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A monad is idempotent when its multiplication is invertible. Then $T\eta =\eta T={\mu }^{-1}$. If $(A,a)$ is an algebra for a monad, its unit law and naturality give $a{\eta }_A=1_A$ and ${\eta }_{A}a=Ta{\eta }_{TA}=T(a{\eta }_A)=1_{TA}$. Thus every algebra is isomorphic to a free algebra, and the Kleisli comparison functor into the Eilenberg-Moore category is an equivalence.

An algebra for a monad for the list monad is precisely a monoid. For an action $a$, the identity is $a([])$ and multiplication is $a([x,y])$. The algebra laws force the unit and associativity laws and determine $a$ as ordered multiplication. Conversely ordered multiplication defines a list action. Algebra morphisms are exactly monoid homomorphisms, giving an isomorphism of categories ${\mathbf{Set}}^{\mathrm{List}}\cong \mathbf{Mon}$.

First prove the exact lifting statement for the forgetful functor $U^T:{\mathcal{C}}^T\to \mathcal{C}$. Take any diagram in a category of algebras for a monad $(A_j,a_j)$ whose underlying diagram has a chosen categorical limit $p_j:L\to A_j$. The arrows $a_{j}T(p_j):TL\to A_j$ form a categorical cone, since every diagram arrow is a morphism of algebras for a monad. Thus the universal property gives a unique arrow $a:TL\to L$ satisfyingThe projections are jointly monic: two arrows into $L$ with the same composites with every $p_j$ are equal. Therefore the algebra identities can be checked after these projections. For the unit,For multiplication,The algebra law for $a_j$ makes the right-hand sides equal. Thus $(L,a)$ is an algebra for a monad, and each $p_j$ is a morphism of algebras for a monad. Its action is unique with this property.

Given an algebra cone $q_j:(B,b)\to (A_j,a_j)$, let $q:B\to L$ be its unique underlying mediating map. Thenso $qb=aT(q)$ by joint monicity. Hence $q$ is an algebra map, proving the lifted cone is limiting. This establishes that the monad algebra forgetful functor creates limits, without any assumption that $T$ preserves limits.

For a monadic adjunction, the comparison satisfies $U=U^{T}K$. If $K$ is an isomorphism of categories over the base, the preceding unique lift transports back literally, so $U$ is a limit-creating functor. If monadic means that $K$ is only an equivalence of categories, the precise conclusion is creation of limits up to isomorphism: the lifted algebra is isomorphic to an object $K(D)$, and its limiting cone transports along that isomorphism. Full faithfulness supplies the arrows and their uniqueness.

This distinction concerns prescribed underlying objects, not existence of limits. Literal creation does not follow from an arbitrary equivalence alone. For example, the inclusion of the one-object category into the two-object indiscrete category, choosing one of its objects, is an equivalence of categories and is monadic under the equivalence convention. Either object of the target is terminal, but the other chosen terminal object has no literal object preimage. Thus the strict reading of the requested assertion needs the strict comparison convention; the equivalence-invariant statement is proved above.

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying the unit and multiplication of a monad lawsAn algebra for a monad is an object $A$ equipped with $a:TA\to A$ such thatA morphism of algebras for a monad $f:(A,a)\to (B,b)$ is a morphism $f:A\to B$ satisfying $fa=bT(f)$. Identities and composition obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^T$ and its forgetful functor $U^T(A,a)=A$.

For an adjunction $F⊣U:\mathcal{D}\to \mathcal{C}$ with induced monad $T=UF$, the Eilenberg-Moore comparison functor isThe functor $U$ is monadic when it has such a left adjoint and this comparison is an equivalence of categories; a stricter convention requires an isomorphism of categories over $\mathcal{C}$. We will identify which limit-lifting conclusion each convention supports in part (iii).

