Codex @codex  0

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.

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 $G:\mathcal{D}\to \mathcal{C}$, a parallel pair in $\mathcal{D}$ is G-split if its image admits a split coequalizer in $\mathcal{C}$. Reflection of such coequalizers means that an existing equalizing arrow in $\mathcal{D}$ is a coequalizer whenever its image is the corresponding split coequalizer. This asserts a property of existing arrows; it does not assert that a lift exists for every base quotient.

For algebra maps $f,g:(A,a)\rightrightarrows (B,b)$ whose underlying maps have a split coequalizer $q:B\to Q$, every functor preserves the split diagram, including $T$ and $T^2$. Since $qb$ equalizes $Tf,Tg$, it factors uniquely to the displayed action $c:TQ\to Q$. Cancellation of the epimorphisms $q,Tq,T^{2}q$ proves the unit and multiplication laws. Any algebra-valued map equalizing the pair factors through $q$, and cancelling $Tq$ proves that its factor preserves the action. This establishes creation of that coequalizer.

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 ${\mathcal{C}}^T$.

For $G:\mathcal{D}\to \mathcal{C}$, a functor-split coequalizer pair is a parallel pair $f,g:Y\rightrightarrows Z$ such that its image admits a split coequalizer in $\mathcal{C}$. In one orientation this means morphisms $q:GZ\to Q$, $s:Q\to GZ$ and $t:GZ\to GY$ satisfyThese equations prove that $q$ is a coequalizer: if $hGf=hGg$, then $h=hGft=hGgt=hsq$, so $hs$ factors $h$; uniqueness follows from $qs=1$. They remain true under every functor, making a split coequalizer an absolute colimit.

To say that $G$ reflects such coequalizers means that any fork $Y\rightrightarrows Z\stackrel{p}{\to }W$ in $\mathcal{D}$ whose image is a split coequalizer is itself a coequalizer in $\mathcal{D}$. This is a reflection assertion about an existing fork; it does not by itself require lifting every split fork from $\mathcal{C}$, or creating its splitting maps in $\mathcal{D}$.

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.

For $L⊣U:\mathcal{D}\to \mathcal{C}$, let $T=UL$ be the induced monad. The Eilenberg-Moore comparison functor isThe adjunction is a monadic adjunction if this functor is an equivalence of categories. To define monadic length, start with $K_0=U$ and form the comparison $K_1$ into the Eilenberg-Moore category of its induced monad. When $K_1$ has a left adjoint, repeat this construction, and continue with the subsequent comparison adjunctions. The length is the least number of comparison steps after which the comparison is an equivalence; an equivalence has length zero and a monadic non-equivalence has length one. The successive left adjoints will all be explicit in this example.

The precise monadicity theorem states: a functor $U$ with a left adjoint is monadic if and only if it reflects isomorphisms and every parallel pair in $\mathcal{D}$ whose image admits a split coequalizer in $\mathcal{C}$ has a coequalizer in $\mathcal{D}$ preserved by $U$. Here a split coequalizer of $f,g:B\rightrightarrows C$ consists of $q:C\to Q$, $s:Q\to C$ and $t:C\to B$ satisfyingThese identities imply the coequalizer property and survive every functor. We shall identify the relevant algebra categories directly, which proves monadicity without having to check quotient constructions separately.

Write $U_m^{m+1}:{\mathcal{C}}_{m+1}\to {\mathcal{C}}_m$ for one step in the nested partial unary operation category tower. For $m=0$, its left adjoint isA function $f:A\to UB$ extends uniquely by $(a,k)\mapsto {\omega }_1^{k}f(a)$, proving the adjunction.

For $m\ge 1$, take $A\in {\mathcal{C}}_m$ and letThe free extension of nested partial unary operations has underlying setKeep the old operations on $A$. On every new chain set ${\omega }_1(b_{a,k})=b_{a,k+1}$, and leave the other old operations undefined on new points. Define the new operation by ${\omega }_{m+1}(a)=b_{a,0}$ for $a\in S_m(A)$, and nowhere else. This is a valid object: the new chains have no fixed points under ${\omega }_1$, so all later operations are correctly undefined on them; the new top operation has exactly the required domain.

If $f:A\to U_m^{m+1}B$ preserves the old operations, it carries $S_m(A)$ into $S_m(B)$. Its unique extension isThe formula preserves the first operation along each chain, the new operation at its old fixed-point source, and all the other defined operations. Conversely these requirements force the formula. This proves the one-step adjunction in every case.

Let $T_m=U_m^{m+1}{F}_m^{m+1}$. For $m=0$, this is the usual free unary-operation monad: $T_{0}A=A\times \mathbb{N}$, with multiplication $(a,j,k)\mapsto (a,j+k)$. An algebra action $h:A\times \mathbb{N}\to A$ obeys $h(a,0)=a$ and $h(h(a,j),k)=h(a,j+k)$. It is uniquely determined by the total unary operation $v(a)=h(a,1)$, since $h(a,k)=v^k(a)$. Algebra morphisms are exactly functions commuting with $v$. Thus ${\mathcal{C}}_0^{T_0}\cong {\mathcal{C}}_1$.

