Let $U\colon \textbf{AbGp}\longrightarrow \textbf{Set}$ be the forgetful functor. By the Crude Monadicity Theorem, it is monadic.

Any reflection is monadic, so $I \colon \textbf{tfAbGp}\longrightarrow \textbf{AbGp}$ is also monadic.

I am trying to show that its composite is not monadic. The left adjoint of $U$ is the free functor and the explanation that I have found is the following: Free abelian groups are torsion free and so the monad on $\textbf{Set}$ induced by the composition is isomorphic so that induced by $F$ and $U$.

However, I do not understand why that contradicts the fact of being monadic. The comparison functor in the first case is $K\colon \textbf{AbGp}\longrightarrow \mathcal{C}^{\mathbb{T}}$ sending $B$ to $(UB, U\epsilon_{B})$, so it is faithful, full and essentially surjective (because the forgetful functor is monadic).

In the second case the comparison functor would have domain $\textbf{tfAbGp}$, so I have been thinking why can't that be faithful, full or essentially surjective. Nevertheless, I do not find any example.

Can anyone help me, please?

  • $\begingroup$ is $U$ going the wrong direction? $\endgroup$ – Andres Mejia Apr 25 '18 at 20:41
  • $\begingroup$ Sorry, it is going the opposite direction. Thank you very much :) $\endgroup$ – Laura Apr 25 '18 at 20:44

The point is that the left adjoint to the forgetful functor $$U'=U \circ I \colon \mathbf{tfAbGp} \to \mathbf{Set}$$ is basically the free abelian group functor, i.e. the left adjoint to $U$ with the codomain restricted to $\mathbf{tfAbGp}$.

This implies that the monad associated to $U'$ is the same monad associated to $U$, i.e. the free abelian group-monad, and the comparison functor from $\mathbf{tfAbGp}$ sends the torsion free groups in torsion free groups, i.e. it is not essentially surjective and so it cannot be an equivalence.


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