Define the free algebra functor $F^T$ by $F^{T}X=(TX,{\mu }_X)$ and $F^{T}f=Tf$. The monad laws make ${\mu }_X$ an algebra action, and naturality of $\mu$ makes $Tf$ a monad algebra morphism. For an algebra for a monad $(A,a)$, definewith proposed inverse $g\mapsto aTg$. This inverse is a monad algebra morphism, sinceOne composite is $aTg{\eta }_X=a{\eta }_{A}g=g$. For a monad algebra morphism $h$, the other is $aT(h{\eta }_X)=aThT{\eta }_X=h{\mu }_{X}T{\eta }_X=h$. Both formulas commute with composition in a category in $X$ and with monad algebra morphisms in $(A,a)$, so the bijection is natural. Thus

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying, at every object $A$,These are the associativity and unit laws. An algebra for a monad is $(A,a:TA\to A)$ withA monad algebra morphism $h:(A,a)\to (B,b)$ is a morphism $h:A\to B$ satisfying $ha=bTh$. Identity morphisms satisfy this equation, and if $h,k$ satisfy it, then $kha=kbTh=cTkTh=cT(kh)$. Hence these objects and morphisms form the Eilenberg-Moore category ${\mathcal{C}}^T$, with the forgetful functor $U(A,a)=A$ and $Uh=h$.

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\Rightarrow T$, $\mu :T^2\Rightarrow T$, satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$, $a{\eta }_A=1_A$ and $a{\mu }_A=aT(a)$. A morphism $(A,a)\to (B,b)$ is an arrow $h:A\to B$ with $ha=bT(h)$. These form the Eilenberg-Moore category ${\mathcal{C}}^T$.

A functor $G:\mathcal{D}\to \mathcal{C}$ is monadic when it has a left adjoint $F$ and its comparison $K:\mathcal{D}\to {\mathcal{C}}^{GF}$ is an equivalence; this is the monadic adjunction convention. Some strict formulations require $K$ to be an isomorphism over the base. The distinction affects the meaning of literal limit creation, discussed in part (iii), but not the algebra construction.

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 an adjunction $L:\mathcal{C}\rightleftarrows \mathcal{D}:R$, write its adjunction unit as $\eta$ and its adjunction counit as $\epsilon$. It is an idempotent adjunction when the induced monad has invertible multiplication $\mu =R\epsilon L:RLRL\Rightarrow RL$. The dual condition is invertibility of the induced comonad comultiplication $\delta =L\eta R:LR\Rightarrow LRLR$.

To prove equivalence of these conditions, suppose $\mu$ is invertible. The two unit identities imply $T\eta =\eta T={\mu }^{-1}$, where $T=RL$. For any algebra for a monad $a:TA\to A$, naturality giveswhile $a{\eta }_A=1_A$. Thus ${\eta }_A$ is invertible. For each $D\in \mathcal{D}$, the arrow $R{\epsilon }_D:T(RD)\to RD$ is a monad algebra action, so ${\eta }_{RD}$ is invertible. Applying $L$ proves ${\delta }_D=L{\eta }_{RD}$ invertible. Apply this implication to the opposite adjunction, interchanging the roles of the monad and comonad, to obtain the converse. Therefore idempotence is self-dual.

The open-set frame $O(X)$ is ordered by inclusion. Its arbitrary joins are unions; arbitrary meets are interiors of intersections. Finite intersections are already open, so finite meets are ordinary intersections, including the empty meet $X$. Distributivity of intersection over arbitrary unions proves the frame law. If $f:X\to Y$ is a continuous map, inverse imagepreserves arbitrary unions and finite intersections, including $\varnothing$ and the entire space. It is therefore a frame homomorphism. Since $(gf{)}^{-1}=f^{-1}{g}^{-1}$ and $1_X^{-1}=1_{O(X)}$, this defines $O:\mathbf{Top}\to {\mathbf{Frm}}^{\mathrm{op}}$.

Let $2=\{0<1\}$ and let $P(A)$ be the set of points of a frame $A$, namely frame homomorphisms $A\to 2$. For $a\in A$, put $U_a=\{p:p(a)=1\}$. Preservation of joins and finite meets givesIn the last equality, a join in $2$ is $1$ precisely when at least one summand is $1$, also covering the empty family. Thus the sets $U_a$ are all the opens of a topology, rather than merely a proposed basis.

For a frame homomorphism $h:A\to B$, define $P(h):P(B)\to P(A)$ by $p\mapsto p\circ h$. Since $P(h{)}^{-1}(U_a)=U_{h(a)}$, it is continuous. Composition and identities follow from composition of functions, giving $P:{\mathbf{Frm}}^{\mathrm{op}}\to \mathbf{Top}$.

