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A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying $\mu T\eta =\mu \eta T=1_T$ and $\mu T\mu =\mu \mu T$. Its Eilenberg-Moore category has objects $(X,a)$ with $a:TX\to X$, $a{\eta }_X=1_X$ and $a{\mu }_X=aTa$. A morphism $f:(X,a)\to (Y,b)$ satisfies $fa=bTf$. An adjunction $F⊣G$, with unit $\eta$ and counit $\epsilon$, induces $T=GF$, $\mu =G\epsilon F$. It is a monadic adjunction if the Eilenberg-Moore comparison functor $K(D)=(GD,G{\epsilon }_D)$ is an equivalence with the Eilenberg-Moore category.

A fork $A\stackrel{p,q}{\rightrightarrows }B\stackrel{r}{\to }C$ is a split coequalizer if $rp=rq$ and there are $s:C\to B$, $t:B\to A$ with $rs=1$, $pt=1$ and $qt=sr$. It is indeed a coequalizer: if $hp=hq$, then $h=hpt=hqt=hsr$, so $h$ factors uniquely through $r$, which is split epic. Every functor preserves these splitting identities, making such coequalizers absolute.

The precise monadicity theorem states: for an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$, the comparison is an equivalence if and only if $G$ reflects isomorphisms and every pair in $\mathcal{D}$ whose $G$-image admits a split coequalizer has a coequalizer which $G$ preserves. Equivalently, the right adjoint creates these split-underlying coequalizers, with creation understood up to the canonical isomorphism of the prescribed underlying coequalizer. Here is a proof.

First consider the forgetful functor from monad algebras. It reflects isomorphisms: the inverse of an invertible algebra morphism also obeys the algebra-morphism identity, by multiplying that identity by the inverses. For an algebra pair with a split underlying coequalizer $r:B\to C$, all powers of $T$ preserve its coequalizer. If $b:TB\to B$ is the algebra structure on $B$, the map $rb$ coequalizes $Tp,Tq$, so there is a unique $c:TC\to C$ withThe unit law follows by composing with the epimorphism $r$: $c{\eta }_{C}r=cTr{\eta }_B=rb{\eta }_B=r$. For associativity, naturality and the algebra law giveCancel the epimorphism $T^{2}r$ to obtain $c{\mu }_C=cTc$. Thus $(C,c)$ is an algebra and $r$ is an algebra morphism. Any algebra morphism coequalizing the pair factors through $r$ on underlying objects, and its factor is an algebra morphism by cancellation of the epimorphism $Tr$. This creates the coequalizer. Transporting these facts through an equivalence proves necessity.

For sufficiency, let $(X,a)$ be a $T$-algebra. In $\mathcal{D}$ consider the free pairIts underlying pair is ${\mu }_X,Ta:T^{2}X\rightrightarrows TX$, with split coequalizer $a:TX\to X$: the splittings are $s={\eta }_X$ and $t={\eta }_{TX}$, using the algebra unit law, the monad unit law and naturality of $\eta$. By hypothesis its coequalizer $q:FX\to D$ exists and is preserved. Identify $GD$ with $X$ by the unique coequalizer isomorphism, so $Gq=a$. Naturality of the counit makes $Kq$ an algebra morphism, hence the comparison structure $c$ on $GD$ satisfies $cTa=a{\mu }_X=aTa$. The map $Ta$ is split epic because $a{\eta }_X=1$, so $c=a$. Every algebra is therefore isomorphic to a comparison algebra.

For an actual object $D$, its counit ${\epsilon }_D:FGD\to D$ is a coequalizer of the analogous free pair. It coequalizes that pair by counit naturality. Compare it with the coequalizer furnished by the hypothesis; applying $G$ gives an isomorphism, since both underlying arrows coequalize the same split pair. Reflection of isomorphisms makes the comparison map an isomorphism. This proves the counit's asserted coequalizer property.

Let $f:KD\to KE$ be an algebra morphism. Its transpose is $h_0={\epsilon }_{E}Ff:FGD\to E$. The two composites of $h_0$ with the free pair have adjunct transposes $G{\epsilon }_{E}Tf$ and $fG{\epsilon }_D$, equal because $f$ is an algebra morphism. Consequently $h_0$ factors uniquely through ${\epsilon }_D$ as $h:D\to E$. Applying $G$ gives $(Gh)G{\epsilon }_D=fG{\epsilon }_D$; the last factor is split epic, so $Gh=f$. Conversely any lift of $f$ must satisfy $h{\epsilon }_D={\epsilon }_{E}Ff$ by naturality and hence is unique. Thus $K$ is full and faithful, and the earlier construction makes it essentially surjective. The comparison is an equivalence, completing the theorem's proof.

Now take $G:\mathbf{CompHaus}\to \mathbf{Set}$. Its left adjoint exists by the permitted topological result; it can be realized as the Stone-Čech compactification of the discrete input set. A bijective continuous map from a compact space to a Hausdorff space is a homeomorphism, so $G$ reflects isomorphisms.

For continuous $f,g:X\rightrightarrows Y$ with a split underlying fork $q,s,t$, let $R$ be the relation of having the same $q$-image. We claimFor the forward inclusion choose $x=t(y)$, $x^{\prime }=t(y^{\prime })$: $ft=1$ and $gt=sq$ supply the equality. For the reverse inclusion use $qf=qg$. The subset $\{(x,x^{\prime }):g(x)=g(x^{\prime })\}\subset X^2$ is closed, because $Y$ is Hausdorff, and compact. Its continuous image $R\subset Y^2$ is compact and hence closed. A closed equivalence relation on a compact Hausdorff space has a compact Hausdorff quotient; its quotient map is a coequalizer here, since it identifies exactly the generated relation. Its underlying set coequalizer is the specified one up to the unique isomorphism. Thus $G$ preserves the needed coequalizers, andThe closedness argument is the closed relation generated by a set-split pair of compact Hausdorff maps lemma; the splitting maps themselves need not be continuous.