= Solution
Assume both unit and counit are monic. The unit criterion makes $F$ faithful. The triangle identity is $\varepsilon_{FA}F\eta_A=1_{FA}$. Since $\varepsilon_{FA}$ is monic, this <split epimorphism> is an <isomorphism>: cancelling it in $\varepsilon_{FA}(F\eta_A\varepsilon_{FA})=\varepsilon_{FA}$ gives $F\eta_A\varepsilon_{FA}=1$. 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 $q=\varepsilon_BFh$ 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 $\varepsilon_B$. More explicitly, if $q$ coequalizes $r,s$, monicity gives $Fh r=Fh s$, so there is $d:B\to FGB$ with $dq=Fh$. Then $\varepsilon_Bdq=q$ implies $\varepsilon_Bd=1$, and monicity gives $d\varepsilon_B=1$. Therefore $B\cong FGB$. The image is closed under regular quotients, proving $\boxed{\text{(i)}\Rightarrow\text{(ii)}}$.
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