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If $\mathcal{C}$ has finite colimits and a monad $T$ preserves reflexive coequalizers, its Eilenberg-Moore category has finite colimits. The forgetful functor creates reflexive coequalizers preserved by $T$. The coproduct presentation for monad algebras yields binary coproducts, and $F0$ is initial. Any pair $a,b:X\rightrightarrows Y$ can then be replaced by the reflexive pair $[a,1],[b,1]:X+Y\rightrightarrows Y$, with identical coequalizers. Finite coproducts and coequalizers give all finite colimits.

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.

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 $\mathcal{C}$.

Take the definition of an elementary topos as a category with finite limits, exponential objects and a subobject classifier $\top :1\hookrightarrow \Omega$. Its power object is $PX={\Omega }^X$, and $Pf:PY\to PX$ is precomposition with $f:X\to Y$. Transposing a predicate on $X\times Y$ in either variable givesThus $P^{\mathrm{op}}:\mathcal{E}\to {\mathcal{E}}^{\mathrm{op}}$ is left adjoint to the contravariant power-object functor $P:{\mathcal{E}}^{\mathrm{op}}\to \mathcal{E}$.

Here are the remaining hypotheses for the crude monadicity theorem, obtained from those topos axioms. The unit of this adjunction is ${\eta }_X:X\to PPX$, internally ${\eta }_X(x)(S)=(x\in S)$. It is monic: if two such evaluations agree, evaluate on the singleton predicate $\{x\}$ classified by the diagonal to obtain $x=y$. If $Pf$ is invertible, naturality $PPf{\eta }_X={\eta }_{Y}f$ therefore makes $f$ monic. The characteristic map of this mono, regarded as a global element of $PY$, pulls back along $f$ to the everywhere-true predicate on $X$. Since $Pf$ is monic, that characteristic map was already everywhere true on $Y$. The classifier pullback then says that $f$ is invertible. Hence $P$ is a conservative functor.

It remains to check preservation of reflexive coequalizers in the opposite category. Equivalently, let $f,g:B\rightrightarrows A$ be a coreflexive pair, with $r:A\to B$ satisfying $rf=rg=1_B$, and let $e:E\hookrightarrow B$ be their equalizer. Both $f$ and $g$ are monic. For any mono $i$, there is a direct-image map ${\exists }_i$ between power objects: a subobject is sent to its composite with $i$. This uses only classification of monos, not the prior existence of general images or colimits. We have $Pi{\exists }_i=1$.

For a predicate $W\hookrightarrow B$, consider its direct image along $f$. Its pullbacks along the two sections areTo verify the second equality, $g(b)=f(w)$ with $w\in W$ implies $b=w$ after applying $r$, and therefore $f(b)=g(b)$; conversely $b\in W\cap E$ supplies that witness. This argument works for parameterized subobjects as well, so it is an equality of arrows between power objects.

Now if $h:PB\to Z$ satisfies $hPf=hPg$, then $h=h{\exists }_{e}Pe$. Hence $h{\exists }_e:PE\to Z$ is its unique factorization through $Pe$, uniqueness following from the section ${\exists }_e$. Thereforeis a coequalizer. Finite limits supply every required coreflexive equalizer. The right adjoint $P$ reflects isomorphisms and preserves the corresponding reflexive coequalizers, so $P$ is monadic. In particular ${\mathcal{E}}^{\mathrm{op}}$ is equivalent to the Eilenberg-Moore category of the double-power-object monad on $\mathcal{E}$.

Let $F:\mathcal{E}\to \mathcal{F}$ be a logical functor, and suppose $L⊣F$. Preservation of exponentials and the classifier gives $F{P}_{\mathcal{E}}\cong P_{\mathcal{F}}{F}^{\mathrm{op}}$, compatibly with the units, counits and resulting monads. Thus $F^{\mathrm{op}}$ is the algebra functor lifting the base functor $F$ through the two monadic power-object functors. The adjoint lifting theorem for monad algebra functors applies to the base adjunction $L⊣F$: its required coequalizers exist in ${\mathcal{E}}^{\mathrm{op}}$, because $\mathcal{E}$ has finite equalizers. It supplies a left adjoint $K⊣F^{\mathrm{op}}$. Taking opposites gives $F⊣K^{\mathrm{op}}$, the required right adjoint to $F$.

Preserving either of the two logical structures by itself is insufficient. For the exponential example take $F:\mathbf{Set}\to \mathbf{Set}$ constant at $1$. The constant-empty functor is its left adjoint, since both relevant hom-sets are singletons. Its canonical exponential comparison is $1\cong 1^1$, so it preserves exponentials. It has no right adjoint: a functor with a right adjoint would preserve the initial object, whereas $F(\varnothing )=1$.

For the classifier example take $G:(\mathbb{Z}/2)\text{-}\mathbf{Set}\to \mathbf{Set}$ to be fixed points. The trivial-action functor is left adjoint to $G$. In a group-action topos the classifier is the trivial-action two-element set, since invariant subsets have ordinary equivariant characteristic maps. Therefore $G$ preserves the classifier, its true arrow and the terminal object. But $G$ does not preserve the coequalizer of the identity and the nontrivial translation on the regular two-element group set: that coequalizer is $1$, while the fixed-point sets of the domain and codomain of the parallel pair are empty. Their set-theoretic coequalizer is empty, not $G(1)=1$. Thus $G$ has no right adjoint.

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 $L⊣K$.

