Disclaimer: this question already has a solution here: Composition of monadic functors may not be monadic . However, I would like to understand how to solve this using another characterisation of monadic functors - Beck's monadicity theorem.

Let TFA denote the full subcategory of torsion-free abelian groups, A of abelian groups, and Set the usual category of sets.

Take it as given that the forgetful functor $$F:$$ A $$\to$$ Set and the inclusion functor $$i:$$ TFA $$\to$$ A are monadic.

Show that, however, the composition $$F \circ i:$$ TFA $$\to$$ Set, is not monadic, using Beck's monadicity theorem.

Recall that Beck's monadicity theorem states that a functor $$G: A \to B$$ is monadic if and only if $$G$$ has a left adjoint, it reflects isomorphisms, and every $$G$$ split pair admits a coequalizer in $$A$$, which $$G$$ preserves.

Where I'm at:

It is clear that $$F \circ i$$ has a left adjoint, and that it reflects isomorphisms, and so it is the last and final condition that must be false.

For it to be false, we must find a pair of group homomorphisms, between torsion-free abelian groups, $$G_1\overset{f}{\underset{g}\rightrightarrows}G_2$$, such that as functions on the respective sets, they admit a split fork:

there is $$C \in$$ Set, $$e: G_2 \rightleftarrows C:s$$ and $$t:G_2 \to G_1$$ s.t:

1. $$e \circ s = id_{C}$$

2. $$g \circ t = s \circ e$$

3. $$f \circ t = id_{G_2}$$

And that this set $$C$$ will not have the canonic structure of a torsion-free abelian group. (Or simply that $$G_1, G_2$$ have no coequalizer in TFA)

Note that, since A is cocomplete, this pair $$f,g$$ will admit a coequalizer (in A), so this set $$C$$ may have the structure of an abelian group.

So my question is, can you help find such an example? Or is my reasoning off?

I've tried several with $$\mathbb{Z}$$, and variants of it, but in order to get that $$C$$ is not torsion free, I need that $$f-g$$ will have a big enough image so that $$\mathbb{Z}/Im(f-g)$$ will be finite. I think this specific example is hopeless though.

Your reasoning is correct, and your idea to look for a case where the coequalizer in $$\mathbf{A}$$ is not torsion-free is the right one. Here's a simple way to obtain this : you know that the group $$\mathbb{Z}/n\mathbb{Z}$$ is the quotient of $$\mathbb{Z}$$ by the subgroup $$n\mathbb{Z}$$. This means that its underlying set is the set of equivalence classes of $$\mathbb{Z}$$ under the relation $$R$$ defined (as a subset of $$\mathbb{Z}\times \mathbb{Z}$$) by $$R=\{(x,y)\in \Bbb Z\times \Bbb Z \mid x-y\in n\mathbb{Z}\};$$ this set of equivalence classes is the coequalizer of the restrictions of the product projections to $$R$$, and it splits in $$\mathbf{Set}$$.
In fact $$R$$ is a subgroup of $$\mathbb{Z}\times \mathbb{Z}$$, and in particular it's a torsion-free group ! Moreover the restriction of the projections are group homomorphisms. So you have two group homomorphisms between torsion-free abelian groups, and their coequalizer in $$\mathbf{Set}$$ (or in $$\mathbf{A}$$) is $$\mathbb{Z}/n\mathbb{Z}$$, which is not torsion-free. As a consequence, the coequalizer of the two morphisms will not be preserved by the forgetful functor $$\mathbf{TFA}\to \mathbf{Set}$$ (but it does exist : in fact it is the trivial group).
• Thank you, I was hovering around this idea, but couldn't make it work. Taking $R$ itself as one of the groups is what makes this work easily, and what I didn't think of doing. – Mariah Jan 15 '19 at 14:25