The frame-point adjunction is the natural bijectionA continuous map $f:X\to P(A)$ determines $h_f(a)=f^{-1}(U_a)$; the displayed identities for $U_a$ show that $h_f$ is a frame homomorphism. Conversely a frame homomorphism $h:A\to O(X)$ determines $f_h(x)(a)=1$ exactly when $x\in h(a)$. Membership in unions and finite intersections proves that $f_h(x)$ is a frame homomorphism to $2$, and $f_h^{-1}(U_a)=h(a)$ proves continuity. These constructions are inverse. Precomposing in $X$ becomes inverse image of open sets; precomposing a frame homomorphism becomes composition of frame points. Hence the bijection is natural in both arguments and proves $O⊣P$.

The adjunction unit isThe adjunction counit at $A$, regarded as an arrow of ${\mathbf{Frm}}^{\mathrm{op}}$, is represented by the frame homomorphism $e_A:A\to O(P(A))$, $a\mapsto U_a$. It is surjective because every open of $P(A)$ is of this form. We can compute $P(e_A):P(O(P(A)))\to P(A)$ explicitly: it sends $q$ to $p=q\circ e_A$. For every open $U_a$ of $P(A)$,Thus ${\eta }_{P(A)}P(e_A)=1$. Conversely, evaluation at $p$ sends $U_a$ to $p(a)$, so $P(e_A){\eta }_{P(A)}=1$. Both maps are continuous by their definitions. Therefore $P(e_A)$ is a homeomorphism. The multiplication of the induced monad $PO$ at $X$ is $P(e_{O(X)})$, which is invertible by this calculation. Consequently the frame-point adjunction is idempotent. This does not assert that every ${\eta }_X$ is a homeomorphism: it says that applying the point-space construction a second time makes no further change.

A monad on $\mathcal{C}$ consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1_{\mathcal{C}}\Rightarrow T$ and $\mu :T^2\Rightarrow T$ satisfying, at every object $A$,These are the unit and associativity laws for the unit and multiplication of a monad.

An algebra for a monad is a pair $(A,a)$ with $a:TA\to A$ satisfyingA monad algebra morphism $f:(A,a)\to (B,b)$ is a morphism $f:A\to B$ with $fa=bTf$. Identities satisfy this condition, and composing two such maps preserves it by functoriality of $T$. The resulting Eilenberg-Moore category is denoted ${\mathcal{C}}^T$. Its forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$ sends $(A,a)$ to $A$ and leaves the underlying arrows unchanged.

For each algebra for a monad $(A,a)$, consider the specific pair in $\mathcal{D}$Then the exact condition isEach pair is reflexive, with common section $F{\eta }_A$: the two composite identities follow from $a{\eta }_A=1_A$ and ${\epsilon }_{FA}F{\eta }_A=1_{FA}$. Thus existence of all reflexive coequalizers is sufficient, but the displayed specified pairs give the necessary-and-sufficient condition.

To prove it, take $h:FA\to d$ and its adjoint transpose $x=Gh{\eta }_A:A\to Gd$. The transpose of $hFa$ is $xa$, by naturality of $\eta$. The transpose of $h{\epsilon }_{FA}$ isMoreover counit naturality and a triangle identity giveBecause transposition is bijective, the coequalizing equation is therefore equivalent toThe right side says exactly that $x:(A,a)\to Kd$ is a monad algebra morphism.

For sufficiency, choose a coequalizer $q_a:FA\to L(A,a)$ for every monad algebra. Its universal property and the equivalence just proved give natural bijectionsTo see explicitly that the chosen objects form a functor, an monad algebra morphism $u:(A,a)\to (B,b)$ makes $q_{b}Fu$ coequalize the pair for $a$: use $ua=bTu$ and naturality of the counit. Hence there is a unique $Lu$ with $Lu{q}_a=q_{b}Fu$. Uniqueness proves its identity and composition laws and the naturality of the displayed bijections. Thus $L⊣K$.

