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A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying $\mu T\eta =\mu \eta T=1_T$ and $\mu T\mu =\mu \mu T$. Its Eilenberg-Moore category has objects $(X,a)$ with $a:TX\to X$, $a{\eta }_X=1_X$ and $a{\mu }_X=aTa$. A morphism $f:(X,a)\to (Y,b)$ satisfies $fa=bTf$. An adjunction $F⊣G$, with unit $\eta$ and counit $\epsilon$, induces $T=GF$, $\mu =G\epsilon F$. It is a monadic adjunction if the Eilenberg-Moore comparison functor $K(D)=(GD,G{\epsilon }_D)$ is an equivalence with the Eilenberg-Moore category.

A fork $A\stackrel{p,q}{\rightrightarrows }B\stackrel{r}{\to }C$ is a split coequalizer if $rp=rq$ and there are $s:C\to B$, $t:B\to A$ with $rs=1$, $pt=1$ and $qt=sr$. It is indeed a coequalizer: if $hp=hq$, then $h=hpt=hqt=hsr$, so $h$ factors uniquely through $r$, which is split epic. Every functor preserves these splitting identities, making such coequalizers absolute.

The precise monadicity theorem states: for an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$, the comparison is an equivalence if and only if $G$ reflects isomorphisms and every pair in $\mathcal{D}$ whose $G$-image admits a split coequalizer has a coequalizer which $G$ preserves. Equivalently, the right adjoint creates these split-underlying coequalizers, with creation understood up to the canonical isomorphism of the prescribed underlying coequalizer. Here is a proof.

First consider the forgetful functor from monad algebras. It reflects isomorphisms: the inverse of an invertible algebra morphism also obeys the algebra-morphism identity, by multiplying that identity by the inverses. For an algebra pair with a split underlying coequalizer $r:B\to C$, all powers of $T$ preserve its coequalizer. If $b:TB\to B$ is the algebra structure on $B$, the map $rb$ coequalizes $Tp,Tq$, so there is a unique $c:TC\to C$ withThe unit law follows by composing with the epimorphism $r$: $c{\eta }_{C}r=cTr{\eta }_B=rb{\eta }_B=r$. For associativity, naturality and the algebra law giveCancel the epimorphism $T^{2}r$ to obtain $c{\mu }_C=cTc$. Thus $(C,c)$ is an algebra and $r$ is an algebra morphism. Any algebra morphism coequalizing the pair factors through $r$ on underlying objects, and its factor is an algebra morphism by cancellation of the epimorphism $Tr$. This creates the coequalizer. Transporting these facts through an equivalence proves necessity.

For sufficiency, let $(X,a)$ be a $T$-algebra. In $\mathcal{D}$ consider the free pairIts underlying pair is ${\mu }_X,Ta:T^{2}X\rightrightarrows TX$, with split coequalizer $a:TX\to X$: the splittings are $s={\eta }_X$ and $t={\eta }_{TX}$, using the algebra unit law, the monad unit law and naturality of $\eta$. By hypothesis its coequalizer $q:FX\to D$ exists and is preserved. Identify $GD$ with $X$ by the unique coequalizer isomorphism, so $Gq=a$. Naturality of the counit makes $Kq$ an algebra morphism, hence the comparison structure $c$ on $GD$ satisfies $cTa=a{\mu }_X=aTa$. The map $Ta$ is split epic because $a{\eta }_X=1$, so $c=a$. Every algebra is therefore isomorphic to a comparison algebra.

For an actual object $D$, its counit ${\epsilon }_D:FGD\to D$ is a coequalizer of the analogous free pair. It coequalizes that pair by counit naturality. Compare it with the coequalizer furnished by the hypothesis; applying $G$ gives an isomorphism, since both underlying arrows coequalize the same split pair. Reflection of isomorphisms makes the comparison map an isomorphism. This proves the counit's asserted coequalizer property.

Let $f:KD\to KE$ be an algebra morphism. Its transpose is $h_0={\epsilon }_{E}Ff:FGD\to E$. The two composites of $h_0$ with the free pair have adjunct transposes $G{\epsilon }_{E}Tf$ and $fG{\epsilon }_D$, equal because $f$ is an algebra morphism. Consequently $h_0$ factors uniquely through ${\epsilon }_D$ as $h:D\to E$. Applying $G$ gives $(Gh)G{\epsilon }_D=fG{\epsilon }_D$; the last factor is split epic, so $Gh=f$. Conversely any lift of $f$ must satisfy $h{\epsilon }_D={\epsilon }_{E}Ff$ by naturality and hence is unique. Thus $K$ is full and faithful, and the earlier construction makes it essentially surjective. The comparison is an equivalence, completing the theorem's proof.

