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Let continuous maps $f,g:X\rightrightarrows Y$ between compact Hausdorff spaces have an underlying split coequalizer $q:Y\to Z$ with set maps $s,t$ satisfying $qs=1$, $ft=1$ and $gt=sq$. Its equivalence relation is the image of the closed subset $\{(x,x^{\prime })\in X^2:g(x)=g(x^{\prime })\}$ under $(f,f)$. One inclusion follows from $qf=qg$; for the other use $x=t(y)$ and $x^{\prime }=t(y^{\prime })$ whenever $q(y)=q(y^{\prime })$. This image is compact and hence closed in the Hausdorff product $Y^2$. The quotient is therefore compact Hausdorff, giving the preserved coequalizer required by the Beck monadicity theorem.

The dual universal family is a coend $Q$ with maps $i_C:T(C,C)\to Q$ satisfying $i_{C}T(f,1_C)=i_{D}T(1_D,f)$ on $T(D,C)$. Every compatible family to another object factors uniquely through $Q$. For small $\mathcal{C}$ in a cocomplete category, it is the coequalizer ${\coprod }_{f:C\to D}T(D,C)\rightrightarrows {\coprod }_{C}T(C,C)\to Q$.

For a reflexive pair $f,g:A\rightrightarrows B$ with common section $r$, form the pushout in a category of $f$ and $g$, with maps $j_1,j_2:B\to P$. The equation $j_{1}f=j_{2}g$, composed with $r$, gives $j_1=j_2=q$. This common map is a coequalizer of $f,g$: every map equalizing them defines a commuting pair into the pushout and hence factors uniquely through $q$.

The cokernel of $f:A\to B$ is the universal morphism $q:B\to Q$ satisfying $qf=0$. It is the coequalizer of $f$ and the zero morphism.

For $F⊣G$ inducing $T=GF$, the Eilenberg-Moore comparison functor has a left adjoint exactly when every algebra $(A,a)$ has a coequalizer $q_a:FA\to L(A,a)$ of $Fa,{\epsilon }_{FA}:FTA\rightrightarrows FA$. Transposition identifies maps equalizing this pair with monad algebra morphisms $(A,a)\to K(D)$, proving both directions of the criterion. The unit is $\lambda =G{q}_a{\eta }_A$ and satisfies $G{q}_a=\lambda a$. In the base category, $a:TA\to A$ is a split coequalizer of $Ta,{\mu }_A:T^{2}A\rightrightarrows TA$, split by ${\eta }_A$ and ${\eta }_{TA}$. Consequently $\lambda$ is invertible exactly when $G{q}_a$ is also a coequalizer. The unit is invertible everywhere precisely when $G$ preserves all these specified coequalizers; preservation of arbitrary coequalizers is not required.

For a finite diagram in a category $D:\mathcal{J}\to \mathcal{C}$, set $X={\coprod }_{j}D(j)$ and $Y={\coprod }_{a:i\to j}D(i)$. The two maps $s,t:Y\rightrightarrows X$ on the $a$-summand are the injection of $D(i)$ and the injection of $D(j)$ after $D(a)$. Their coequalizer is the colimit of $D$. The pair $[s,1_X],[t,1_X]:Y⨿X\rightrightarrows X$ is a reflexive pair with the same coequalizer. Thus finite coproduct in a category constructions and coequalizers of reflexive pairs suffice, including the empty coproduct for the empty diagram. Finite products cannot replace coproducts here: the poset with elements $0,a,b,u_0,u_1,\dots$, order $0<a,b<u_{n+1}<u_n$, and incomparable $a,b$ has finite meets and top $u_0$, hence finite categorical products. Every reflexive parallel pair is an equal pair and has its identity as coequalizer, but $a,b$ have no least upper bound and therefore no coproduct.

For algebras for a monad $(A,\alpha ),(B,\beta )$ and underlying binary coproducts in a category, put $\kappa =[T{\nu }_1,T{\nu }_2]:TA+TB\to T(A+B)$. If the algebra pair $F(\alpha +\beta ),{\mu }_{A+B}F\kappa :F(TA+TB)\rightrightarrows F(A+B)$ has a coequalizer, that coequalizer is their algebra coproduct. An algebra map from $F(A+B)$ transposes to $[a,b]:A+B\to C$; equalizing the pair is exactly the two algebra-morphism equations for $a,b$. The pair is reflexive through $F({\eta }_A+{\eta }_B)$.

Every epimorphism in an abelian category is the coequalizer of its kernel pair. Its kernel-pair square is a pullback in a category along an epimorphism and therefore also a pushout. A map equalizing the two projections supplies the same map on both pushout legs and so factors uniquely through the original epimorphism. In particular, all such epimorphisms are regular epimorphisms.

