About the characterization of categories of model of algebraic theories

Question

So, in his Handbook of categorical algebra Vol 2, Borceux states a theorem (the 3.9.1, page 158) that says that:

Given a category $C$ with a functor $U:C \to Set$, $C$ is the category of models of a (single sorted) algebraic theory with U the forgetful functor to set if and only if the following condition holds:

1. $C$ has coequalizer and Kerner pairs.
2. $U$ has a left adjoint F.
3. $U$ reflects isomorphism.
4. $U$ preserves regular epimorphisms
5. $U \circ F$ preserve filtered colimits.

I've read that theorem many years ago and always thought it was true (I had in mind something about having a projective generator of finite type, but that that's essentially the same) but, I tried proving it today because I was planning to talk about it in a lecture, and now a little confused by it...

Is this theorem actually correct? I think I have a counterexample and that condition 4. need to be replaced with the stronger condition: "4'. U preserves reflexive co-equalizers". I haven't seen any corrigendum or errata to the book nor any mention of this kind of mistake in the book.

Here is what I think is a counter-example:

Take $C$ to be the category of torsion free abelian groups. $C$ is a reflective subcategory of $\text{Ab}$ the category of abealian group as every abelalian group $G$ has a a Torsion free quotient $G/G^{tor}$. Moreover the unit of the reflection is a regular epimorphism in $\text{Ab}$.

It follows that $C$ has all limits and colimits. Colimits in $C$ are computed by taking the colimits in abelian group and then quotient out by the torsion elements that may have appeared in the colimit. In particular, the map from the colimit in $\text{Ab}$ to the colimit in $C$ is a regular epimorphism.

Take $U$ to be the forgetful functor to set, it has a left adjoint because free groups are torsion-free, condition 3. is immediate, condition 4. follows from the description of colimits, and condition 5 is also immediate (more generally, U preserve filtered colimits because torsion-free groups are closed under filtered colimits so they are just colimits in Ab).

But $U$ is not even monadic over set as the monad induced by this adjunction is just the ordinary free abelian group monad.

Am I missing something? (Obviously, if the counter-example is wrong that is a good answer, but if it justs look correct, comments saying so would be appreciated!).

Answer 1

To me, your reformulation using reflexive coequalizers is more natural anyway (this is basically the crude monadicity theorem). As an aside, algebraicity can include infinitary operations, so I personally would discard 5., but that's not important here.

You're certainly correct that the forgetful functor from torsionfree abelian groups to sets is not monadic. This is mentioned for example in Toposes, Triples and Theories by Barr and Wells. I just wanted to independently verify that 4. holds in your example. A regular epi $q: A \to B$ here is the same as a map that is the coequalizer of its kernel pair $p_1, p_2: K \rightrightarrows A$, and we compute the coequalizer by composing the coequalizer $Q$ of $(p_1, p_2)$ in abelian groups with its projection to the quotienting out by its torsion subgroup

$$A \twoheadrightarrow Q \twoheadrightarrow Q/\mathrm{Tors}(Q),$$

and sure, each of these is surjective, so the composite is also surjective, i.e., a (regular) epi when interpreted in $\mathbf{Set}$. So yes, it seems that your counterexample holds up.