary meets. (https://ncatlab.org/nlab/show/complete+small+category#in_classical_logic)
2. By the adjoint functor theorem, any preorder with arbitrary meets also has arbitrary joins. (https://ncatlab.org/nlab/show/suplattice).

More concretely, if you have all meets, you can define an arbitrary join of a set $S$ to be the meet of all the upper bounds of the set $\bigvee S = \bigwedge \{y\in X | \forall s \in S. y \geq s \}$.

Of course if you change "all" to mean something more reasonable like "finite" or indexed by some other small cardinal, you can get many examples.

Alternatively you can look at constructive mathematics, where you can add as an axiom that such a non-trivial small complete category exists. This is modeled by e.g., realizability toposes (https://ncatlab.org/nlab/show/realizability+topos).

Answer 7

Opposites of Kleisli categories.

If C is a category with coproducts, and T a monad on C, then it is easy to see that the Kleisli category Kl(T) will inherit the coproducts from C. On the other hand, Kleisli categories generally don't have products. Try your favorite monads on Set, e.g. lists, probability distributions. The crucial idea is that products of free algebras aren't generally free. Now dualize this construction to answer your question.

Why should you care about opposites of Kleisli categories? This seems rather contrived after all. Here are some ideas:

• Lawvere theories are essentially opposites of Kleisli categories (restricted to finite sets). This fits the context of this question well: a Lawvere theory is freely generated under products from a single sort $1$. We should generally not expect this to have coproducts.

• Opposites of Kleisli categories of continuation monads are prime examples of Selinger's control categories