= Solution
The <crude monadicity theorem> in its reflexive-coequalizer form says: if $F\dashv G:\mathcal D\to\mathcal C$, $G$ is a <conservative functor>, $\mathcal D$ has <reflexive coequalizers>, and $G$ preserves them, then the <Eilenberg-Moore comparison functor> $K:\mathcal D\to\mathcal C^T$, for $T=GF$, is an <equivalence of categories>. Requiring all <coequalizers> to exist and be preserved is a stronger sufficient form.
For an algebra $(A,a)$, form in $\mathcal D$ the <coequalizer>
$$
FTA\mathrel{\substack{\xrightarrow{Fa}\\\xrightarrow[\varepsilon_{FA}]{} }}FA\xrightarrow{q}L(A,a).
$$
The pair is a <reflexive pair> with common section $F\eta_A$: both composites are the identity by the algebra unit law and the triangle identity. An arrow $u:FA\to B$ transposes to $v=Gu\eta_A:A\to GB$. The two composites $uFa$ and $u\varepsilon_{FA}$ transpose to $va$ and $G\varepsilon_BTv$, respectively. Indeed the latter transpose is $Gu\mu_A\eta_{TA}=Gu$, while $G\varepsilon_BTv=Gu\mu_AT\eta_A=Gu$ by counit naturality. Thus $u$ equalizes the pair exactly when $v$ is an <monad algebra morphism> $(A,a)\to K(B)$. The <coequalizer> property gives natural <bijections>
$$
\mathcal D(L(A,a),B)\cong\mathcal C^T((A,a),K(B)).
$$
These define $L$ on arrows by uniqueness, so $L\dashv K$. Its unit has underlying arrow $\beta=Gq\eta_A:A\to GL(A,a)$.
By preservation, $Gq$ coequalizes $Ta$ and $\mu_A$ in $\mathcal C$. The action $a:TA\to A$ is a <split coequalizer> of that pair: take $s=\eta_A$ and $t=\eta_{TA}$, with $\mu_At=1_{TA}$ and $Ta\,t=\eta_Aa$. Thus there is a unique <isomorphism> $\alpha:GL(A,a)\to A$ with $\alpha Gq=a$. We have $\alpha\beta=a\eta_A=1_A$. Also
$$
Gq=Gq\mu_A\eta_{TA}=GqTa\eta_{TA}=Gq\eta_Aa=\beta a.
$$
Consequently $\beta\alpha Gq=Gq$, and epimorphic cancellation gives $\beta\alpha=1$. The adjunction unit $\beta$ is an <monad algebra morphism>; its invertible underlying arrow has an algebra-morphism inverse. Thus the unit of $L\dashv K$ is an <isomorphism>.
For its counit $c_B:LK(B)\to B$, the triangle identity gives $K(c_B)\beta_{K(B)}=1_{K(B)}$. Hence $K(c_B)$, and therefore $G(c_B)$, is an <isomorphism>. Since $G$ is a <conservative functor>, $c_B$ is an <isomorphism> too. Both unit and counit of $L\dashv K$ are invertible, which proves the claimed equivalence. This proof uses only reflexive <coequalizers> in $\mathcal D$ and split <coequalizers> in $\mathcal C$.
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