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Given the two natural transformations, defineNaturality of $\eta$ and the second triangle identity for an adjunction giveNaturality of $\epsilon$ and the first triangle identity for an adjunction giveThe maps are natural in $C,D$, since the defining compositions commute with precomposition in $C$ and postcomposition in $D$ by naturality of $\eta ,\epsilon$. They therefore define an adjunction $F⊣G$. Their values on identity morphisms recover precisely $\eta ,\epsilon$. Conversely any adjunction with these unit and counit of an adjunction must have the same transpose formulas, so it is unique with the prescribed unit and counit.

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.

Write $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$. The characterization concerns an adjunction with the specified unit and counit of an adjunction $\eta$ and $\epsilon$. Its triangle identities for an adjunction areAssume these identities. Define the hom-set mapsNaturality of $\eta$ and $\epsilon$ makes these maps natural in both objects. Naturality and the triangle identities for an adjunction giveThus they are inverse bijections and define $F⊣G$.

Conversely, from the natural hom-set bijections of an adjunction, define ${\eta }_A=\Phi (1_{FA})$ and ${\epsilon }_D={\Phi }^{-1}(1_{GD})$. Naturality gives the same formulas for $\Phi$ and $\Psi$ above. Applying $\Psi \Phi$ to $1_{FA}$ and $\Phi \Psi$ to $1_{GD}$ gives the two triangle identities for an adjunction. Therefore these identities are exactly the compatibility conditions on the specified unit and counit. If the printed equivalence were read as mere existence of some adjunction, independently of the supplied transformations, its only-if direction would be false: on the category of abelian groups, $F=G=1$ are adjoint, but choosing both transformations to be zero does not satisfy either triangle on a nonzero object.

Use the orientation $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$. An adjunction $F⊣G$ is equivalently specified by the natural transformationscalled the unit and counit of an adjunction, satisfying the triangle identities for an adjunctionThe corresponding natural bijection is $\mathcal{D}(FC,D)\cong \mathcal{C}(C,GD)$, with $f\mapsto G(f){\eta }_C$ and inverse $g\mapsto {\epsilon }_{D}F(g)$. The two triangular equations are the required compatibility conditions. No proof of equivalence of the formulations is needed here.

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 balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism. A faithful functor reflects both of these cancellation properties: for instance, $fu=fv$ implies $FfFu=FfFv$, so monicity of $Ff$ and faithfulness imply $u=v$; the dual argument applies to epimorphisms. If $Ff$ is invertible, it is both monic and epic. Thus a faithful functor from a balanced category reflects isomorphisms.

For the adjunction $F⊣G$, the unit and counit of an adjunction satisfyIf $F$ is a faithful functor and ${\eta }_{A}u={\eta }_{A}v$, applying $F$ and composing with ${\epsilon }_{FA}$ gives $Fu=Fv$, hence $u=v$. Thus each ${\eta }_A$ is a monomorphism. Conversely, if every ${\eta }_A$ is monic and $Fu=Fv$ for $u,v:X\to A$, naturality givesso $u=v$. This proves the faithful left adjoint criterion.

Now assume both $\eta$ and $\epsilon$ are pointwise monic. The first triangle identity makes ${\epsilon }_{FA}$ a split epimorphism; since it is also a monomorphism, it is invertible and $F{\eta }_A$ is its inverse. The faithful left adjoint criterion says that $F$ is faithful, and the balanced category argument above then reflects the invertibility of $F{\eta }_A$ to that of ${\eta }_A$.

For completeness, an invertible unit makes $F$ a full and faithful functor. For $h:FA\to F{A}^{\prime }$, set $k={\eta }_{A^{\prime }}^{-1}G(h){\eta }_A$. The triangle identity gives $F{\eta }_A={\epsilon }_{FA}^{-1}$, so naturality of $\epsilon$ yieldsFaithfulness was already proved.