Now take $G:\mathbf{CompHaus}\to \mathbf{Set}$. Its left adjoint exists by the permitted topological result; it can be realized as the Stone-Čech compactification of the discrete input set. A bijective continuous map from a compact space to a Hausdorff space is a homeomorphism, so $G$ reflects isomorphisms.

For continuous $f,g:X\rightrightarrows Y$ with a split underlying fork $q,s,t$, let $R$ be the relation of having the same $q$-image. We claimFor the forward inclusion choose $x=t(y)$, $x^{\prime }=t(y^{\prime })$: $ft=1$ and $gt=sq$ supply the equality. For the reverse inclusion use $qf=qg$. The subset $\{(x,x^{\prime }):g(x)=g(x^{\prime })\}\subset X^2$ is closed, because $Y$ is Hausdorff, and compact. Its continuous image $R\subset Y^2$ is compact and hence closed. A closed equivalence relation on a compact Hausdorff space has a compact Hausdorff quotient; its quotient map is a coequalizer here, since it identifies exactly the generated relation. Its underlying set coequalizer is the specified one up to the unique isomorphism. Thus $G$ preserves the needed coequalizers, andThe closedness argument is the closed relation generated by a set-split pair of compact Hausdorff maps lemma; the splitting maps themselves need not be continuous.

Take $A=1$. By assumption $E=(1\downarrow F)$ has finite limits, so $E^{\mathrm{op}}$ has finite colimits. A filtered category is nonempty, admits a cocone over every pair of objects, and admits a coequalizing arrow for every parallel pair. The initial object of $E^{\mathrm{op}}$ gives nonemptiness, its binary coproducts give the required two-object cocones, and its coequalizers equalize parallel arrows. Consequently $\boxed{\text{(ii)}\Rightarrow \text{(iii)}}$. These statements also handle the otherwise possible empty category of elements.

Assume both unit and counit are monic. The unit criterion makes $F$ faithful. The triangle identity is ${\epsilon }_{FA}F{\eta }_A=1_{FA}$. Since ${\epsilon }_{FA}$ is monic, this split epimorphism is an isomorphism: cancelling it in ${\epsilon }_{FA}(F{\eta }_A{\epsilon }_{FA})={\epsilon }_{FA}$ gives $F{\eta }_A{\epsilon }_{FA}=1$. Thus $F{\eta }_A$ is invertible. As $\mathcal{C}$ is balanced, the reflection result in the root solution makes ${\eta }_A$ invertible. Hence $F$ is full and faithful.

Let $q:FA\to B$ be a regular epimorphism. Its adjunct factorization is $q={\epsilon }_{B}Fh$ for $h:A\to GB$. A regular epimorphism is extremal: if a coequalizer $q$ factors through a monomorphism $m$, its other factor also coequalizes the defining pair, descends through $q$, and gives a right inverse for $m$; monicity then makes $m$ invertible. Apply this to the monic ${\epsilon }_B$. More explicitly, if $q$ coequalizes $r,s$, monicity gives $Fhr=Fhs$, so there is $d:B\to FGB$ with $dq=Fh$. Then ${\epsilon }_{B}dq=q$ implies ${\epsilon }_{B}d=1$, and monicity gives $d{\epsilon }_B=1$. Therefore $B\cong FGB$. The image is closed under regular quotients, proving $\boxed{\text{(i)}\Rightarrow \text{(ii)}}$.

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.

Assume the coequalizers from part (b) exist and write them $q_a:FA\to L(A,a)$. The underlying arrow of the comparison-adjunction unit isWe first relate it to the monad algebra action. Naturality of $\eta$, the coequalizer equation and a monad unit law giveIn $\mathcal{C}$, $a:TA\to A$ is itself a split coequalizer of $Ta,{\mu }_A:T^{2}A\rightrightarrows TA$. Indeed,For an explicit universal-property proof, if $r:TA\to X$ satisfies $rTa=r{\mu }_A$, then $s=r{\eta }_A$ obeysIts uniqueness follows by composing any equation $sa=r$ with ${\eta }_A$.