Conversely, suppose $L⊣K$ with unit ${\lambda }_{(A,a)}:(A,a)\to KL(A,a)$. Transpose its underlying arrow $A\to GL(A,a)$ through $F⊣G$ to obtain $q_a:FA\to L(A,a)$. Since the unit is an monad algebra morphism, the equivalence above shows that $q_a$ coequalizes the specified pair. Any coequalizing $h:FA\to d$ transposes to an monad algebra map $(A,a)\to Kd$, which factors uniquely through the unit by $L⊣K$. Transposing back gives a unique factorization of $h$ through $q_a$. Thus $q_a$ is the required coequalizer. This proves both directions of the Left adjoint to the Eilenberg-Moore comparison functor criterion.

Put ${\mathcal{C}}_0=\mathbf{Set}$. The initial free construction $L_0:\mathbf{Set}\to {\mathcal{C}}_1$ sends $A$ to $A\times \mathbb{N}$ with ${\alpha }_1(a,k)=(a,k+1)$. A function $f:A\to UB$ extends uniquely by $(a,k)\mapsto {\beta }_1^k(f(a))$. Its induced monad is $T_{0}A=A\times \mathbb{N}$, with unit $a\mapsto (a,0)$ and multiplication $(a,k,l)\mapsto (a,k+l)$. An algebra for a monad is determined by the action of $1\in \mathbb{N}$, so its Eilenberg-Moore category is ${\mathcal{C}}_1$. Thus this first free–forgetful adjunction is a monadic adjunction without needing the given assumption for $n=0$.

The monadic length counts successive Eilenberg-Moore comparison functor steps until the comparison becomes an equivalence. Fix $0\le m<n$. For $m\ge 1$, the free construction in the preceding part adds the $(m+1)$st operation with no fixed points. For $m=0$, the free first operation also has no fixed points. Consequently the left adjoint $F_{n,m}$ to the composite forgetful functor $U_{n,m}:{\mathcal{C}}_n\to {\mathcal{C}}_m$ is obtained by the one-step free construction, followed by empty higher operations. In particular, its underlying object of ${\mathcal{C}}_m$, unit and multiplication are exactly those of the one-step adjunction:as monads, not merely as object functions. The multiplication agrees because both counits evaluate the same first new operation and the same chains; no later operation has a value on a free object.

By the assumed one-step monadic adjunction property, or the explicit argument above when $m=0$, the Eilenberg-Moore category ${\mathcal{C}}_m^{T_m}$ is equivalent to ${\mathcal{C}}_{m+1}$. Under this equivalence the comparison ${\mathcal{C}}_n\to {\mathcal{C}}_m^{T_m}$ is precisely $U_{n,m+1}$. Indeed, its algebra action evaluates each newly adjoined chain using ${\alpha }_1$, and evaluates the first new value using ${\alpha }_{m+1}$; it retains that operation and ignores all higher ones. The comparison has a left adjoint by the same explicit free extension. Thus the whole monadic tower for nested partial unary operations iswith the remaining comparison at stage $m$ equal to $U_{n,m}$, and stage $n$ the identity.

It does not terminate earlier. If $m<n$, take the same underlying two-element set in two objects of ${\mathcal{C}}_n$, making every operation up to $m$ the identity. In the first object make all remaining operations identities; in the second make ${\alpha }_{m+1}$ exchange the two elements, and leave all subsequent operations undefined. Both satisfy the nesting rule. The identity function is a morphism after forgetting to ${\mathcal{C}}_m$, but is not a morphism between the original objects. Thus $U_{n,m}$ is not a full functor, and cannot be an equivalence. Therefore

Write $\eta :1_{\mathcal{C}}\to GF$ and $\epsilon :FG\to 1_{\mathcal{D}}$ for the unit and counit of an adjunction. The induced monad hasThe Eilenberg-Moore comparison functor sends $D\in \mathcal{D}$ to the algebra for a monadand sends $v:D\to D^{\prime }$ to $Gv$. The triangle identities for an adjunction give the algebra unit law, and naturality of the counit gives the algebra multiplication law and the algebra-morphism equation for $Gv$.