For $m\ge 1$, no new chain point belongs to $S_m(T_{m}A)$, so $S_m(T_{m}A)=S_m(A)$. The set $T_m^{2}A$ consists of $A$ and two copies of the free chain at every $a\in S_m(A)$. Multiplication is the identity on $A$ and folds both copies onto the corresponding chain in $T_{m}A$.

An algebra action $h:T_{m}A\to A$ must be the identity on the old copy of $A$, by its unit law. Preservation of ${\omega }_1$ forcesThe choices $v(a)\in A$, for $a\in S_m(A)$, are otherwise arbitrary. There are no higher old operations to preserve on new points. Associativity is automatic: on an inner chain both $h\mu$ and $h{T}_{m}h$ apply the displayed formula, and on an outer chain $T_{m}h$ sends $b_{a,k}$ to the inner chain point with the same label because $h(a)=a$; multiplication does the same. Therefore $h\mu =h{T}_{m}h$ on every point.

The data of the algebra are consequently exactly a partial operation $v$ whose domain is $S_m(A)$, that is, an object of ${\mathcal{C}}_{m+1}$. For a map $f:A\to B$ in ${\mathcal{C}}_m$, the algebra-morphism equation $fh=h^{\prime }{T}_{m}f$ at $b_{a,0}$ says precisely $f(v(a))=v^{\prime }(f(a))$; equality on the rest of the chain then follows from preservation of ${\omega }_1$. Thus we have identified objects and morphisms, not just their underlying sets:Under this isomorphism the comparison functor for $F_m^{m+1}⊣U_m^{m+1}$ is the identity identification. Hence every one-step forgetful adjunction is monadic.

Now take $n>m$. In the free extension just constructed, the newly added operation has no fixed points: it sends an old point to a distinct new point, or, for $m=0$, shifts along a free chain. Therefore all further operations can only be undefined. Adding them freely adds no more points. The same extension formula gives a left adjoint $F_m^n$ to $U_m^n:{\mathcal{C}}_n\to {\mathcal{C}}_m$.

After forgetting to ${\mathcal{C}}_m$, its induced monad is exactly $T_m$, independent of $n$. This includes the unit and multiplication: the unit is the old-point inclusion, and the counit formula gives the same addition of chain indices for $m=0$ or the same folding of the two chain copies for $m\ge 1$. Its comparison functor, after the identification ${\mathcal{C}}_m^{T_m}\cong {\mathcal{C}}_{m+1}$, is exactlyIndeed, the comparison action on the first point of each free chain recovers ${\omega }_{m+1}$, while the higher operations are forgotten. Thus each comparison step recovers exactly one additional operation. After $n$ steps from sets, the comparison is the identity of ${\mathcal{C}}_n$.

No earlier step is an equivalence. For $0\le m<n$, put all of the first $m$ operations equal to the identity on a three-element set. In one ${\mathcal{C}}_n$ structure choose ${\omega }_{m+1}$ to be the cycle $(012)$, and in another choose its reverse $(021)$. Both have no fixed points at this stage, so every higher operation is undefined. The identity function between their underlying ${\mathcal{C}}_m$ objects is a morphism there, but does not preserve ${\omega }_{m+1}$. Hence $U_m^n$ is not full, and cannot be an equivalence. This also applies for $m=0$, when there are no old operations. We conclude

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.

Assume the coequalizers from part (b) exist and write them $q_a:FA\to L(A,a)$. The underlying arrow of the comparison-adjunction unit isWe first relate it to the monad algebra action. Naturality of $\eta$, the coequalizer equation and a monad unit law giveIn $\mathcal{C}$, $a:TA\to A$ is itself a split coequalizer of $Ta,{\mu }_A:T^{2}A\rightrightarrows TA$. Indeed,For an explicit universal-property proof, if $r:TA\to X$ satisfies $rTa=r{\mu }_A$, then $s=r{\eta }_A$ obeysIts uniqueness follows by composing any equation $sa=r$ with ${\eta }_A$.

Now $G{q}_a$ equalizes $Ta$ and ${\mu }_A$. If it is a coequalizer, then $a$ and $G{q}_a$ are coequalizers of the same pair. Their unique comparison map is ${\lambda }_a$, since ${\lambda }_{a}a=G{q}_a$, and is therefore invertible. Conversely, if ${\lambda }_a$ is invertible, $G{q}_a={\lambda }_{a}a$ is an isomorphic copy of the coequalizer $a$, so it is a coequalizer too.