Now $G{q}_a$ equalizes $Ta$ and ${\mu }_A$. If it is a coequalizer, then $a$ and $G{q}_a$ are coequalizers of the same pair. Their unique comparison map is ${\lambda }_a$, since ${\lambda }_{a}a=G{q}_a$, and is therefore invertible. Conversely, if ${\lambda }_a$ is invertible, $G{q}_a={\lambda }_{a}a$ is an isomorphic copy of the coequalizer $a$, so it is a coequalizer too.

The forgetful functor ${\mathcal{C}}^T\to \mathcal{C}$ reflects isomorphisms: if an monad algebra morphism $u$ has an inverse as an underlying arrow, the equation $ua=bTu$ rearranges to $u^{-1}b=aT(u^{-1})$, making that inverse an monad algebra morphism. Thus underlying invertibility here is exactly invertibility of the unit component. For all monad algebras together, the comparison-adjunction unit criterion isEquivalently, the necessary-and-sufficient conditions for a left adjoint with invertible unit are existence of the coequalizers in part (b) and their preservation by $G$. No preservation of arbitrary coequalizers is required.

Use the elementary, or weak, topos axioms: finite limits and a power object $PA$ for every object $A$, naturally representing $\mathrm{Sub}(A\times X)$. No arbitrary products or coproducts are assumed. In particular $\Omega =P1$ is the subobject classifier. Inverse image of the membership relation defines $Pf:PB\to PA$ for $f:A\to B$.

Interchanging the factors in a relation gives natural bijectionsThus the contravariant power-object functor $P:{\mathcal{E}}^{\mathrm{op}}\to \mathcal{E}$ has left adjoint $P^{\mathrm{op}}$. Its adjunction unit ${\eta }_A:A\to PPA$ sends a generalized element $a$ to the predicate on $PA$ given by membership of $a$.

This unit is monic. If $x,y:T\to A$ have the same image under ${\eta }_A$, membership of $x$ and $y$ agrees in every $T$-parameterized subobject of $A$. Apply this to the graph of $x$ in $A\times T$: membership of $x$ is true, so membership of $y$ is true, which means $y=x$. This argument uses parameterized relations, not an assumption that global elements detect morphisms.

The functor $P$ reflects isomorphisms. Suppose $Pf$ is invertible. Then $PPf$ is invertible, and naturality $PPf{\eta }_A={\eta }_{B}f$ together with monicity of ${\eta }_A$ shows that $f$ is monic. Regard $f$ as a subobject of $B$. Its inverse image along itself is the whole of $A$, just as for the whole subobject of $B$. Those two subobjects correspond to maps $1\to PB$; since $Pf$ is monic, they coincide. Therefore the subobject $f$ is all of $B$, and $f$ is an isomorphism.

We now prove the preservation condition for the crude monadicity theorem. A reflexive pair in ${\mathcal{E}}^{\mathrm{op}}$ is a coreflexive pair $f,g:B\rightrightarrows A$ in $\mathcal{E}$, with $rf=rg=1_B$ for some $r:A\to B$. Let $e:E\hookrightarrow B$ be its equalizer. For any monomorphism $m$, define ${\exists }_m$ on power objects by composing a parameterized subobject with $m$; no general image construction is needed because $m$ is monic. Inverse image followed by this extension obeys $Pm{\exists }_m=1$.

Here both $f,g$ are monic. For a parameterized subobject $S$ of $B$, an element $b$ belongs to $g^{-1}(f(S))$ exactly when $gb=f{b}^{\prime }$ for some $b^{\prime }\in S$. Applying $r$ gives $b=b^{\prime }$, so this condition is precisely $b\in S$ and $fb=gb$. Consequently, as morphisms of power objects,If $h:PB\to Z$ equalizes $Pf,Pg$, these identities giveIt therefore factors uniquely through $Pe$, which is a split epimorphism with section ${\exists }_e$. Since $fe=ge$, $Pe$ also equalizes $Pf,Pg$. Hence $Pe$ is their coequalizer. This proves explicitly that power objects turn coreflexive equalizers into coequalizers.

