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A logical functor between elementary toposes preserves finite limits, exponential objects and the subobject classifier, with their canonical comparison maps. It therefore preserves power objects and the double-power-object monad. If it has a left adjoint, power-object monadicity and the adjoint lifting theorem for monad algebra functors give it a right adjoint as well.

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

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 elementary topos, existence of products in a category indexed by a fixed set is equivalent to existence of the corresponding coproducts in a category. The power-object monadicity equivalence ${\mathcal{E}}^{\mathrm{op}}\simeq {\mathcal{E}}^{PP}$ gives the forward direction because Eilenberg-Moore categories create limits. For the converse, the diagonal into the pointwise product category is a logical functor. A coproduct is its left adjoint; the adjoint lifting theorem for monad algebra functors, applied through power-object monadicity, then gives its right adjoint, namely the product. The required reflexive coequalizers on the algebra side are opposites of finite equalizers in the topos.