Composition of monadic functors isn't monadic

Question

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.

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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.

Answer 1

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).