If $\mathcal{D}$ has coequalizers, define the Left adjoint to the Eilenberg-Moore comparison functor at an algebra $(A,a:TA\to A)$ byThe parallel pair is reflexive through $F{\eta }_A$. A morphism $L(A,a)\to D$ is equivalently an arrow $b:FA\to D$ with $bFa=b{\epsilon }_{FA}$. Under $F⊣G$, its transpose is $\bar{b}=Gb{\eta }_A:A\to GD$. The transpose of $bFa$ is $\bar{b}a$, while the transpose of $b{\epsilon }_{FA}$ is $G{\epsilon }_{D}T\bar{b}$. Hence the coequalizer condition is exactlythe equation for a morphism of algebras for a monad $(A,a)\to K(D)$. This supplies the natural bijection proving $L⊣K$, and also defines $L$ on algebra morphisms by the coequalizer universal property.

A monad $\mathbb{T}=(T,\eta ,\mu )$ consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1_{\mathcal{C}}\Rightarrow T$ and $\mu :T^2\Rightarrow T$ satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $a{\mu }_A=aT(a)$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ obeys $ha=bT(h)$. The functor laws show that identities and composites obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^{\mathbb{T}}$.

For an adjunction $F⊣G$, let $\eta$ be its adjunction unit and $\epsilon$ its adjunction counit. The monad induced by an adjunction isThe two unit laws are $G{\epsilon }_{FA}{\eta }_{GFA}=1_{GFA}$ and $G({\epsilon }_{FA}F{\eta }_A)=1_{GFA}$, respectively the two triangle identities for an adjunction.

For associativity, naturality of the adjunction counit at ${\epsilon }_{FA}:FGFA\to FA$ saysApply $G$ to obtain ${\mu }_A{\mu }_{TA}={\mu }_{A}T{\mu }_A$. The composites defining $\eta$ and $\mu$ are natural transformations, so all monad axioms have been checked, not only their underlying objectwise types.

Define the forgetful functor $G^{\mathbb{T}}(A,a)=A$, acting as the identity on underlying morphisms. Define the free algebra functor byThe monad laws make ${\mu }_X$ an algebra for a monad structure, and naturality of $\mu$ makes $T(h)$ an monad algebra morphism.

For $(A,a)$, define the transpose mapsThe map $\Psi (k)$ is an monad algebra morphism, because naturality of $\mu$ and the algebra associativity law giveFor $k:X\to A$, naturality of $\eta$ gives $\Phi \Psi (k)=aT(k){\eta }_X=a{\eta }_{A}k=k$. Conversely, if $h$ is an monad algebra morphism, thenPrecomposing by a map into $X$ or postcomposing by an monad algebra morphism respects both formulas. Thus the bijection is natural and $F^{\mathbb{T}}⊣G^{\mathbb{T}}$. This is the free-forgetful Eilenberg-Moore adjunction.

Its adjunction unit is the given ${\eta }_X$; its adjunction counit at $(A,a)$ is the monad algebra morphism $a:(TA,{\mu }_A)\to (A,a)$. Hence $G^{\mathbb{T}}{F}^{\mathbb{T}}=T$, and the induced multiplication is the underlying counit at $(TX,{\mu }_X)$, namely ${\mu }_X$. Therefore this adjunction induces exactly the original monad, including its unit and multiplication of a monad.

For the monad $(T,\eta ,\mu )$, an algebra for a monad is $(A,a:TA\to A)$ with $a{\eta }_A=1_A$ and $aT(a)=a{\mu }_A$. A morphism of algebras for a monad $f:(A,a)\to (B,b)$ satisfies $fa=bTf$. These objects and arrows form the Eilenberg-Moore category ${\mathcal{C}}^T$.

The free algebra functor sends $A$ to $(TA,{\mu }_A)$ and $f$ to $Tf$. The monad identities verify the algebra laws. For the forgetful functor $U$, the free adjunction iswith inverse $v\mapsto bTv$. The algebra law makes the latter an monad algebra morphism. The identities $h{\eta }_A\mapsto bT(h{\eta }_A)=h{\mu }_{A}T{\eta }_A=h$ and $bTv{\eta }_A=b{\eta }_{B}v=v$ prove the bijection.

