Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 26 / 4 / i

Assume both unit and counit are monic. The unit criterion makes $F$ faithful. The triangle identity is ${\epsilon }_{FA}F{\eta }_A=1_{FA}$. Since ${\epsilon }_{FA}$ is monic, this split epimorphism is an isomorphism: cancelling it in ${\epsilon }_{FA}(F{\eta }_A{\epsilon }_{FA})={\epsilon }_{FA}$ gives $F{\eta }_A{\epsilon }_{FA}=1$. Thus $F{\eta }_A$ is invertible. As $\mathcal{C}$ is balanced, the reflection result in the root solution makes ${\eta }_A$ invertible. Hence $F$ is full and faithful.

Let $q:FA\to B$ be a regular epimorphism. Its adjunct factorization is $q={\epsilon }_{B}Fh$ for $h:A\to GB$. A regular epimorphism is extremal: if a coequalizer $q$ factors through a monomorphism $m$, its other factor also coequalizes the defining pair, descends through $q$, and gives a right inverse for $m$; monicity then makes $m$ invertible. Apply this to the monic ${\epsilon }_B$. More explicitly, if $q$ coequalizes $r,s$, monicity gives $Fhr=Fhs$, so there is $d:B\to FGB$ with $dq=Fh$. Then ${\epsilon }_{B}dq=q$ implies ${\epsilon }_{B}d=1$, and monicity gives $d{\epsilon }_B=1$. Therefore $B\cong FGB$. The image is closed under regular quotients, proving $\boxed{\text{(i)}\Rightarrow \text{(ii)}}$.