If each standard counit presentation is a coequalizer, the Eilenberg-Moore comparison functor is full and faithful. An monad algebra morphism $\alpha :GB\to GC$ makes ${\epsilon }_{C}F\alpha$ equalize $FG{\epsilon }_B$ and ${\epsilon }_{FGB}$; its unique descent is the required morphism $B\to C$. The triangle identities for an adjunction make $G{\epsilon }_B$ a split epimorphism, so descent has underlying arrow exactly $\alpha$. Epimorphic cancellation at ${\epsilon }_B$ proves faithfulness.

If $e=ir$ and $ri=1$, then $r$ is the coequalizer of $(e,1)$: an arrow $h$ with $he=h$ factors uniquely as $(hi)r$. Conversely a coequalizer $q$ makes $e$ factor as $iq=e$; epimorphic cancellation gives $qi=1$. Thus the splitting of an idempotent morphism is equivalent to this particular coequalizer, without assuming arbitrary coequalizers exist.

For the nested partial unary operation category tower, each one-step forgetful functor is a monadic adjunction right adjoint: it reflects isomorphisms and creates its split coequalizers. The Beck monadicity theorem identifies the first Eilenberg-Moore category with the next category in the tower. A long composite has that same first monad because free extension of nested partial unary operations leaves no new top fixed points. Repeating the comparison forgets one fewer operation at each step. The remaining forgetful functor is not full until no operation remains to forget, giving monadic length $m-n$.

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 $(A,a)$, consider the reflexive free-algebra presentation of a monad algebraBoth parallel morphisms are monad algebra morphisms: ${\mu }_A$ satisfies the monad associativity law, while $Ta$ is the image under $F^T$ of $a$. The map $a$ is a monad algebra morphism because $a{\mu }_A=aTa$, and the same equation says it equalizes the pair.

Suppose $h:F^T(A)\to (B,b)$ is a monad algebra morphism with $h{\mu }_A=hTa$. Put $k=h{\eta }_A:A\to B$. Naturality of $\eta$ and the unit law giveMoreover, since $h$ is a monad algebra morphism,Thus $k$ is a monad algebra morphism and factors $h$ through $a$. If $k^{\prime }a=h$, then $k^{\prime }=k^{\prime }a{\eta }_A=h{\eta }_A=k$, proving uniqueness. Therefore every monad algebra is the stated coequalizer of free algebras.

The underlying fork is a split coequalizer: take $q=a$, $s={\eta }_A$ and $t={\eta }_{TA}$. Then $qs=1_A$, ${\mu }_{A}t=1_{TA}$, and $(Ta)t=sa$. Its splitting maps need not be monad algebra morphisms; the preceding argument separately proves the coequalizer in ${\mathcal{C}}^T$.

Write the adjunction unit and adjunction counit of $F⊣G$ as $\eta ,\epsilon$. Its induced monad is $T=GF$, with multiplication $\mu =G{\epsilon }_F$. The Eilenberg-Moore comparison functor isThe unit law for $a_d$ is $G{\epsilon }_d{\eta }_{Gd}=1_{Gd}$, a triangle identity for an adjunction. Naturality of $\epsilon$ at ${\epsilon }_d$ gives ${\epsilon }_{d}FG{\epsilon }_d={\epsilon }_d{\epsilon }_{FGd}$; applying $G$ gives $a_{d}T{a}_d=a_d{\mu }_{Gd}$. Naturality at $h$ gives the monad algebra morphism equation for $Gh$. Thus $K$ is well-defined.

For each $d$, consider the fork in $\mathcal{D}$It commutes by the naturality equation just used. Its image under $G$ is the fork $T^2(Gd)\rightrightarrows T(Gd)\to Gd$ of the preceding question, with parallel morphisms ${\mu }_{Gd},T{a}_d$ and final map $a_d$. It is split using $s={\eta }_{Gd}$ and $t={\eta }_{T(Gd)}$. By the assumed reflection, ${\epsilon }_d$ is the coequalizer of this pair in $\mathcal{D}$.

For faithfulness, if $Gh=Gk$ for $h,k:d\to e$, naturality gives $h{\epsilon }_d={\epsilon }_{e}FGh={\epsilon }_{e}FGk=k{\epsilon }_d$. A coequalizer is an epimorphism, so $h=k$.

For fullness, let $u:K(d)\to K(e)$ be a monad algebra morphism. Thus $u:Gd\to Ge$ satisfies $u{a}_d=a_{e}Tu$. Set $v={\epsilon }_{e}Fu:FGd\to e$. We verify explicitly that $v$ equalizes the two morphisms of the displayed fork. The monad algebra morphism equation givesNaturality of $\epsilon$ first at $Fu$ and then at ${\epsilon }_e$ givesThe coequalizer therefore supplies a unique $h:d\to e$ with $h{\epsilon }_d=v$. Applying $G$ yields $Gh{a}_d=Gv=a_{e}Tu=u{a}_d$. The map $a_d$ is split epic with section ${\eta }_{Gd}$, so cancellation gives $Gh=u$. This proves fullness, and the earlier cancellation proves faithfulness. HenceNo claim that $K$ is essentially surjective is needed; that stronger conclusion needs the corresponding existence hypotheses.