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If $e=ir$ and $ri=1$, then $r$ is the coequalizer of $(e,1)$: an arrow $h$ with $he=h$ factors uniquely as $(hi)r$. Conversely a coequalizer $q$ makes $e$ factor as $iq=e$; epimorphic cancellation gives $qi=1$. Thus the splitting of an idempotent morphism is equivalent to this particular coequalizer, without assuming arbitrary coequalizers exist.

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.

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 $F$.