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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

A functor-split coequalizer pair for $G$ is a parallel pair $f,g:A\rightrightarrows B$ in $\mathcal{D}$ whose image has a split coequalizer in $\mathcal{C}$. Explicitly there are $q:GB\to Q$, $s:Q\to GB$, $t:GB\to GA$ withThese equations give the coequalizer property: if $hGf=hGg$, then $h=h(Gf)t=h(Gg)t=hsq$, and $q$ is split epic so the factor through it is unique. Every functor preserves such a split diagram, since it preserves these equations.

Reflection means that, for such a pair, any existing arrow $e:B\to D$ whose image is a coequalizer is itself a coequalizer in $\mathcal{D}$. A coequalizer of the image pair is isomorphic to the specified split one and inherits its splitting. This is reflection of an existing quotient, not an assertion that every base quotient has a lift.

Broadly in favour, if “trivial” means transparent once the correct structure has been identified. The value of category theory is often to isolate a universal property that makes a repeated argument automatic.

For example, two objects with the same universal property have unique connecting maps in each direction. Their composites and the identities satisfy the same defining conditions, so uniqueness makes those composites identities. This one argument explains the uniqueness of categorical products, categorical limits, adjoint functors and many other constructions. It also specifies which isomorphism is canonical, a point that an arbitrary objectwise choice would miss.

The Yoneda lemma makes this perspective systematic. A natural transformation out of $\mathcal{C}(c,-)$ is determined by its value at $1_c$. Once that evaluation principle is recognized, questions about representable functors reduce to compositions in the original category. Similarly, the Beck monadicity theorem organizes many free-and-forgetful constructions around monad algebras, reflection of isomorphisms and specified coequalizers. The proof supplies a reusable explanation of why the algebraic structure can be reconstructed from the forgetful functor.

The substantial work lies in finding suitable definitions and verifying their hypotheses. The Freyd general adjoint functor theorem, for instance, needs completeness, local smallness and a solution set; none follows merely from calling a construction natural. The partial-operation tower below even has a monadic length that records how many distinct reconstruction steps are needed. These examples support the view that abstraction clarifies the source of an argument's simplicity. They also show why that simplicity is an achievement requiring proof.

An adjunction $L⊣U$ induces the monad $T=UL$ with unit $\eta$ and multiplication $U\epsilon L$. The Eilenberg-Moore comparison functor sends $B$ to $(UB,U{\epsilon }_B)$. A monadic adjunction is one for which that comparison is an equivalence of categories. Iterating comparisons, when their left adjoints exist, gives a tower; its monadic length is the number of steps until equivalence, with length zero for an equivalence and length one for a monadic adjunction that is not already an equivalence.

The precise monadicity theorem says that a right adjoint is monadic exactly when it reflects isomorphisms and creates coequalizers of pairs whose images admit split coequalizers.

Write $U_r:{\mathcal{C}}_r\to {\mathcal{C}}_{r-1}$ for one-step forgetting. For $r=1$, the free extension of nested partial unary operations is the familiar set $A\times \mathbb{N}$, with ${\alpha }_1(a,j)=(a,j+1)$ and unit $a\mapsto (a,0)$. A map into a set with endomorphism ${\beta }_1$ extends uniquely by $(a,j)\mapsto {\beta }_1^{j}f(a)$.

For $r\ge 2$, letKeep $A$ and all its old operations, and adjoin the disjoint set $S(A)\times \mathbb{N}$. On the new points define ${\alpha }_1(a,j)=(a,j+1)$; every old operation ${\alpha }_i$ with $i\ge 2$ is undefined there. Define the new top operation only at the old eligible points, by ${\alpha }_r(a)=(a,0)$ for $a\in S(A)$.

These domains satisfy the defining rules of the nested partial unary operation category: the new points have no ${\alpha }_1$-fixed points, so none has any higher operation defined. Given $f:A\to U_{r}B$, extend it byThe expression is defined because $f$ sends eligible points to eligible points and ${\beta }_1$ is total. It preserves every defined operation; preservation of ${\alpha }_r$ at old points determines $j=0$, and preservation of ${\alpha }_1$ determines the whole chain. Thus it is the unique extension, proving $L_r⊣U_r$.

