# Category where not every mono is regular

I'm looking for a category with the titular property. Since $$f : A \to B$$ is supposed to be mono, the desired factorisation is automatically unique (correct?). Thus, I would need to find a category in which the factorisation does not exist, i.e., there are no arrows ending at $$A$$. Therefore, I went to look at preorders, where there is sometimes no arrow ending in $$A$$. However, unwinding the definitions, I came up empty again because the fact that parallel arrows $$k, l : B \to C$$, say) are equal here, makes $$f$$ into an isomorphism, giving an arrow back that does let $$f$$ fit into an equaliser diagram.

Another category I thought of (which I know to be somewhat pathological) is Field, where all morphisms are injective and thus mono (correct?). Since fields are also rings, where oftentimes morphisms to an object don't exists, I thought this might be a nice example. However, the condition I need to prove is quite strong. Sure, I can cook up some monos that don't fit into certain equaliser diagrams, but to be regular, they should fit in no equaliser diagrams.

I have also tried some 'toy' categories with just a few objects and arrows, but then many things are vacuously true, and, again, I come up empty.

Is there anything in my attempts above that I could pursue further? Or do I really need a very exotic category that I won't come up with myself? (Admission: since in Cat, the morphisms are functors, and there are so many things to check, I haven't spent much time on that. Let alone functor categories $$[\mathcal{C}, \mathcal{D}]$$, where the natural transformations are the morphisms.)

I do notice that in this sort of quest, one learns many things, and finds that one does not know the answer to many seemingly simple questions (e.g., what are the monos in preorders, what are the (relevant) roles of initial and terminal objects here, can finite categories work at all, etc.? Any additional (if tangential) information on these questions would be greatly appreciated! =D)

EDIT

LOL, I look two exercises ahead: "characterise the regular monos in Pos". Clearly, the category of posets should be an example (or this would be a very trivial exercise..) So that's what I'm going to focus on now!

## 5 Answers

One example is the category of topological spaces. The regular monomorphisms are subspaces, while subobjects can have a topology that is finer than the subspace topology. So concretely, the “identity map” $$2 \to 2$$, where the domain has the discrete topology and the codomain has the indiscrete topology, is a monomorphism which is not regular.

• Thanks! By 'the "identity map" $2 → 2$', do you mean the map from a topological space to itself where $|2| = 2$, i.e., one with just two points? (Also, I'm not quite sure what a 'subobject' means exactly..) Commented Jan 22, 2023 at 6:41
• @JosvanNieuwman Fix a 2-element set, and call it $2$. I am abusing notation a little bit here, as the domain and codomain are different topological spaces on the underlying set $2$. A subobject of an object $C$ is a pair $(B, g)$, where $g : B \to C$ is mono, typically considered up to isomorphism. A monomorphism $g : B \to C$ in the category of topological spaces is always, up to isomorphism, the canonical map induced by the inclusion $B \subseteq C$, but $B$ may have a topology that is not the subset topology. Commented Jan 22, 2023 at 6:46
• Thanks! But oooh wait, I think I made a grave mistake at the beginning. I need a category with a mono that is not regular. That is, for any (candidate) equaliser diagram, it must not be an equaliser diagram. But that means that it is sufficient that for some object $C$ and $c : C → B$, there is no unique factorisation $h : C → A$ s.t. $c = f ∘ h$, correct? Because in that case, I have already found many examples indeed! (I kept trying to prove that for any $C, c$, there should be a unique factorisation, but that would make it an equaliser diagram! Which is what I didn't want! Commented Jan 22, 2023 at 7:00

I think I interpreted this whole thing wrongly from the start. $$f : A → B$$ being not regular means that there is no equaliser diagram in which it fits. That is, given a candidate equaliser diagram, there must be some object and arrow from it to $$B$$, that does not allow for a factorisation of $$f$$.

Given the error I made in my thinking, I think I may have a simple example already. Consider the preordered set $$A = \{Z, Y, X, W\}$$ with $$Z ≤ X ≤ W$$ and $$Y ≤ X ≤ W$$, considered as a category. That is, there are arrows $$f : Z → X$$, $$g : Y → X$$, $$k : X → W$$, but no arrow $$Y → Z$$. At first I was happy, because the 'parallel arrows' $$k, k : X → W$$ made the equalising quite trivial :p, and for $$g : Y → X$$, even though $$k ∘ g = k ∘ g$$, there is still no arrow $$Y → Z$$. Then I got sad again, because I realised that for $$f : Z → Z$$, there certainly was a (unique) arrow (id$$_{Z}$$) making the triangle commute: $$f = f ∘ \text{id}_Z$$. So it was not true that for every map to $$X$$, there did not exist a factorisation of $$f$$. But that only means that this is not an equaliser! Which was exactly what I was trying to prove: there is a mono $$f : Z → X$$ s.t. for all equaliser diagram candidates (the only ones here being $$Z → X → W$$ and $$Y → X → W$$, the conditions $$k ∘ f = k ∘ f$$ and $$k ∘ g = k ∘ g$$, did not give an arrow $$Y → Z$$ (at all, let alone with the property of making the triangle commute)!

So if I haven't scrambled a negation in there somewhere, I think I had fixed it all along!

Checking that a monomorphism is not regular is a bit awkward; it's easier to instead prove that in any category with pullbacks, a monomorphism is regular iff it's effective. Then to check that a monomorphism is not effective you only have to compute its cokernel pair and then compute the equalizer of the cokernel pair. You can see some more details in this blog post.

It also happens that in practice it's easier to find categories in which epimorphisms are not effective; once we've done that we can take opposite categories. For example, to check whether an epimorphism $$f : A \to B$$ of commutative rings is effective we have to compute the coequalizer of its kernel pair, which turns out to be the quotient $$A/\text{ker}(f)$$ of $$f$$ by its kernel in the usual sense. So the effective epimorphisms of commutative rings are exactly the surjective homomorphisms, by the isomorphism theorems. But there are non-surjective epimorphisms; for example any localization. So any such epimorphism, such as the map $$\mathbb{Z} \to \mathbb{Q}$$, is a non-effective epimorphism in $$\text{CRing}$$, and so its opposite

$$\text{Spec } \mathbb{Q} \to \text{Spec } \mathbb{Z}$$

is a non-effective monomorphism in the opposite category $$\text{Aff}$$, the category of affine schemes.

One example is the category of sets with a binary relation, i.e pairs $$(V,E)$$ with $$E\subseteq V\times V$$, where a morphism from $$(V,E)$$ to $$(V',E')$$ is a map $$f\colon V\to V'$$ satisfying $$(v,w)\in E \Longrightarrow (f(v),f(w))\in E'$$ for all $$v,w\in V$$.

These are the same as directed graphs without parallel arcs. Posets are a special case of this, where the relation is relexive, symmetric and transitive.

For two morphisms $$f,g\colon (V,E)\to (V',E')$$, their equalizer can be computed as the inclusion of $$(W,E\cap (W\times W))$$ into $$(V,E)$$ where $$W=\{v\in V\mid f(v)=g(v)\}$$. Such subgraphs are "induced" (=full) in the sense that they already include any possible edge from $$E$$ between two of their vertices.

So any noninduced subgraph provides an example of a mono that cannot be an equalizer.

Consider the category generated by the following graph:

Just to make sure that we are on the same page, this category has 3 objects and 8 non-identity morphisms, namely $$p, q, f, g, fp, fq, gp, gq$$.

In this category, it is easy to verify that $$p$$ is not an equalizer, but it is vacuously a mono.