Unit and counit of an adjunction ($\eta ,\epsilon$)

For $F⊣G$, the unit $\eta :1\to GF$ and counit $\epsilon :FG\to 1$ are the natural transformations corresponding to identity morphisms under the adjunction. They satisfy ${\epsilon }_{F}F\eta =1_F$ and $G\epsilon {\eta }_G=1_G$.

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.

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.

For $F⊣G$ with adjunction unit $\eta$ and adjunction counit $\epsilon$, the identities are ${\epsilon }_{FC}F{\eta }_C=1_{FC}$ and $G{\epsilon }_D{\eta }_{GD}=1_{GD}$. They make the transpose maps $f\mapsto Gf\eta$ and $g\mapsto \epsilon Fg$ inverse.

Let $F:\mathcal{C}\to \mathcal{D}$, $G:\mathcal{D}\to \mathcal{C}$, $\eta :1\Rightarrow GF$ and $\epsilon :FG\Rightarrow 1$ be natural transformations. If $G\epsilon {\eta }_G=1_G$, then $e={\epsilon }_{F}F\eta :F\Rightarrow F$ is an idempotent morphism in the functor category. Naturality implies $Ge\eta =\eta$ and $\epsilon e_G=\epsilon$; these absorption identities prove $e^2=e$. The omitted triangle measures exactly whether $e=1_F$.

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 right adjoint $G$ is full and faithful exactly when the counit $FG\to 1$ is an isomorphism. Dually, $F$ is full and faithful exactly when the unit $1\to GF$ is an isomorphism.

For $F⊣G⊣H$ with full and faithful $G$, write $\eta ,\epsilon$ for the first unit and counit and $\alpha ,\beta$ for the second. The two compositescoincide. Their common value is the double-adjoint comparison transformation $H\to F$.

For $F⊣G$, the left adjoint $F$ is faithful exactly when every unit component ${\eta }_A:A\to GFA$ is a monomorphism. Under the adjunction, equality after ${\eta }_A$ corresponds precisely to equality after applying $F$.

Suppose $\mathcal{C}$ is balanced and every morphism of $\mathcal{D}$ factors as a strong epimorphism followed by a monomorphism. For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$, both unit and counit are pointwise monic exactly when $F$ is full and faithful and its essential image is closed under strong quotients.

Let $\mathcal{C}=\{0<1\}$ and let $\mathcal{D}$ be the terminal category. The unique functor $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to the functor selecting $1$. Every arrow in a poset is monic, so the unit and counit are pointwise monic, but $F$ is not full because the unique arrow $F1\to F0$ has no preimage $1\to 0$. The category $\mathcal{C}$ is not balanced.