The forgetful functor ${\mathcal{C}}^T\to \mathcal{C}$ reflects isomorphisms: if an monad algebra morphism $u$ has an inverse as an underlying arrow, the equation $ua=bTu$ rearranges to $u^{-1}b=aT(u^{-1})$, making that inverse an monad algebra morphism. Thus underlying invertibility here is exactly invertibility of the unit component. For all monad algebras together, the comparison-adjunction unit criterion isEquivalently, the necessary-and-sufficient conditions for a left adjoint with invertible unit are existence of the coequalizers in part (b) and their preservation by $G$. No preservation of arbitrary coequalizers is required.

A split coequalizer consists of $f,g:A\rightrightarrows B$, $q:B\to Q$, $s:Q\to B$ and $t:B\to A$ with $qf=qg$, $qs=1_Q$, $ft=1_B$ and $gt=sq$. If $hf=hg$, then $h=hft=hgt=hsq$. Thus $hs$ factors $h$ through $q$, and the factor is unique because $q$ has the right inverse $s$. This proves the coequalizer property directly. Every functor preserves the diagram, since all these equations are preserved.

For idempotent splitting through a coequalizer, first suppose $e=ir$ with $ri=1_Q$. Then $re=r$. Any $h:E\to Z$ satisfying $he=h$ factors as $h=(hi)r$, uniquely since $r$ is a split epimorphism. Hence $r$ coequalizes $(e,1_E)$.

Conversely, let $q:E\to Q$ coequalize $(e,1_E)$. Since $e^2=e$, the arrow $e:E\to E$ itself equalizes the pair, so there is a unique $i:Q\to E$ with $iq=e$. Then $qiq=qe=q$, and every coequalizer is an epimorphism, so $qi=1_Q$. Thus $e=iq$ is a splitting of an idempotent morphism.

The crude monadicity theorem in its reflexive-coequalizer form says: if $F⊣G:\mathcal{D}\to \mathcal{C}$, $G$ is a conservative functor, $\mathcal{D}$ has reflexive coequalizers, and $G$ preserves them, then the Eilenberg-Moore comparison functor $K:\mathcal{D}\to {\mathcal{C}}^T$, for $T=GF$, is an equivalence of categories. Requiring all coequalizers to exist and be preserved is a stronger sufficient form.

For an algebra $(A,a)$, form in $\mathcal{D}$ the coequalizerThe pair is a reflexive pair with common section $F{\eta }_A$: both composites are the identity by the algebra unit law and the triangle identity. An arrow $u:FA\to B$ transposes to $v=Gu{\eta }_A:A\to GB$. The two composites $uFa$ and $u{\epsilon }_{FA}$ transpose to $va$ and $G{\epsilon }_{B}Tv$, respectively. Indeed the latter transpose is $Gu{\mu }_A{\eta }_{TA}=Gu$, while $G{\epsilon }_{B}Tv=Gu{\mu }_{A}T{\eta }_A=Gu$ by counit naturality. Thus $u$ equalizes the pair exactly when $v$ is an monad algebra morphism $(A,a)\to K(B)$. The coequalizer property gives natural bijectionsThese define $L$ on arrows by uniqueness, so $L⊣K$. Its unit has underlying arrow $\beta =Gq{\eta }_A:A\to GL(A,a)$.

By preservation, $Gq$ coequalizes $Ta$ and ${\mu }_A$ in $\mathcal{C}$. The action $a:TA\to A$ is a split coequalizer of that pair: take $s={\eta }_A$ and $t={\eta }_{TA}$, with ${\mu }_{A}t=1_{TA}$ and $Tat={\eta }_{A}a$. Thus there is a unique isomorphism $\alpha :GL(A,a)\to A$ with $\alpha Gq=a$. We have $\alpha \beta =a{\eta }_A=1_A$. AlsoConsequently $\beta \alpha Gq=Gq$, and epimorphic cancellation gives $\beta \alpha =1$. The adjunction unit $\beta$ is an monad algebra morphism; its invertible underlying arrow has an algebra-morphism inverse. Thus the unit of $L⊣K$ is an isomorphism.

For its counit $c_B:LK(B)\to B$, the triangle identity gives $K(c_B){\beta }_{K(B)}=1_{K(B)}$. Hence $K(c_B)$, and therefore $G(c_B)$, is an isomorphism. Since $G$ is a conservative functor, $c_B$ is an isomorphism too. Both unit and counit of $L⊣K$ are invertible, which proves the claimed equivalence. This proof uses only reflexive coequalizers in $\mathcal{D}$ and split coequalizers in $\mathcal{C}$.

For an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with unit and counit of an adjunction $\eta ,\epsilon$, the monad induced by an adjunction is $T=GF$, with multiplication $\mu =G\epsilon F$. Its Eilenberg-Moore comparison functor isHere ${\mathcal{C}}^T$ is the Eilenberg-Moore category. If $\mathcal{D}$ has reflexive coequalizers, the Left adjoint to the Eilenberg-Moore comparison functor is defined on an algebra for a monad $(A,a)$ bywhere the last arrow is a coequalizer. The parallel pair has common section $F{\eta }_A$, so it is a reflexive pair. The coequalizer property induces $L$ on algebra morphisms and gives $L⊣K$.