= Solution
A <monad> on a <category> $\mathcal C$ consists of an <endofunctor> $T$, a unit <natural transformation> $\eta:1_{\mathcal C}\Rightarrow T$, and a multiplication $\mu:T^2\Rightarrow T$, satisfying
$$
\mu\,T\mu=\mu\,\mu T,\qquad \mu\,T\eta=1_T=\mu\,\eta T.
$$
An <algebra for a monad> is $(A,a:TA\to A)$ with $a\eta_A=1_A$ and $aTa=a\mu_A$. A <morphism of algebras for a monad> $h:(A,a)\to(B,b)$ satisfies $ha=bTh$. These objects and arrows form the <Eilenberg-Moore category> $\mathcal C^T$.
For an <adjunction> $F:\mathcal C\rightleftarrows\mathcal D:U$ with $F\dashv U$, the <monad induced by an adjunction> is $T=UF$, with $\mu=U\varepsilon F$. The <Eilenberg-Moore comparison functor> is $K(D)=(UD,U\varepsilon_D)$ and $K(h)=Uh$. The <functor> $U$ is monadic when this comparison is an <equivalence of categories>; this is a <monadic adjunction>.
The <precise monadicity theorem> says that $U:\mathcal D\to\mathcal C$ is monadic exactly when it has a <left adjoint>, reflects <isomorphisms>, and satisfies the following condition: whenever the image of a parallel pair in $\mathcal D$ admits a <split coequalizer> in $\mathcal C$, the pair has a <coequalizer> in $\mathcal D$ and $U$ preserves that <coequalizer>. Equivalently, these <coequalizers> are created up to the canonical <isomorphism> identifying their underlying <coequalizer> objects. Here a <split coequalizer> of $f,g:A\rightrightarrows B$ consists of $q:B\to Q$, $s:Q\to B$, and $t:B\to A$ with
$$
qf=qg,\qquad qs=1_Q,\qquad ft=1_B,\qquad gt=sq.
$$
These equations prove its <universal property>: if $hf=hg$, then $h=hft=hgt=hsq$, and $hs$ is the unique factor through $q$. Every <functor> preserves these equations, so every <functor> preserves <split coequalizers>.
First consider the forgetful <functor> $U^T:\mathcal C^T\to\mathcal C$. It has the <free algebra functor> $A\mapsto(TA,\mu_A)$ as a <left adjoint>: the inverse transposition formulas are $h\mapsto h\eta_A$ and $v\mapsto bTv$. An <monad algebra morphism> whose underlying arrow is invertible has an algebra-morphism inverse, by rearranging its structure equation. Thus $U^T$ reflects <isomorphisms>.
Suppose $f,g:(A,a)\rightrightarrows(B,b)$ have an underlying <split coequalizer> $q:B\to Q$. Applying $T$ gives a <coequalizer> of $Tf,Tg$. The equation $qbTf=qfa=qga=qbTg$ therefore gives a unique $c:TQ\to Q$ with $cTq=qb$. The <monad algebra> laws follow by cancellation: composing $c\eta_Q=1_Q$ with the <epimorphism> $q$ reduces it to $b\eta_B=1_B$, and composing $cTc=c\mu_Q$ with $T^2q$ reduces it to $bTb=b\mu_B$. Both $Tq$ and $T^2q$ are <epimorphisms> because they remain split <coequalizers>. Thus $(Q,c)$ is an <algebra for a monad> and $q$ is an <monad algebra morphism>. If an algebra map $h:(B,b)\to(R,r)$ equalizes $f,g$, its underlying factor $v:Q\to R$ obeys $vc=rTv$, since this equation becomes $hb=rTh$ after composition with $Tq$. This proves that $U^T$ creates the required <coequalizers>. An <equivalence of categories> transfers these properties to any monadic $U$, proving necessity.