Let $q:FA\to B$ be a strong epimorphism, and putThen naturality gives ${\epsilon }_{B}t=q$. The square with left edge $q$, right edge the monomorphism ${\epsilon }_B$, top edge $t$, and bottom edge $1_B$ has a diagonal $s:B\to FGB$ by the defining lifting property of a strong epimorphism. Thus ${\epsilon }_{B}s=1_B$. A monic split epimorphism is invertible, so $B\cong FGB$. The essential image of $F$ is therefore closed under strong quotients.

An adjunction $F⊣G$ is equivalently specified by a unit and counit of an adjunctionsatisfying the triangle identities

Let $F_i⊣G_i$ have units ${\eta }_i$ and counits ${\epsilon }_i$. The mate correspondence sends $\alpha :F_1\to F_2$ toConversely, $\beta :G_2\to G_1$ givesNaturality and the triangle identities show that the two constructions are inverse, yielding the required bijection of natural transformations.

Now let the endofunctor $F$ carry a monad $(F,\eta ,\mu )$ and let $F⊣G$. Taking right mates turnsinto a counit $ϵ:G\to 1$ and comultiplication $\delta :G\to G^2$. Since mates reverse composition, the monad unit and associativity laws become the comonad counit and coassociativity laws. Moreover, a map $FA\to A$ corresponds under the adjunction to a map $A\to GA$, and the algebra axioms correspond exactly to the coalgebra axioms. This is the monad on a left adjoint induces a comonad on its right adjoint construction and gives an isomorphism of the two structure categories.

For a monoid $M$, the free-$M$-set functor is $F(X)=M\times X$, and its algebras are precisely left $M$-sets, the objects of $[M,\mathbf{Set}]$. Its right adjoint is $G(X)=X^M$. The preceding isomorphism identifies $[M,\mathbf{Set}]$ with the Eilenberg-Moore category of coalgebras for the induced comonad on sets, compatibly with the forgetful functor. Therefore

For an adjoint functor pair $F:\mathcal{D}\rightleftarrows \mathcal{C}:G$, the unit and counit of an adjunction areUnder the adjunction bijection, ${\eta }_A$ corresponds to $1_{FA}$ and ${\epsilon }_B$ corresponds to $1_{GB}$. They satisfy the triangle identitiesConversely, natural transformations with these identities recover the adjunction through the mutually inverse maps

The fully faithful adjoint criterion gives the first equivalence directly. If $F$ is full and faithful, there is a unique $r_A:GFA\to A$ with $F(r_A)={\epsilon }_{FA}$; the triangle identity and faithfulness show that $r_A$ and ${\eta }_A$ are inverse, so the unit is an isomorphism. If the unit is an isomorphism, the displayed adjunction bijection shows thatis bijective, so $F$ is full and faithful. This also proves that either condition gives a natural isomorphism $GF\cong 1_{\mathcal{D}}$.

For the remaining direction, suppose merely that $GF$ is naturally isomorphic to the identity. Transport the monad induced by an adjunction along this isomorphism. Its underlying endofunctor is then the identity, its unit is a natural endomorphism $u:1\Rightarrow 1$, and its multiplication is a natural endomorphism $m:1\Rightarrow 1$ with $mu=1$. Naturality makes $u_A$ commute with $m_A$, so also $um=1$. Hence the transported unit, and therefore $\eta$, is an isomorphism. The three conditions are equivalent.

Now let $F⊣G⊣H$. If $F$ is full and faithful, then $GF\cong 1_{\mathcal{D}}$. For $A,B\in \mathcal{D}$, the two adjunctions give natural bijectionsThe Yoneda lemma therefore gives $GHA\cong A$, naturally in $A$. The fully faithful adjoint criterion applied to $G⊣H$ shows that $H$ is full and faithful. Conversely, if $H$ is full and faithful, then $GH\cong 1_{\mathcal{D}}$ andAnother application of the Yoneda lemma gives $GFA\cong A$, so $F$ is full and faithful. Thus $F$ is full and faithful exactly when $H$ is.

