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For an adjunction $F⊣G$, applying $G$ to its standard counit presentation gives $T^{2}GD\rightrightarrows TGD\to GD$, split by the monad units. Reflection therefore makes every counit presentation a coequalizer and gives full faithfulness by descent. Conversely, monad algebra forgetful functor creates split coequalizers makes an existing G-split quotient a coequalizer in the Eilenberg-Moore category. Fullness and faithfulness of the comparison lift its universal factorization back to the original category.

A form of Beck's monadicity theorem compatible with equivalence-based monadic adjunctions is this. A functor $U:\mathcal{D}\to \mathcal{C}$ is monadic exactly when it has a left adjoint, is a conservative functor, and $\mathcal{D}$ has coequalizers of all pairs whose images under $U$ admit a split coequalizer, with $U$ preserving these coequalizers. In the strict lifting formulation, one states creation of these coequalizers. Such a pair is a functor-split coequalizer pair; the splitting is required in the base category, not necessarily upstairs.

A split coequalizer fork $A\begin{array}{c}\stackrel{r}{\to }\\ \underset{s}{\to }\end{array}B\stackrel{q}{\to }Q$ has maps $u:B\to A$ and $t:Q\to B$ such thatEvery functor preserves this coequalizer: if $hr=hs$, then $h=hru=hsu=htq$, and $q$ is split epic, so the factor through $q$ is unique. The same equations remain true after applying any functor.

For necessity, let $U^T:{\mathcal{C}}^T\to \mathcal{C}$ be the monad algebra forgetful functor. It has the free-algebra left adjoint. It reflects isomorphisms, because if an algebra map $f$ is invertible in $\mathcal{C}$, the equation $fa=bTf$ rearranges to $f^{-1}b=aT(f^{-1})$.

If two algebra maps $r,s:(A,a)\rightrightarrows (B,b)$ have an underlying split coequalizer $q:B\to Q$, then $Tq$ and $T^{2}q$ are also coequalizers. Since $qb$ equalizes $Tr,Ts$, there is a unique $c:TQ\to Q$ satisfyingAfter composing with the epimorphism $q$, the equation $c{\eta }_Q=1_Q$ reduces to the unit law for $b$. After composing with the epimorphism $T^{2}q$, the equation $cT(c)=c{\mu }_Q$ reduces to $bT(b)=b{\mu }_B$. Thus $(Q,c)$ is an algebra and $q$ an algebra map. If an algebra map $h:(B,b)\to (Z,z)$ equalizes $r,s$, its unique underlying factor $k:Q\to Z$ satisfies $kc=zTk$: this follows by precomposing with the epimorphism $Tq$. Therefore this is a coequalizer upstairs as well. This proves that the monad algebra forgetful functor creates split coequalizers. The same properties transport through the comparison equivalence of categories, with the object-lifting convention distinguished in Question 9.

For sufficiency, suppose the three conditions hold, choose $F⊣U$, and let $T=UF$ with comparison $K$. For each algebra $(X,a)$, consider the pairIts image has coequalizer $a:TX\to X$. It is split: take $r={\mu }_X$, $s=T(a)$, $q=a$, $u={\eta }_{TX}$ and $t={\eta }_X$. The monad and algebra laws giveBy hypothesis the pair has a coequalizer $q:FX\to D_X$ and $Uq$ is also a coequalizer. Identify $U{D}_X$ with $X$ using the unique coequalizer isomorphism, so $Uq=a$. Since $q$ is carried by $K$ to an algebra map from the free algebra, the transported action $b:TX\to X$ satisfiesBut $T(a)$ is a split epimorphism, with right inverse $T({\eta }_X)$, so $b=a$. Hence $(X,a)\cong K(D_X)$: the comparison is essentially surjective.

Next, for every $D\in \mathcal{D}$, put $a_D=U{\epsilon }_D$. The counit of an adjunction coequalizesThis follows from naturality of the counit. The image pair has the split coequalizer $a_D$, as above. If $q:FUD\to Q$ is the hypothesized preserved coequalizer, the induced map $h:Q\to D$ has $Uh$ invertible because both $Uq$ and $a_D$ coequalize the same pair. Conservativity makes $h$ invertible. Therefore ${\epsilon }_D$ itself is a coequalizer.

