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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).

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.

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.

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.

The free algebra functor isThe monad laws make ${\mu }_A$ an algebra action, and naturality of $\mu$ makes $Tf$ a morphism of algebras for a monad. Let $U:{\mathcal{C}}^T\to \mathcal{C}$ forget the action. DefineThe proposed inverse is an algebra morphism becauseNaturality of $\eta$ and the algebra unit law give $(bTf){\eta }_A=f$. If $h$ is an algebra morphism, thenThe formulas are natural in both variables, so $F⊣U$. Its unit is $\eta$, and its counit at $(B,b)$ has underlying map $b:TB\to B$. Thus the monad induced by an adjunction has endofunctor $UF=T$, unit $\eta$, and multiplication $U{\epsilon }_F=\mu$. It is exactly the original monad, not merely a monad with the same endofunctor.

For the monad $\mathbb{T}=(T,\eta ,\mu )$ define the free algebra functor byThe unit and associativity identities of the monad make $(TX,{\mu }_X)$ an algebra for a monad, and naturality of $\mu$ makes $Tf$ a morphism of algebras for a monad. Let $G^{\mathbb{T}}$ be the forgetful functor. For $(A,a)$ in the Eilenberg-Moore category, setThe inverse candidate is a monad algebra morphism, sinceThe two composites are identities:These use respectively the unit law for a monad algebra, the algebra-morphism equation, and a unit identity of the monad. The formulas commute with precomposition in $X$ and postcomposition by monad algebra morphisms, so they form a natural bijection. Therefore the free-algebra functor is left adjoint to forgetting:Its adjunction unit is ${\eta }_X$, and its adjunction counit at $(A,a)$ is $a:(TA,{\mu }_A)\to (A,a)$.

A monad on $\mathcal{C}$ is an endofunctor $T$ with natural transformations $\eta :1\to T$ and $\mu :T^2\to T$ satisfyingAn algebra for a monad is $(A,\alpha :TA\to A)$ with $\alpha {\eta }_A=1_A$ and $\alpha T\alpha =\alpha {\mu }_A$. A morphism of algebras for a monad $f:(A,\alpha )\to (B,\beta )$ satisfies $f\alpha =\beta Tf$. These form the Eilenberg-Moore category ${\mathcal{C}}^T$. Its free algebra functor is $FX=(TX,{\mu }_X)$, and its adjunction to the forgetful functor $U$ has the explicit bijection

Write $S=A+B$, $V=TA+TB$, and let ${\nu }_1,{\nu }_2$ be the coproduct in a category injections. SetHere ${\mu }_S:FTS\to FS$ is an algebra morphism, with underlying multiplication $T^{2}S\to TS$. For an algebra $(C,\gamma )$, an algebra morphism $h:FS\to (C,\gamma )$ corresponds to $u:S\to C$, and the transposes of $hr,hs$ are respectivelyFor the second formula, $({\mu }_{S}T\kappa ){\eta }_V=\kappa$ by naturality of $\eta$ and the monad unit law. Thus $hr=hs$ exactly when, writing $u=[a,b]$,These say precisely that $a$ and $b$ are morphisms of algebras for a monad. Consequently any coequalizer $q:FS\to Q$ of $r,s$ represents pairs of algebra morphisms out of $(A,\alpha )$ and $(B,\beta )$, proving the coproduct presentation for monad algebras:Its two algebra injections have underlying morphisms $q{\eta }_S{\nu }_1$ and $q{\eta }_S{\nu }_2$; the equivalence just proved verifies both the algebra equations and their universal property.

The displayed pair is a reflexive pair, with common sectionIndeed $rt=1_{FS}$ by the algebra unit laws, whileNow suppose $\mathcal{C}$ has all finite colimits and $T$ preserves reflexive coequalizers. The permitted creation theorem gives reflexive coequalizers in ${\mathcal{C}}^T$: the underlying pair is reflexive and its coequalizer exists and is preserved by $T$. The preceding construction therefore gives binary algebra coproducts in a category. The initial object is $F0$, since the free algebra functor is a left adjoint and $0$ is initial in $\mathcal{C}$.

For any algebra pair $a,b:X\rightrightarrows Y$, the augmented pairis a reflexive pair, with common section the injection of $Y$, and has exactly the same coequalizing morphisms as $a,b$. Hence all coequalizers exist in ${\mathcal{C}}^T$. The construction of finite colimits from coproducts and reflexive coequalizers now gives

For $L⊣R$ inducing $T$, the Kleisli comparison functor is always full and faithful, since the adjunction givesTherefore it is part of an equivalence of categories if and only if every object of $\mathcal{D}$ is isomorphic to $LA$ for some $A$.

For an idempotent monad, multiplication is invertible. The unit laws imply $T\eta =\eta T={\mu }^{-1}$. If $(A,a)$ is an algebra for a monad, then $a{\eta }_A=1_A$, while naturality of $\eta$ givesThus $a:TA\to A$ is an isomorphism with inverse ${\eta }_A$. The algebra law $a{\mu }_A=aT(a)$ says that $a$ is a morphism of algebras for a monad from the free algebra $(TA,{\mu }_A)$ to $(A,a)$, so every object of the Eilenberg-Moore category is isomorphic to a free algebra. The criterion above yields