Assume henceforth that $G$ is full and faithful. Then the counit $\epsilon :FG\to 1_{\mathcal{C}}$ and unit $\alpha :1_{\mathcal{C}}\to HG$ are natural isomorphisms. ConsiderApplying the faithful functor $G$, then using naturality and the four triangle identities, turns both composites intoTherefore ${\theta }_1={\theta }_2$; denote their common value by $\theta$, the double-adjoint comparison transformation.

The pointwise monicity criterion is clearest from the following natural square, in which both vertical maps are bijections:The left vertical map is the adjunction $G⊣H$, while the right one precomposes with the isomorphism ${\epsilon }_B:FGB\to B$. Thus every ${\theta }_A$ is a monomorphism exactly when $F$ is faithful on all morphisms $GB\to A$, namely morphisms whose domains lie in the image of $G$.

Dually, the natural squarehas bijective vertical maps, using $F⊣G$ on the left and the isomorphism ${\alpha }_B:B\to HGB$ on the right. Hence every ${\theta }_A$ is an epimorphism exactly when $H$ is faithful on all morphisms $A\to GB$, namely morphisms whose codomains lie in the image of $G$.

For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with $F⊣G$, the unit and counit of an adjunction areobtained by transposing identity morphisms. They satisfy the triangular identities ${\epsilon }_{F}F\eta =1_F$ and $G\epsilon {\eta }_G=1_G$.

If $G$ is full and faithful, fullness gives a map $u:B\to FGB$ with $G(u)={\eta }_{GB}$; the second triangular identity and faithfulness show that $u$ is inverse to ${\epsilon }_B$. Conversely, if $\epsilon$ is an isomorphism, then for $h:GB\to G{B}^{\prime }$ the unique arrow whose transpose is $h$ iswhich proves that $G$ is full, and the triangular identities prove faithfulness. Hence (i) and (ii) are equivalent. Condition (ii) immediately gives (iii). Conversely, a natural isomorphism $FG\cong 1_{\mathcal{D}}$ combines with the adjunction bijection to show naturally that $\mathcal{D}(B,B^{\prime })\to \mathcal{C}(GB,G{B}^{\prime })$ is bijective; by naturality and the triangular identities this is the map induced by $G$ up to invertible natural conjugation, so $G$ is full and faithful. Thus all three conditions are equivalent.

Suppose $L⊣F⊣R$. If $L$ is full and faithful, the unit $1\to FL$ is an isomorphism. For $A,B$ in the codomain of $F$, the two adjunctions then give natural bijectionsBy the Yoneda lemma, the counit $FRB\to B$ is an isomorphism, so the fully faithful adjoint criterion makes $R$ full and faithful. The converse is dual.

Now assume $F$ is full and faithful. For $L⊣F$, its counit $ϵ:LF\to 1$ is invertible; for $F⊣R$, its unit $\alpha :1\to RF$ is invertible. DefineApplying $F$, then using naturality and the triangular identities, reduces this composite to the same map asSince $F$ is faithful, the two displayed composites are equal.

For a morphism $u:FB\to A$, let $\hat{u}=Ru{\alpha }_B:B\to RA$ be its transpose under $F⊣R$. The identities just proved giveIf every ${\theta }_A$ is monic and $Lu=Lv$, this equation gives $\hat{u}=\hat{v}$, hence $u=v$; thus $L$ is faithful on arrows whose domains are in the image of $F$. Conversely, if ${\theta }_{A}f={\theta }_{A}g$ for $f,g:B\to RA$, naturality of $ϵ$ and the second formula for $\theta$ giveBoth maps have domain $FB$, so the assumed faithfulness gives ${\beta }_{A}Ff={\beta }_{A}Fg$. These are the transposes of $f$ and $g$, hence $f=g$. Therefore $\theta$ is pointwise monic exactly under the stated faithfulness condition.