Let $f:K(D)\to K(E)$ be an algebra map, so $f{a}_D=a_{E}Tf$. Set $g={\epsilon }_{E}F(f):FUD\to E$. Under the adjunction transpose bijection, the two composites of $g$ with the displayed presentation pair correspond respectively to $f{a}_D$ and $a_{E}Tf$. They are equal, so $g$ factors uniquely through ${\epsilon }_D$, giving $h:D\to E$ with $h{\epsilon }_D=g$. Applying $U$ yieldsSince $a_D$ is split epic, $Uh=f$. Thus $K$ is full. If $Uh=Uk$, the transposes of $h{\epsilon }_D$ and $k{\epsilon }_D$ are equal, so those composites agree. The counit is epic, hence $h=k$; thus $K$ is faithful.

The comparison is full, faithful and essentially surjective, so it is an equivalence of categories. This proves monadicity. The argument also explains the role of each hypothesis: split base presentations construct all algebras, and reflection of isomorphisms turns the counit presentations into actual coequalizers upstairs.

For $T=GF$, define the Eilenberg-Moore comparison functorThe algebra unit law is a triangle identity. Naturality of the counit at ${\epsilon }_D$ gives its algebra associativity law; naturality at $h:D\to E$ gives $Gh{a}_D=a_{E}T(Gh)$, making $K(h)$ an algebra morphism.

Suppose first that $G$ reflects G-split coequalizers. For every $D$, consider the counit presentationIt equalizes the pair by counit naturality. After applying $G$ it becomes $T^{2}GD\rightrightarrows TGD\stackrel{a_D}{\to }GD$, with parallel maps ${\mu }_{GD},T(a_D)$. This is split: take $s={\eta }_{GD}$ and $t={\eta }_{TGD}$. The identities $a_{D}s=1$, ${\mu }_{GD}t=1$ and $T(a_D)t={\eta }_{GD}{a}_D$ follow from the algebra law, a monad unit law, and naturality of $\eta$. Reflection therefore makes ${\epsilon }_D$ a coequalizer in $\mathcal{D}$.

For faithfulness, if $Gh=Gk$ then naturality gives $h{\epsilon }_D={\epsilon }_{E}FGh={\epsilon }_{E}FGk=k{\epsilon }_D$. Coequalizers are epic, so $h=k$.

For fullness, let $\alpha :GD\to GE$ be an algebra morphism, so $\alpha a_D=a_{E}T\alpha$. Put $h_0={\epsilon }_{E}F\alpha :FGD\to E$. Its two composites with the counit-presentation pair are equal: by counit naturality they are ${\epsilon }_{E}F(a_{E}T\alpha )$ and ${\epsilon }_{E}F(\alpha a_D)$. Hence there is a unique $h:D\to E$ with $h{\epsilon }_D=h_0$. Applying $G$ gives $Gh{a}_D=a_{E}T\alpha =\alpha a_D$. Since $a_D$ is split epic, $Gh=\alpha$. This proves full faithfulness directly by descent.

Conversely suppose $K$ is full and faithful. We need the following explicit algebra lifting fact. For algebra morphisms $f,g:(A,a)\rightrightarrows (B,b)$ with a split underlying coequalizer $q:B\to Q$, the arrow $qb:TB\to Q$ equalizes $Tf,Tg$. The split diagram is preserved by $T$, so $Tq$ is their coequalizer and there is a unique $c:TQ\to Q$ with $cTq=qb$. Precomposing the unit equation with the epimorphism $q$ gives $c{\eta }_{Q}q=q$; precomposing the associativity equation with the epimorphism $T^{2}q$ gives equality by $bT(b)=b{\mu }_B$. Thus $c$ is an algebra action. If an algebra map $h:B\to (Z,z)$ equalizes $f,g$, its unique underlying factor $k:Q\to Z$ satisfies $kcTq=hb=zTh=zTkTq$. Cancelling $Tq$ shows $kc=zTk$. This proves monad algebra forgetful functor creates split coequalizers.

Now take any existing equalizing arrow $e:B\to D$ in $\mathcal{D}$ whose image is such a coequalizer. The lifting fact makes the unique compatible action on $GD$ its already given action $a_D$, since $Ke$ preserves that action. Hence $Ke$ is a coequalizer of $Kf,Kg$ in the algebra category. For any arrow $h:B\to Z$ equalizing the pair, its unique algebra factor $KD\to KZ$ lifts to an arrow $D\to Z$ by fullness of $K$; faithfulness gives its factorization equation and uniqueness. Therefore $e$ is a coequalizer in $\mathcal{D}$. We have proved the full comparison and reflection of split coequalizers equivalence