Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 119 / 4

For an adjunction $F⊣U:\mathcal{D}\to \mathcal{C}$ with induced monad $T=UF$, the Eilenberg-Moore comparison functor isThe adjunction is monadic when $K$ is an equivalence.

The Crude monadicity theorem states that a right adjoint is monadic if it reflects isomorphisms, its source has coequalizers of reflexive pairs, and it preserves those coequalizers. The dual statement is the crude comonadicity theorem.

Assume the free functor reflects isomorphisms. If $f,g:X\rightrightarrows Y$ satisfy $Ff=Fg$, let $q:Y\to Q$ be their coequalizer in sets. As a left adjoint, $F$ preserves it. Since the coequalizer of an equal pair is an identity, $Fq$ is an isomorphism. Reflection makes $q$ an isomorphism, so $f=g$. Thus $F$ is faithful.

Assume $F$ is faithful. If ${\eta }_X(x)={\eta }_X(y)$, regard $x,y$ as maps $1\rightrightarrows X$. Their free extensions are $F(x),F(y):F1\rightrightarrows FX$, and the adjunction identifies these with ${\eta }_{X}x,{\eta }_{X}y$. They are equal, so faithfulness gives $x=y$. Hence every unit component ${\eta }_X:X\to TX$ is monic.

Pointwise monicity of the unit immediately implies thatis monic.

If ${\eta }_2$ is monic, then ${\eta }_2(0)\ne {\eta }_2(1)$. The free algebra $(T2,{\mu }_2)$ therefore has more than one element, proving the existence of a nontrivial $T$-algebra.

Let $(A,a)$ be a $T$-algebra with at least two elements, and suppose $Ff:FX\to FY$ is an isomorphism. For every $T$-algebra $A$, precomposition with $Ff$ gives a bijectionBy the free-forgetful adjunction this isFor one set $UA$ with at least two elements, bijectivity of this precomposition forces $f$ to be bijective: surjectivity detects injectivity of $f$, and injectivity detects surjectivity. Thus $f$ is an isomorphism. The free functor reflects isomorphisms, completing the cycle of equivalences.

Assume the equivalent conditions and that $T$ preserves finite coproducts. The free functor already reflects isomorphisms. By the dual Crude monadicity theorem, it remains to preserve the relevant coreflexive equalizers.

A coreflexive equalizer diagramwith $A\ne \varnothing$ can be equipped with the extra sections making it a split equalizer: choose a point of $A$ and use the common retraction to define the missing splitting maps on the complementary fibres. Every functor preserves split equalizers. If $A=\varnothing$, preservation follows from preservation of the initial object, which follows from preservation of finite coproducts. Consequently $F$ preserves all required coreflexive equalizers, and the adjunction is comonadic.