# A category which direct limits but no general colimits

I am looking for a (at best, real life) category that has direct limits, but no general small colimits, or a category that has inverse limits, but no general small limits. Are there any interesting examples that are not too obviously made to be an example for this?

I am asking this because I wonder why many lectures include the construction of direct/inverse limits as an exercise, instead of general small (co)limits. In particular, I wonder this for topology and algebra classes.

Consider the category with two objects and only identity arrows. Or more generally, any poset which has least upper bounds for all chains, but not arbitrary joins (like the disjoint union of two copies of $$\mathbb{R}\cup\{\infty\}$$).

If you insist that these are not "real life" categories, you might be more satisfied with the example of the category of fields, which has directed colimits but does not have coproducts or an initial object.

Consider any nontrivial group as a 1-object category. Then it has all filtered (co)limits (exercise: if all the morphisms in a filtered diagram are isomorphisms, then any object in the diagram is a (co)limit by taking an appropriate composition of the isomorphisms and their inverses). However, it does not have a (co)equalizer of any two distinct morphisms, or a (co)product of any number of copies of the unique object besides 1.

This is not really an answer, as I don’t know examples, but I think I might have a reason for why they are considering them separately.

Filtered colimits (I always get confused over directed / inverse) are particularly nice in concrete categories like $$\mathsf{Set}, \mathsf{Ab}, \mathsf{Mod}_R, \mathsf{Top}$$ and alike. There is an explicit formula for computing and dealing with them in $$\mathsf{Set}$$, which lifts to similar formulas in other concrete categories. From this formula one can deduce for example that filtered colimits commute with finite products (only for good categories!), which does not hold for arbitrary colimits! They may have even more special properties.

Long story short, often we are not interested in dealing with arbitrary shapes of colimits but only want to work with nice ones like coproducts, quotients, pushouts, gluing constructions or filtered colimits, of which we might know more than just „they are colimits“.

Part of the reason might be as well that most people don’t want to be bothered with abstract nonsense, but rather like to work with these things implicitly...

1- Consider a first order functional (meaning it only has function symbols) language $$L$$, and a first order theory $$T$$.

If $$T$$ is especially simple, e.g. it's an equational theory, then the category of models of $$T$$ has all colimits, it's even presentable.

If we make $$T$$ more complicated, it might not have all colimits; but if we keep it simple enough, it'll still have filtered colimits. Alex Kruckman's example of fields is a particular example of this situation, where the "extra" axiom (on top of the equational ones) is (for instance) $$\forall x, \exists y,( xy = 1 \lor x= 0)$$

Now let's take this example and generalize it to find other examples. Suppose you have formula $$\varphi(x_1,...,x_n,y_1,...,y_m)$$ built from atomic formulas using only $$\land, \lor$$. Then the models of $$\forall x_1,...,\forall x_n, \exists y_1,...\exists y_m, \varphi(x_1,...,y_m)$$ has all filtered colimits, and they're computed as in sets. This is quite easy to prove: just prove by structural induction on the formula that it is satisfied in the filtered colimit structure for a particular tuple $$(a_1,...,a_n,b_1,...,b_m)$$ if and only if it is at some finite stage.

But they might not have arbitrary colimits. For instance one can encode a cardinality bound with such a formula ($$\forall x_1,...,x_n, \bigvee_{i,j}x_i= x_j$$ encodes a bound of cardinality $$). So "groups of cardinal $$\leq n$$" for a fixed $$n$$ are an instance which of course doesn't have all colimits.

Another somewhat less silly example is the category of groups of exponent "either $$2$$ or $$3$$", which you can encode with the formula $$\forall x, \forall y (x^2 =1 \lor y^3 = 1)$$. Now this doesn't have coproducts (can you figure out why ?).

2- Another example I like is an example which relates the question of having (co)limits and preserving (co)limits for a functor. There are nice examples of functors that commute with filtered colimits but not all colimits (e.g. taking fixed points of a $$G$$-action in $$G$$-sets, or taking global sections of a sheaf, etc.). I claim that these examples provide examples of categories which don't have all colimits, but that have filtered ones.

Indeed let $$F:C\to D$$ be a functor, and let $$E$$ be the category whose objects are $$Ob(C)\coprod Ob(D)$$, where an arrow between objects of $$C$$ is an arrow in $$C$$, an arrow between objects in $$D$$ is an arrow in $$D$$, and an arrow $$c\to d$$ is an arrow $$F(c)\to d$$ (and there are no arrows from an object in $$D$$ to an object in $$C$$).

Let $$K: I\to C$$ be a diagram, and assume it has a colimit in $$C$$. Then $$F$$ preserves that colimit if and only if the obvious diagram $$I\to C\to E$$ has a colimit. Indeed, let $$Q, j_i: K(i)\to Q$$ be a colimit for $$D$$ in $$C$$; and assume $$F(Q),F(j_i)$$ forms a colimit in $$D$$. Then $$Q, j_i$$ forms a colimit in $$E$$. Indeed the universal property is obviously satisfied for objects of $$C$$, and for objects of $$D$$ by preservation.

Conversely, assume $$I\to C\to E$$ has a colimit. Then, since it maps to the objects in the diagram,it must be in $$C$$. Then by analyzing the universal property, one can easily see that it has to be a colimit in $$C$$, and that it must be preserved by $$F$$.

In particular, suppose $$C,D$$ are cocomplete, then $$F:C\to D$$ preserves (filtered) colimits if and only if $$E$$ has them. So if $$F$$ preserves filtered colimits but not general ones, as in the examples I mentioned (let me mention some others: the functor taking a category to its core groupoid, the functor taking a ring $$R$$ to its $$K_0$$, most forgetful functors from algebra to sets,...), $$E$$ has filtered colimits, but not general ones.

As Kevin Arlin points out in the comment below, it's extremely frequent for right adjoints between presentable categories to preserve filtered colimits (it's equivalent to the left adjoint preserving compact objects), but much less frequent for them to preserve arbitrary colimits.

• Cool answer! It might be worth abstracting the examples of functors preserving filtered colimits: this is a thing right adjoint do, when their left adjoints preserve finitely presentable objects, which is exceedingly common. Commented Nov 12, 2020 at 17:58
• @KevinArlin : good point, I added a few words ! Commented Nov 12, 2020 at 18:07