Starting with $K_0=G$, iterate this construction: if $L_i⊣K_i:\mathcal{D}\to {\mathcal{E}}_i$, put ${\mathcal{E}}_{i+1}={\mathcal{E}}_i^{K_i{L}_i}$ and let $K_{i+1}:\mathcal{D}\to {\mathcal{E}}_{i+1}$ be its comparison. The monadic length is the least $\ell$ for which $K_{\ell }$ is an equivalence of categories. Thus an equivalence has length zero, and an already monadic adjunction whose right adjoint is not an equivalence has length one. The assumed reflexive coequalizers ensure the needed left adjoints; in the example we identify them explicitly.

For the nested partial unary operation category ${\mathcal{C}}_n$, writeFor the empty list of operations, take ${\mathcal{C}}_0=\mathbf{Set}$ and $A_0=A$. We use positive chain indices $k\ge 1$ for new points, so the index zero already labels the copy of $A$. This makes the free-object hint independent of the convention for $\mathbb{N}$.

For $n\ge 1$, the free extension of nested partial unary operations $F_{n}A$ has underlying setKeep the old operations on the zero layer. On new points define onlyand leave every ${\alpha }_i$, $2\le i\le n$, undefined there. Define the new top operation byand leave it undefined elsewhere. These are precisely the permitted domains: the first unary operation on every new point moves it, so none of the higher operations is defined there. When $n=0$, take $F_{0}A=A\times {\mathbb{N}}_{\ge 0}$ with ${\alpha }_1(a,k)=(a,k+1)$. In either case the unit is $a\mapsto (a,0)$.

For $B\in {\mathcal{C}}_{n+1}$, let its operations be ${\beta }_i$. Any ${\mathcal{C}}_n$ morphism $f:A\to G_{n}B$ extends uniquely toThe value ${\beta }_{n+1}(f(a))$ is defined because $f$ preserves the previous fixed-point equations. The extension respects the top operation at old fixed points and respects the first operation along each new chain; there are no other defined operations on those chains to check. Uniqueness follows from the same forced formulas. For $n=0$ the formula is $\bar{f}(a,k)={\beta }_1^k(f(a))$. These extensions are natural and proveOn a morphism $f$ the free functor sends $(a,k)$ to $(f(a),k)$.

Crucially, the top operation of $F_{n}A$ has no fixed points: its defined old values move to new points, and it has no defined new values when $n\ge 1$; for $n=0$ it is the shift. The next free extension therefore adds no points and equips the object with an everywhere-undefined next operation. The same remains true at every subsequent free extension. Hence, for $m>n$, the composite left adjoint $F_{m-1}\cdots F_n$ has the same first $n+1$ operations and underlying set as $F_n$.

This agreement includes the monad structure, not just its underlying endofunctor. The units are the same zero-layer inclusions. The composite counit on an object of ${\mathcal{C}}_m$ is the same forced extension formula, depending only on its first $n+1$ operations. Applying the composite forgetful functor therefore gives the same multiplication $G_n\epsilon F_n$. Thus

Next prove creation of the required split coequalizers. Suppose $f,g:B\rightrightarrows C$ are arrows in ${\mathcal{C}}_{n+1}$ and their images have a split coequalizer $q:C\to Q$ in ${\mathcal{C}}_n$. Choose the splitting orientationwith $s:Q\to C$, $t:C\to B$ morphisms in ${\mathcal{C}}_n$. Denote the top operations on $B,C$ by $b,c$. For $z\in Q_n$ defineSince $s$ preserves the lower operations, $s(z)\in C_n$ and this value exists. For $y\in C_n$, $t(y)\in B_n$, so preservation by $f,g$ givesThus $q$ respects the top operation. The splitting $s$ and $qs=1$ also show that this is the unique operation on its prescribed domain $Q_n$ for which $q$ is a morphism. For $n=0$ all these domains mean the whole underlying set.

If $u:C\to D$ is a ${\mathcal{C}}_{n+1}$ morphism equalizing $f,g$, its unique factor $v:Q\to G_{n}D$ in ${\mathcal{C}}_n$ satisfies, on $Q_n$,where $d$ is the top operation of $D$. Hence $v$ is a ${\mathcal{C}}_{n+1}$ morphism. This proves $G_n$ creates coequalizers of $G_n$-split pairs.

Moreover, $G_n$ reflects isomorphisms: an isomorphism of the first $n$ operations reflects their fixed-point conditions, hence the domain of the next operation; the inverse of a bijective top-operation-preserving map then preserves that operation too. The Beck monadicity theorem now givesUnder this equivalence, the first Eilenberg-Moore comparison functor for the long composite is exactly forgetting all operations after the $(n+1)$st: its algebra structure is the top-operation extension formula already computed. Repeating the argument identifies its monadic tower with the successive categoriesAt each stage the corresponding comparison has the remaining composite of the already constructed free extensions as a left adjoint.

No earlier stage can be an equivalence of categories. For $j<m$, take the two-point set with its first $j$ operations all identities. One extension to ${\mathcal{C}}_m$ has every later operation the identity; another has its $(j+1)$st operation interchange the two points, and all subsequent operations undefined. Both have the same image in ${\mathcal{C}}_j$, but the identity function between those images is not a morphism between the extensions. Thus forgetting from ${\mathcal{C}}_m$ to ${\mathcal{C}}_j$ is not full. The final comparison at $j=m$ is an equivalence. This proves the monadic tower for nested partial unary operations has

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