Conversely assume the three conditions and write $T=UF$. Given $(A,a)\in\mathcal C^T$, take the pair
$$
FTA\mathrel{\substack{\xrightarrow{\varepsilon_{FA}}\\[-2pt]\xrightarrow[Fa]{}}}FA.
$$
Its image is $\mu_A,Ta:T^2A\rightrightarrows TA$, with <split coequalizer> $a:TA\to A$: the splittings are $s=\eta_A$ and $t=\eta_{TA}$, because
$$
a\eta_A=1_A,\qquad \mu_A\eta_{TA}=1_{TA},\qquad Ta\,\eta_{TA}=\eta_Aa.
$$
Let $q:FA\to D_a$ be its <coequalizer> in $\mathcal D$. Preservation gives a canonical <isomorphism> $UD_a\cong A$ under which $Uq=a$. Transporting the algebra action of $K(D_a)$ along this <isomorphism> gives $\alpha:TA\to A$. Naturality of the <adjunction counit> says that $K(q)$ is an <monad algebra morphism>, hence $\alpha Ta=a\mu_A$. But $aTa=a\mu_A$ too, and $Ta$ is a <split epimorphism>, with section $T\eta_A$. Therefore $\alpha=a$. Every algebra is consequently isomorphic to some $K(D_a)$.
It remains to establish full faithfulness, rather than merely construct objects. For $B\in\mathcal D$, put $b=U\varepsilon_B$. The <adjunction counit> $\varepsilon_B:FUB\to B$ coequalizes $\varepsilon_{FUB},Fb$. The <coequalizer> supplied by the hypothesis has underlying <coequalizer> $b:TUB\to UB$. The induced arrow from that <coequalizer> object to $B$ is therefore sent by $U$ to an <isomorphism>; reflection of <isomorphisms> makes it invertible. Thus $\varepsilon_B$ itself is this <coequalizer>.
Now let $h:K(B)\to K(C)$ be an <monad algebra morphism>, and put $c=U\varepsilon_C$. The arrow $v=\varepsilon_CFh:FUB\to C$ coequalizes the two arrows in the preceding presentation. To check this directly, use transposition along $F\dashv U$: the transposes of $vFb$ and $v\varepsilon_{FUB}$ are respectively $hb$ and $cTh$, which agree by the algebra-morphism equation. Hence a unique $\bar h:B\to C$ satisfies $\bar h\varepsilon_B=v$. Applying $U$ gives $(U\bar h)b=cTh=hb$; cancellation of the <split epimorphism> $b$ yields $U\bar h=h$. Finally, any arrow $r:B\to C$ satisfies $r\varepsilon_B=\varepsilon_CFU r$ by naturality, so two arrows with equal image under $U$ are equal by the <coequalizer> property. The comparison is <full and faithful> and essentially surjective, hence an <equivalence of categories>. This proves the theorem.
For the application, the free <compact Hausdorff space> on a <set> $S$ is the <Stone-Čech compactification> $\beta(S_{\mathrm{disc}})$ of the corresponding <discrete space>. Its standard extension property says that every set map $S\to UX$, for compact Hausdorff $X$, extends uniquely to a <continuous map> $\beta S\to X$. This constructs the required <left adjoint>. Also a continuous bijection from a compact space to a Hausdorff space is a <homeomorphism>, so $U$ reflects <isomorphisms>.
Take <continuous maps> $f,g:X\rightrightarrows Y$ whose underlying functions have a <split coequalizer> $q:Y\to Q$ with splittings $s,t$. The splittings need not be continuous. Let
$$
E=\{(f(x),g(x)):x\in X\}\subseteq Y\times Y,\qquad R=\{(y,z):q(y)=q(z)\}.
$$
The set $E$ is compact and hence closed. Since $qf=qg$, it is contained in $R$. Conversely $(y,sq(y))=(ft(y),gt(y))\in E$. Thus
$$
R=\{(y,z):\exists w\in Y,\ (y,w)\in E,\ (z,w)\in E\}.
$$
The set of triples on the right is closed in the <compact Hausdorff space> $Y^3$, so its projection onto the first two coordinates is compact and closed. This proves that the <equivalence relation> $R$ is closed, without assuming continuity of either splitting.
Give $Q\cong Y/R$ the <quotient topology>. The allowed closed-relation criterion makes it a <compact Hausdorff space>. If a <continuous map> $h:Y\to Z$ equalizes $f,g$, the split-coequalizer property makes it constant on the fibres of $q$; its unique factor $Q\to Z$ is continuous by the quotient <topology>. Hence this is a <coequalizer> in <compact Hausdorff spaces>, preserved by $U$. Its <topology> is forced uniquely, since any continuous surjection from a compact space to a Hausdorff space is a quotient map. All three conditions hold, and therefore \b[the forgetful functor from compact Hausdorff spaces to sets is monadic].
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