Finite equalizers in $\mathcal{E}$ give all coequalizers of reflexive pairs in ${\mathcal{E}}^{\mathrm{op}}$, and we have shown that $P$ preserves them and reflects isomorphisms. The crude monadicity theorem now yields

Fix a set $I$ of cardinality $\kappa$. If $\mathcal{E}$ has $I$-indexed products in a category, its Eilenberg-Moore category for the double-power monad has them, since the forgetful functor creates existing limits. The displayed equivalence then gives $I$-products in ${\mathcal{E}}^{\mathrm{op}}$, hence $I$-coproducts in a category in $\mathcal{E}$.

For the converse, suppose those coproducts exist. The diagonal $\Delta :\mathcal{E}\to {\mathcal{E}}^I$ is a logical functor, because all the elementary topos structure in ${\mathcal{E}}^I$ is pointwise, and it has left adjoint ${\coprod }_I$. Apply power-object monadicity to both toposes. Under their monadic identifications, ${\Delta }^{\mathrm{op}}$ is the lift of $\Delta$ to the double-power Eilenberg-Moore categories. The base functor $\Delta$ is a right adjoint to ${\coprod }_I$, so the adjoint lifting theorem for monad algebra functors gives a left adjoint to its algebra lift. Its required reflexive coequalizers exist in ${\mathcal{E}}^{\mathrm{op}}$, since they are opposites of finite equalizers in $\mathcal{E}$. Taking opposites gives a right adjoint to $\Delta$, which is exactly the $I$-product functor. Thus the products and coproducts of fixed cardinality in a topos satisfy

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 regular category has finite limits, coequalizers of kernel pairs, and regular epimorphisms stable under pullback. Equivalently, every morphism factors as a regular epimorphism followed by a monomorphism, and these factorizations are stable under pullback. A regular functor preserves finite limits and regular epimorphisms, hence also preserves these image factorizations.

For this question, the capital regular category convention is the one in which every well-supported object is a well-pointed object in a category. Here an object $A$ is well-supported if $A\to 1$ is a regular epimorphism. It is well-pointed if a subobject $m:B\hookrightarrow A$ through which every point $1\to A$ factors must be the whole object. This convention is documented in Definitions 1.1–1.3. It imposes no such point-detection requirement on objects whose support is a proper subobject of $1$. In particular, it should not be replaced here by the stronger assertion that $\mathcal{A}(1,-)$ reflects every isomorphism.

First prove two facts about a capital regular category $\mathcal{A}$. Every well-supported object $A$ has a point. Otherwise $A\times A$ is well-supported and has no points, so its diagonal $A\hookrightarrow A\times A$ contains all points and is invertible. That makes $A$ a subterminal object; the regular epimorphism $A\to 1$ is then also monic and hence invertible, contradicting the absence of a point. If $e:X\to Y$ is a regular epimorphism and $y:1\to Y$ is a point, its pullback in a category is a well-supported object over $1$, so it has a point lifting $y$. Thus the representable functorpreserves regular epimorphisms; it already preserves all existing limits. Hence $\Gamma$ is a regular functor. Second, if $\Gamma X$ is a singleton, its unique point $x:1\to X$ is a monomorphism, $X$ is well-supported, and $x$ contains all points of $X$. Well-pointedness makes $x$ invertible. Thus $\Gamma$ reflects terminal objects, even though it need not reflect arbitrary isomorphisms.

Now let $\mathcal{C}$ be a small category that is a regular category. For each object $B$, its slice category $\mathcal{C}/B$ is small and regular: finite limits, images and the stability of regular epimorphisms are computed using the corresponding constructions in $\mathcal{C}$. By the permitted capitalization result choose a conservative functor that is regular,with ${\mathcal{A}}_B$ a small capital regular category. The base-change functor $B^{\ast }:\mathcal{C}\to \mathcal{C}/B$, given by $X\mapsto (X\times B\to B)$, preserves finite limits and regular epimorphisms. So doesTaking these as the coordinates defines a regular functorThe exponent is a set of indices, so this is a power of the Category of sets, with finite limits and regular epimorphisms computed coordinatewise.

