= Solution
We prove <full faithfulness from counit coequalizers>. Let $B,C\in\mathcal D$ and let $\alpha:K(B)\to K(C)$ be an <monad algebra morphism>. Thus
$$
\alpha G\varepsilon_B=G\varepsilon_CGF\alpha.
$$
Set $u=\varepsilon_CF\alpha:FGB\to C$. Its composites with the two arrows of the printed presentation are equal: naturality of $\varepsilon$ gives
$$
u\,FG\varepsilon_B
=\varepsilon_CFG\varepsilon_CFGF\alpha
=\varepsilon_C\varepsilon_{FGC}FGF\alpha
=u\,\varepsilon_{FGB}.
$$
In the last equality we used naturality at $F\alpha$. The <coequalizer> property therefore gives a unique $v:B\to C$ satisfying $v\varepsilon_B=u$.
Apply $G$ to that identity. The algebra-morphism equation gives $Gv\,G\varepsilon_B=\alpha G\varepsilon_B$. The triangle identity makes $G\varepsilon_B$ a <split epimorphism> with section $\eta_{GB}$, so $Gv=\alpha$. This proves fullness of $K$.
If $v,w:B\to C$ have $Gv=Gw$, naturality gives $v\varepsilon_B=\varepsilon_CFGv=\varepsilon_CFGw=w\varepsilon_B$. The counit is epic since it is a <coequalizer>, so $v=w$. Thus $K$ is a <faithful functor>. Both parallel arrows matter: the converted TeX loses the second one, $\varepsilon_{FGB}$, which is visible in the original PDF.
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