For monadicity, $U_r$ reflects isomorphisms: an isomorphism of the lower structures takes $S(A)$ bijectively to $S(B)$, so the inverse of a map preserving the new operation preserves it too.

Next prove the split-coequalizer lifting for a fixed-point-domain operation. Let $f,g:A\rightrightarrows B$ be ${\mathcal{C}}_r$-morphisms and suppose their lower images have a split coequalizer $q:U_{r}B\to Q$, with $s:Q\to U_{r}B$, $t:U_{r}B\to U_{r}A$ satisfyingEquip $Q$ with ${\beta }_Q(z)=q{\beta }_B(sz)$ on its required eligible domain. The lower-structure maps $s,t$ preserve eligibility. For eligible $x\in B$,so $q$ preserves the top operation. If $h:B\to D$ coequalizes $f,g$, its lower factor $\overline{h}:Q\to U_{r}D$ satisfiesThus it is a morphism of the full structures. Uniqueness of the lifted operation follows from the eligible section $s$ and surjectivity of $q$. This creates the specified coequalizer, so the Beck monadicity theorem proves each one-step adjunction is monadic.

Finally, every free top operation just constructed has no fixed points: for $r=1$ it shifts the free chain, while for $r\ge 2$ it sends old eligible points to new distinct points and is undefined on the new points. Freely adjoining any further operation therefore changes nothing, since its required domain is empty. For $n>m$, the monad induced on ${\mathcal{C}}_m$ by the long free-and-forgetful adjunction is consequently the same monad, including unit and multiplication, as that for ${\mathcal{C}}_{m+1}\to {\mathcal{C}}_m$.

The first comparison category is thus equivalent to ${\mathcal{C}}_{m+1}$. Its algebra action evaluates the new formal value by ${\alpha }_{m+1}$, so the comparison from ${\mathcal{C}}_n$ retains precisely that next operation and is the remaining forgetful functor. Iteration recovers one additional operation at each step. These remaining comparisons are not equivalences while an operation is still forgotten: take a two-element set with all retained operations equal to the identity, and choose the next operation to be the identity in one extension and the transposition in another. Higher operations can be identities in the first case and undefined in the second. The identity on the retained structure is not a morphism between these extensions, so the forgetful functor is not full. This includes ${\mathcal{C}}_1\to \mathbf{Set}$.

The monadic tower for nested partial unary operations therefore has exactly the claimed length:

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 $\mu :T^2\Rightarrow T$, satisfyingAn 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 $F⊣U$, the monad induced by an adjunction is $T=UF$, with $\mu =U\epsilon F$. The Eilenberg-Moore comparison functor is $K(D)=(UD,U{\epsilon }_D)$ 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 $t:B\to A$ withThese 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 $T^{2}q$ reduces it to $bTb=b{\mu }_B$. Both $Tq$ and $T^{2}q$ 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 pairIts image is ${\mu }_A,Ta:T^{2}A\rightrightarrows TA$, with split coequalizer $a:TA\to A$: the splittings are $s={\eta }_A$ and $t={\eta }_{TA}$, becauseLet $q:FA\to D_a$ be its coequalizer in $\mathcal{D}$. Preservation gives a canonical isomorphism $U{D}_a\cong A$ 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 $b=U{\epsilon }_B$. The adjunction counit ${\epsilon }_B:FUB\to B$ coequalizes ${\epsilon }_{FUB},Fb$. 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 ${\epsilon }_B$ itself is this coequalizer.

Now let $h:K(B)\to K(C)$ be an monad algebra morphism, and put $c=U{\epsilon }_C$. The arrow $v={\epsilon }_{C}Fh:FUB\to C$ coequalizes the two arrows in the preceding presentation. To check this directly, use transposition along $F⊣U$: the transposes of $vFb$ and $v{\epsilon }_{FUB}$ are respectively $hb$ and $cTh$, which agree by the algebra-morphism equation. Hence a unique $\bar{h}:B\to C$ satisfies $\bar{h}{\epsilon }_B=v$. 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 $r{\epsilon }_B={\epsilon }_{C}FUr$ 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. LetThe set $E$ is compact and hence closed. Since $qf=qg$, it is contained in $R$. Conversely $(y,sq(y))=(ft(y),gt(y))\in E$. ThusThe 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 the forgetful functor from compact Hausdorff spaces to sets is monadic.