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Write the adjunction unit and adjunction counit of $F⊣G$ as $\eta ,\epsilon$. Its induced monad is $T=GF$, with multiplication $\mu =G{\epsilon }_F$. The Eilenberg-Moore comparison functor isThe unit law for $a_d$ is $G{\epsilon }_d{\eta }_{Gd}=1_{Gd}$, a triangle identity for an adjunction. Naturality of $\epsilon$ at ${\epsilon }_d$ gives ${\epsilon }_{d}FG{\epsilon }_d={\epsilon }_d{\epsilon }_{FGd}$; applying $G$ gives $a_{d}T{a}_d=a_d{\mu }_{Gd}$. Naturality at $h$ gives the monad algebra morphism equation for $Gh$. Thus $K$ is well-defined.

For each $d$, consider the fork in $\mathcal{D}$It commutes by the naturality equation just used. Its image under $G$ is the fork $T^2(Gd)\rightrightarrows T(Gd)\to Gd$ of the preceding question, with parallel morphisms ${\mu }_{Gd},T{a}_d$ and final map $a_d$. It is split using $s={\eta }_{Gd}$ and $t={\eta }_{T(Gd)}$. By the assumed reflection, ${\epsilon }_d$ is the coequalizer of this pair in $\mathcal{D}$.

For faithfulness, if $Gh=Gk$ for $h,k:d\to e$, naturality gives $h{\epsilon }_d={\epsilon }_{e}FGh={\epsilon }_{e}FGk=k{\epsilon }_d$. A coequalizer is an epimorphism, so $h=k$.

For fullness, let $u:K(d)\to K(e)$ be a monad algebra morphism. Thus $u:Gd\to Ge$ satisfies $u{a}_d=a_{e}Tu$. Set $v={\epsilon }_{e}Fu:FGd\to e$. We verify explicitly that $v$ equalizes the two morphisms of the displayed fork. The monad algebra morphism equation givesNaturality of $\epsilon$ first at $Fu$ and then at ${\epsilon }_e$ givesThe coequalizer therefore supplies a unique $h:d\to e$ with $h{\epsilon }_d=v$. Applying $G$ yields $Gh{a}_d=Gv=a_{e}Tu=u{a}_d$. The map $a_d$ is split epic with section ${\eta }_{Gd}$, so cancellation gives $Gh=u$. This proves fullness, and the earlier cancellation proves faithfulness. HenceNo claim that $K$ is essentially surjective is needed; that stronger conclusion needs the corresponding existence hypotheses.

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 $\mathcal{D}$.

Let $T:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathcal{D}$. A categorical end is an object $E$ with morphisms $e_A:E\to T(A,A)$ satisfying, for every $f:A\to B$,It is universal among such families: any $x_A:X\to T(A,A)$ satisfying the same equations is uniquely $e_{A}h$ for one $h:X\to E$. This is a universal dinatural transformation from the constant value $E$, and is written $E={\int }_{A}T(A,A)$.

Dually a categorical coend is $Q$ with morphisms $j_A:T(A,A)\to Q$ satisfyingAny family $x_A:T(A,A)\to X$ satisfying these equations is uniquely $h{j}_A$ for one $h:Q\to X$. Write $Q={\int }^{A}T(A,A)$. For small $\mathcal{C}$, the categorical end can be constructed as the equalizer of ${\prod }_{A}T(A,A)\rightrightarrows {\prod }_{f:A\to B}T(A,B)$ when these objects exist. The categorical coend is the coequalizer of ${\coprod }_{f:A\to B}T(B,A)\rightrightarrows {\coprod }_{A}T(A,A)$ when those objects exist. The two morphisms impose exactly the displayed equations for a dinatural transformation.

A regular category has finite limits, factorizations of every arrow as a regular epimorphism followed by a monomorphism, and pullback-stable regular epimorphisms. Equivalently one may require coequalizers of kernel pairs and pullback stability of the resulting quotients. A regular epimorphism is a coequalizer of some parallel pair; an extremal epimorphism is an epimorphism $e$ such that $e=mh$, with $m$ monic, forces $m$ to be invertible.

First, every regular epimorphism is extremal without assuming regularity of the whole category. Suppose $e:A\to B$ coequalizes $r,s$ and $e=mh$. Since $m$ is monic, $hr=hs$, so the universal property gives $k:B\to \mathrm{dom}(m)$ with $ke=h$. Then $mke=e$; cancellation of the epimorphism $e$ gives $mk=1_B$. Monicity of $m$ now gives $km=1$, so $m$ is an isomorphism. Conversely, in a regular category, factor an extremal $e$ as $mq$ with $q$ regular epic and $m$ monic. Extremality makes $m$ invertible, so $e$ is regular epic. Thus covers and regular epimorphisms coincide.

For a counterexample, work in the category of small categories, which has finite limits constructed on objects and arrows. Let $C$ be two disjoint walking arrows $a\stackrel{u}{\to }b$ and $c\stackrel{v}{\to }d$. Let $D$ be the walking isomorphism, with two objects $0,1$ and inverse arrows $p:0\to 1$, $q:1\to 0$. Define $F:C\to D$ by $a,d\mapsto 0$, $b,c\mapsto 1$, $u\mapsto p$, $v\mapsto q$. These images generate all of $D$, so two functors agreeing after $F$ agree everywhere; thus $F$ is epic.

A monic functor is injective on objects and faithful, as tested by functors from the terminal category and from the walking arrow. If $F$ factors through such a functor $m:D^{\prime }\to D$, its image contains both objects and both arrows $p,q$. Faithfulness forces their lifted composites to be the identities, so the image contains every arrow of $D$ and $m$ is an isomorphism. Hence $F$ is extremal.