To prove reflection of isomorphisms, suppose $R(f)$ is invertible for $f:A\to B$. In $\mathcal{C}/B$, the object $(A,f)$ is the pullback of $B^{\ast }f:(A\times B\to B)\to (B\times B\to B)$ along the point given by the diagonal $B\to B\times B$. Applying ${\Gamma }_B{J}_B$ shows that ${\Gamma }_B{J}_B(A,f)$ is the fibre of the bijection $R_B(f)$ over this diagonal point, and is a singleton. The preceding lemma gives $J_B(A,f)\cong 1$ through its canonical arrow to $1$. Since $J_B$ reflects isomorphisms, $(A,f)\to (B,1_B)$ is invertible. Its underlying morphism is $f$. Therefore

In the small-category setting of this representation result, the elementary condition for replacing the power by one copy of $\mathbf{Set}$ is:This is almost total support for a regular category. “Strict initial” means initial and every arrow into it is invertible; the condition allows the case in which all objects are well-supported and no initial object exists.

For necessity, let $V:\mathcal{C}\to \mathbf{Set}$ be regular and conservative. If $VA\ne \varnothing$, factor $A\to 1$ through its support of an object in a regular category $S\hookrightarrow 1$. Preservation of images gives $VS=1$. Conservativity makes $S\to 1$ invertible, so $A$ is well-supported. If $VA=\varnothing$, then for every $X$ the projection $A\times X\to A$ is sent to the bijection $\varnothing \to \varnothing$, hence is invertible. Its inverse followed by the other projection supplies an arrow $A\to X$. For any two such arrows, their equalizer is sent to a bijection onto $VA$, hence is invertible, proving uniqueness. Thus $A$ is initial. If $X\to A$ exists, then $VX\to \varnothing$ exists, so $VX=\varnothing$ and that arrow is sent to a bijection. Conservativity makes it invertible. Hence $A$ is strict initial.

For sufficiency, choose the permitted conservative regular functor $J:\mathcal{C}\to \mathcal{A}$ into a small capital regular category, and put $V=\Gamma J$. This is regular. A well-supported object of $\mathcal{C}$ is sent to a well-supported object of $\mathcal{A}$, which has a point. A non-well-supported object is strict initial by the condition, and is consequently a proper subterminal object. The functor $J$ preserves subterminality and reflects the noninvertibility of its arrow to $1$; its image is a proper subterminal object and has no points. Thus $VA$ is empty exactly on the non-well-supported objects.

Suppose $Vf$ is a bijection for $f:A\to B$. If both sets are empty, $A$ and $B$ are strict initial objects and $f$ is invertible. Otherwise $A,B$ are well-supported. Factor $Jf$ as a regular epimorphism followed by a monomorphism $m:I\hookrightarrow JB$. Surjectivity of $Vf$ makes every point of $JB$ factor through $m$; capitality of $\mathcal{A}$ makes $m$ invertible. Hence $Jf$ is a regular epimorphism. Its kernel pair $P=JA{\times }_{JB}JA$ is well-supported, because $P\to JA$ and $JA\to 1$ are regular epimorphisms. Injectivity of $Vf$ implies that every point of $P$ has equal coordinates, and therefore factors through the diagonal $JA\hookrightarrow P$. Since $P$ is well-pointed, this diagonal is invertible. Thus $Jf$ is also monic, hence an isomorphism, and conservativity of $J$ makes $f$ invertible. This proves sufficiency with actual point and image arguments, rather than asserting that global sections of every capital regular category is conservative.

A split coequalizer consists of $f,g:A\rightrightarrows B$, $q:B\to Q$, $s:Q\to B$ and $t:B\to A$ with $qf=qg$, $qs=1_Q$, $ft=1_B$ and $gt=sq$. If $hf=hg$, then $h=hft=hgt=hsq$. Thus $hs$ factors $h$ through $q$, and the factor is unique because $q$ has the right inverse $s$. This proves the coequalizer property directly. Every functor preserves the diagram, since all these equations are preserved.

For idempotent splitting through a coequalizer, first suppose $e=ir$ with $ri=1_Q$. Then $re=r$. Any $h:E\to Z$ satisfying $he=h$ factors as $h=(hi)r$, uniquely since $r$ is a split epimorphism. Hence $r$ coequalizes $(e,1_E)$.

Conversely, let $q:E\to Q$ coequalize $(e,1_E)$. Since $e^2=e$, the arrow $e:E\to E$ itself equalizes the pair, so there is a unique $i:Q\to E$ with $iq=e$. Then $qiq=qe=q$, and every coequalizer is an epimorphism, so $qi=1_Q$. Thus $e=iq$ is a splitting of an idempotent morphism.