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Torsion-free abelian groups give a conservative right adjoint that is not monadic

Statement refuted

The assertion that every conservative right adjoint is monadic is false. Torsion-free abelian groups give a conservative right adjoint that is not monadic.

Facts & Assumptions

Given: The category $T F A b$ of torsion-free abelian groups and group homomorphisms, with underlying-set functor $U : T F A b \to S e t$ .

[L1]

The full subcategory of torsion-free abelian groups is reflective in $A b$ (Torsion-free abelian groups form a reflective full subcategory of abelian groups).

[L2]

For $R = Z$ , the Eilenberg–Moore category of the free-module monad is isomorphic over $S e t$ to the category of abelian groups (For a unital ring R, the free-R-module monad on sets has left R-modules as its Eilenberg–Moore algebras).

[L3]

A $Z$ -module is torsion-free when no nonzero integer annihilates a nonzero element (Annihilators, torsion elements and the torsion subset of a module).

[L4]

The quotient group $(Z, +) / 2 Z$ is $(Z / 2, +)$ on the same underlying congruence classes (For every $n \in N$ , the congruence-class group $(Z / n, +)$ is the quotient group $(Z, +) / n Z$ ).

Counterexample

technique · direct

1.1

Free abelian groups are torsion-free, so the usual free-abelian-group functor lands in $T F A b$ and is left adjoint to $U$ . Equivalently, this adjunction is obtained by combining the free-abelian adjunction with the reflective inclusion in [L1].

L1construct

1.2

A bijective homomorphism of torsion-free abelian groups has an inverse that preserves addition, so it is an isomorphism. Hence $U$ reflects isomorphisms and is conservative.

givenalgebra

2.1

The induced monad is the usual free-abelian-group monad: applying the torsion-free reflector to a free abelian group changes nothing. By [L2], its full Eilenberg–Moore category is $A b$ over $S e t$ .

step 1.1L2

3.1

The comparison from $T F A b$ to $A b$ is the inclusion and misses the group $Z / 2 Z$ in [L4]. The nonzero class of $1$ is killed by the nonzero integer $2$ , so this group is not torsion-free by [L3].

step 2.1L3L4construct

4.1

Thus the comparison is not essentially surjective and is not an equivalence, so $U$ is not monadic, while step 1.2 shows it is conservative.

step 1.2step 3.1∎

Depends on

Used by

FALSE: every conservative right adjoint is monadic False statement

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. Mehrle, Category Theory Part III, Example 5.20(d) (standard reference, not scraped)
E. Riehl, Category Theory in Context, 2nd ed., monadicity examples in Section 5.5 (standard reference, not scraped)