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An adjunction is comonadic when the comparison from its left-hand category to coalgebras for the induced comonad is an equivalence. The dual Beck theorem tests this using reflected isomorphisms and suitable equalizers.

The relevant criterion is the Beck comonadicity theorem, dual to Beck's monadicity theorem.

An adjunction is monadic when its comparison functor from the right-hand category to the Eilenberg-Moore category of the induced monad is an equivalence.

The induced monad determines this comparison; Beck's monadicity theorem gives a criterion for it to be an equivalence of categories.

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.

Put $P(A)={\Omega }^A$, the power object, with $P(f)$ acting by inverse image of predicates. Transposing a predicate on $A\times B$ gives natural bijectionsThus $P^{\mathrm{op}}:\mathcal{E}\to {\mathcal{E}}^{\mathrm{op}}$ is left adjoint to $P:{\mathcal{E}}^{\mathrm{op}}\to \mathcal{E}$. Its unit ${\eta }_A:A\to PPA$ sends $a$ to evaluation at $a$, namely the predicate $S\mapsto (a\in S)$. This unit is monic: two generalized elements having the same evaluation on all predicates agree by testing the singleton predicate of the first, or equivalently its graph in the relevant parameter product.

We verify a sufficient form of Beck's monadicity theorem: a right adjoint is monadic if it reflects isomorphisms and its domain has, and it preserves, coequalizers of reflexive pairs. Suppose $Pf$ is invertible. Then $PPf$ is invertible, and naturality of the monic unit shows that $f$ is monic. Regard it as the subobject $A\hookrightarrow B$. Its characteristic predicate and the constantly true predicate have the same restriction to $A$, so injectivity of $Pf$ makes them equal. Thus this subobject is all of $B$, and $f$ is invertible. Hence $P$ reflects isomorphisms.

A reflexive pair in ${\mathcal{E}}^{\mathrm{op}}$ is a coreflexive pair $f,g:B\rightrightarrows A$ in $\mathcal{E}$, with a common retraction $r$ such that $rf=rg=1_B$. Its equalizer $e:E\hookrightarrow B$ exists. We show that $Pe:PB\to PE$ is the coequalizer of $Pf,Pg:PA\rightrightarrows PB$. Direct image of subobjects along the mono $f$ defines ${\exists }_f:PB\to PA$, and similarly ${\exists }_e:PE\to PB$. At every parameter object the identities areFor the middle identity, an equality $f(b)=g(b^{\prime })$ forces $b=b^{\prime }$ after applying $r$; their common point must therefore lie in the equalizer. This proves the identity for parameterized predicates, not merely for global elements. If $h:PB\to Z$ satisfies $hPf=hPg$, composing with ${\exists }_f$ gives $h=h{\exists }_{e}Pe$. Thus it factors through $Pe$, and the last displayed identity gives uniqueness. This proves that power objects turn coreflexive equalizers into coequalizers. All hypotheses of the monadicity criterion have now been checked, so $P$ is monadic, with

A logical functor $L:\mathcal{E}\to \mathcal{F}$ preserves finite limits, exponentials and the classifier, hence gives compatible natural isomorphisms $LP\cong PL$ and $LPP\cong PPL$. They respect units and multiplications, because these are defined by evaluation and transposition. Under the displayed equivalences, $L^{\mathrm{op}}$ is therefore the induced functor on double-power Eilenberg-Moore categories. Their forgetful functors create finite limits, and the underlying $L$ preserves them. Consequently the induced functor preserves finite limits, so $L^{\mathrm{op}}$ does too. Reversing arrows proves every logical functor preserves finite colimits.

For the first counterexample, the constant-terminal functor $F:\mathbf{Set}\to \mathbf{Set}$ has the constant-empty functor as left adjoint: both sets of maps in the adjunction are singletons. It preserves exponentials, since $F(B^A)=1=1^1=(FB{)}^{FA}$ with the canonical comparison. It does not preserve the initial object, because $F(0)=1$. Thus preservation of exponentials alone is insufficient.

For the second, take the fixed-point functor $\Gamma :G\text{-}\mathbf{Set}\to \mathbf{Set}$ for $G=\mathbb{Z}/2$. Its left adjoint gives a set the trivial action. The classifier of $G$-sets is the two-element set with trivial action, so $\Gamma$ preserves it and its truth map. Let $X$ be the free two-element $G$-set. The coequalizer of its identity and its swapping automorphism is the singleton $G$-set. Applying $\Gamma$ gives two maps $0\rightrightarrows 0$ followed by $1$, whereas their coequalizer in sets is $0$. Thus $\Gamma$ preserves the classifier but not all finite colimits. These examples also explain why both requirements in logicalness matter.