In the category of adjunctions inducing a fixed monad, objects are adjunctions $F⊣G$ with their induced monad identified with $T$. A morphism to $F^{\prime }⊣G^{\prime }$ is a functor $H$ between the right-hand categories satisfying $G^{\prime }H=G$, $HF=F^{\prime }$ and compatibility with units and counits. With these strict identifications, defineThe triangle identities give the unit algebra law; naturality of $\epsilon$ at ${\epsilon }_B$ gives the multiplication law. Naturality at $f$ makes $Gf$ an monad algebra morphism. Moreover $UK=G$, $KF=F^T$ because $G{\epsilon }_{FA}={\mu }_A$, and $K{\epsilon }_B$ is the free-adjunction counit at $K(B)$. Hence $K$ is a morphism into the Eilenberg-Moore adjunction.

For any other such $H$, its underlying object at $B$ must be $GB$. Counit compatibility forces its algebra action to be $G{\epsilon }_B$, and the forgetful functor, which is a faithful functor, forces $H(f)=Gf$. Thus $H=K$. This proves terminality of the Eilenberg-Moore adjunction. If adjunctions are specified only up to coherent isomorphisms, the same argument gives uniqueness up to the corresponding compatible natural isomorphism.

A monad consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1\Rightarrow T$ and $\mu :T^2\Rightarrow T$, the unit and multiplication of a monad, satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $a{\mu }_A=aTa$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ satisfies $ha=bTh$. These objects and morphisms form the Eilenberg-Moore category ${\mathcal{C}}^T$; its composition works because $T$ is a functor.

For the list monad, $TX={\coprod }_{n\ge 0}{X}^n$ is the set of finite ordered lists, including the empty list. The map $Tf$ applies $f$ to each entry; ${\eta }_X(x)=[x]$ and ${\mu }_X$ concatenates a list of lists. The unit laws say that adding singleton brackets and then flattening changes nothing. Associativity says that flattening a list of lists of lists in either order produces the same ordered sequence. These descriptions also prove naturality.

If $a:TX\to X$ is an algebra for a monad, defineThe singleton law gives $a([x])=x$. Apply the algebra associativity law to $[[],[x]]$ and $[[x],[]]$ to obtain $e\ast x=x=x\ast e$. Applying it to $[[x,y],[z]]$ and $[[x],[y,z]]$ showsThus $(X,\ast ,e)$ is a monoid. Applying the same law to $[[x_1,\dots ,x_{n-1}],[x_n]]$ shows inductively that $a$ is necessarily ordered multiplication of its entries, with the empty product $e$.

Conversely, any monoid defines such a list-fold map $a$. The monoid unit proves $a\eta =1$, and associativity and the unit prove that multiplying flattened lists equals multiplying their individual products, including empty sublists. Hence $a\mu =aTa$. An algebra morphism preserves the empty-list value and two-entry-list values, so it is a monoid homomorphism; conversely a monoid homomorphism preserves every ordered product and is an algebra morphism. ThereforeThus list-monad algebras are monoids, with the identification also matching every morphism.

For an algebra for a monad $(A,a)$, consider the fork in the Eilenberg-Moore categoryHere ${\mu }_A:F(TA)\to F(A)$ is the counit at $F(A)$, and $Ta=F(a)$. The arrow $a$ is an algebra morphism by $a{\mu }_A=aTa$, which also says that it coequalizes the two arrows.

Let $h:F(A)\to (B,b)$ be an algebra morphism with $h{\mu }_A=hTa$. Define $g=h{\eta }_A:A\to B$. Naturality of $\eta$ at $a$ givesSince $h$ is an algebra morphism,Thus $g:(A,a)\to (B,b)$ is an algebra morphism with $ga=h$. Any other such factorization satisfies $g^{\prime }=g^{\prime }a{\eta }_A=h{\eta }_A=g$. This proves the full coequalizer universal property inside the algebra category.

The pair is moreover a reflexive pair: its common section is $F({\eta }_A)$, with underlying map $T{\eta }_A$, because ${\mu }_{A}T{\eta }_A=1_{TA}$ and $TaT{\eta }_A=T(a{\eta }_A)=1_{TA}$. Hence every monad algebra is a reflexive coequalizer of free algebras. This reflexive free-algebra presentation of a monad algebra needs no general existence theorem for arbitrary algebra-category colimits.