It is not regular. Its kernel pair identifies the endpoint pairs $a,d$ and $b,c$ but has no relation forcing the composites of the images of $u,v$ to be identities. Concretely, send all four objects to the one-object category with endomorphism monoid $\{1,e\}$, $e^2=e\ne 1$, and send both $u,v$ to $e$. This functor equalizes the kernel pair: corresponding identities agree, and the only nonidentity arrow pairs in the kernel pair are an arrow paired with itself. It cannot factor through $D$, because images of $p,q$ would have to be inverse, whereas $e^2\ne 1$. In a category with kernel pairs, every regular epimorphism is the coequalizer of its kernel pair: its original coequalized pair factors through that kernel pair, giving the universal property. Therefore $F$ is an extremal but nonregular epimorphism, the promised nonregular extremal epimorphism in the category of categories.

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.

Let $T:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathcal{D}$. An end of a functor is an object $E$ with maps $e_C:E\to T(C,C)$ satisfying $T(1_C,f)e_C=T(f,1_D)e_D$ for every $f:C\to D$, universal among all such families. Thus every compatible family with vertex $A$ factors uniquely through $E$. Write it as ${\int }_{C}T(C,C)$. These are the wedge equations for a dinatural transformation from a constant functor to $T$.

Dually, a coend of a functor is an object $Q$ with maps $i_C:T(C,C)\to Q$ satisfying $i_{C}T(f,1_C)=i_{D}T(1_D,f)$ as maps from $T(D,C)$, universal among compatible families with a common target. Write it as ${\int }^{C}T(C,C)$. These are cowedge equations, and every cowedge to $A$ factors uniquely through $Q$.

For small $\mathcal{C}$, if the needed products and equalizers exist, the end is the equalizer of ${\prod }_{C}T(C,C)\rightrightarrows {\prod }_{f:C\to D}T(C,D)$. If the needed coproducts and coequalizers exist, the coend is the coequalizer of ${\coprod }_{f:C\to D}T(D,C)\rightrightarrows {\coprod }_{C}T(C,C)$. The two arrows in each construction are precisely the two sides of the compatibility equation. This explains both the mixed variance and the universal property, rather than treating the integral notation as an ordinary numerical integral.

For $T=GF$, define the Eilenberg-Moore comparison functorThe algebra unit law is a triangle identity. Naturality of the counit at ${\epsilon }_D$ gives its algebra associativity law; naturality at $h:D\to E$ gives $Gh{a}_D=a_{E}T(Gh)$, making $K(h)$ an algebra morphism.

Suppose first that $G$ reflects G-split coequalizers. For every $D$, consider the counit presentationIt equalizes the pair by counit naturality. After applying $G$ it becomes $T^{2}GD\rightrightarrows TGD\stackrel{a_D}{\to }GD$, with parallel maps ${\mu }_{GD},T(a_D)$. This is split: take $s={\eta }_{GD}$ and $t={\eta }_{TGD}$. The identities $a_{D}s=1$, ${\mu }_{GD}t=1$ and $T(a_D)t={\eta }_{GD}{a}_D$ follow from the algebra law, a monad unit law, and naturality of $\eta$. Reflection therefore makes ${\epsilon }_D$ a coequalizer in $\mathcal{D}$.

For faithfulness, if $Gh=Gk$ then naturality gives $h{\epsilon }_D={\epsilon }_{E}FGh={\epsilon }_{E}FGk=k{\epsilon }_D$. Coequalizers are epic, so $h=k$.

For fullness, let $\alpha :GD\to GE$ be an algebra morphism, so $\alpha a_D=a_{E}T\alpha$. Put $h_0={\epsilon }_{E}F\alpha :FGD\to E$. Its two composites with the counit-presentation pair are equal: by counit naturality they are ${\epsilon }_{E}F(a_{E}T\alpha )$ and ${\epsilon }_{E}F(\alpha a_D)$. Hence there is a unique $h:D\to E$ with $h{\epsilon }_D=h_0$. Applying $G$ gives $Gh{a}_D=a_{E}T\alpha =\alpha a_D$. Since $a_D$ is split epic, $Gh=\alpha$. This proves full faithfulness directly by descent.

Conversely suppose $K$ is full and faithful. We need the following explicit algebra lifting fact. For algebra morphisms $f,g:(A,a)\rightrightarrows (B,b)$ with a split underlying coequalizer $q:B\to Q$, the arrow $qb:TB\to Q$ equalizes $Tf,Tg$. The split diagram is preserved by $T$, so $Tq$ is their coequalizer and there is a unique $c:TQ\to Q$ with $cTq=qb$. Precomposing the unit equation with the epimorphism $q$ gives $c{\eta }_{Q}q=q$; precomposing the associativity equation with the epimorphism $T^{2}q$ gives equality by $bT(b)=b{\mu }_B$. Thus $c$ is an algebra action. If an algebra map $h:B\to (Z,z)$ equalizes $f,g$, its unique underlying factor $k:Q\to Z$ satisfies $kcTq=hb=zTh=zTkTq$. Cancelling $Tq$ shows $kc=zTk$. This proves monad algebra forgetful functor creates split coequalizers.

Now take any existing equalizing arrow $e:B\to D$ in $\mathcal{D}$ whose image is such a coequalizer. The lifting fact makes the unique compatible action on $GD$ its already given action $a_D$, since $Ke$ preserves that action. Hence $Ke$ is a coequalizer of $Kf,Kg$ in the algebra category. For any arrow $h:B\to Z$ equalizing the pair, its unique algebra factor $KD\to KZ$ lifts to an arrow $D\to Z$ by fullness of $K$; faithfulness gives its factorization equation and uniqueness. Therefore $e$ is a coequalizer in $\mathcal{D}$. We have proved the full comparison and reflection of split coequalizers equivalence