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For $F⊣G$, the counit is the natural transformation $\epsilon :FG\Rightarrow 1$ corresponding to the identity of $GY$ under the adjunction at each $Y$. It participates in the triangle identities for an adjunction.

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.

Put $e_A={\epsilon }_{FA}F{\eta }_A$. This is a natural transformation $F\Rightarrow F$. Only the $G$-triangle $G{\epsilon }_D{\eta }_{GD}=1_{GD}$ is assumed. Naturality gives two useful absorption identities:For the second equality in the last line use naturality of $\epsilon$ at ${\epsilon }_D$, and then the assumed $G$-triangle. Naturality of $\epsilon$ at $F{\eta }_A$ and of $\eta$ at ${\eta }_A$ now givesHence $e$ is an idempotent in the functor category: this is the one-triangle adjunction idempotent.

For the splitting of an idempotent morphism, suppose this idempotent morphism splits as natural transformations $r:F\Rightarrow H$ and $s:H\Rightarrow F$, with $sr=e$ and $rs=1_H$. DefineThe first absorption identity givesFor the other triangle, naturality of $s$ at ${\eta }_A^{\prime }$ and of $\epsilon$ at $r_A$ givesThus the triangle identities for an adjunction prove $H⊣G$.

Conversely, suppose $H⊣G$, with unit $\psi :1_{\mathcal{C}}\Rightarrow GH$ and counit $\phi :HG\Rightarrow 1_{\mathcal{D}}$. The unit has target $GH$, as its type requires. DefineThese are natural transformations. Transposition under $H⊣G$ gives $G{s}_A{\psi }_A={\eta }_A$. Independently, naturality of $\eta$ at ${\psi }_A$ and the assumed $G$-triangle giveThe transpose of $r_A{s}_A$ is therefore $G{r}_{A}G{s}_A{\psi }_A={\psi }_A$, the transpose of $1_{HA}$. Injectivity of the hom-set bijection implies $r_A{s}_A=1_{HA}$. Naturality of $\epsilon$ at $s_A$ givesConsequently $G$ has a left adjoint if and only if $e$ splits. This is the criterion for splitting a one-triangle adjunction idempotent. The argument gives both the explicit splitting and the new unit and counit, without assuming the other triangle for $F$.

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.

Write $K:\mathcal{B}\to \mathcal{C}$, $J:\mathcal{C}\to \mathcal{B}$, with adjunction unit $\rho$ and adjunction counit $\epsilon$. The adjunction givesFor $f:C\to C^{\prime }$, precomposition with ${\epsilon }_C$ followed by this bijection givesby the triangle identities for an adjunction. Thus $J$ is fully faithful exactly when precomposition with ${\epsilon }_C$ is bijective for every $C,C^{\prime }$.

A morphism $e:P\to Q$ with this property is an isomorphism. Surjectivity for target $P$ supplies $u:Q\to P$ with $ue=1_P$. Since $(eu)e=e=1_{Q}e$, injectivity for target $Q$ gives $eu=1_Q$. Conversely, precomposition with an isomorphism is always bijective. Applied to every ${\epsilon }_C$, this proves the fully faithful right-adjoint criterion:When the adjunction counit is invertible, the inverse to $f\mapsto Jf$ is explicitly $h\mapsto {\epsilon }_{C^{\prime }}K(h){\epsilon }_C^{-1}$.

Assume the hom-set condition. Its surjectivity at $B^{\prime }=B$ gives $u:KB\to B$ with $u{\rho }_B=1_B$. We show that this splitting is two-sided.

Because $J$ is fully faithful, ${\epsilon }_{KB}$ is invertible. Both triangle identities for an adjunction give, after suppressing $J$,Applying $K$ to $u{\rho }_B=1_B$ yields $Ku{\rho }_{KB}=1_{KB}$. Naturality of the adjunction unit at $u$ gives ${\rho }_{B}u=Ku{\rho }_{KB}$, hence ${\rho }_{B}u=1_{KB}$. Consequently the hom-set condition forces the unit to be invertible:Only surjectivity at $B^{\prime }=B$ was needed for this direction; the given family of bijections certainly supplies it.

The one-triangle adjunction idempotent splits if and only if $G$ has a left adjoint. From a splitting $F\stackrel{r}{\to }H\stackrel{s}{\to }F$, define the new unit $Gr\eta$ and counit $\epsilon s_G$; the absorption identities imply both triangle identities for an adjunction. Conversely, for $H⊣G$ with unit $\psi :1\Rightarrow GH$ and counit $\phi :HG\Rightarrow 1$, the splitting maps are $r={\epsilon }_{H}F\psi$ and $s={\phi }_{F}H\eta$. Their composites are $rs=1_H$ and $sr=e$.

For $F⊣G$, the unit is the natural transformation $\eta :1\Rightarrow GF$ corresponding to the identity of $FX$ under the adjunction at each $X$. It participates in the triangle identities for an adjunction.