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A monad is an endofunctor $T:\mathcal{C}\to \mathcal{C}$ with natural transformations $\eta :1_{\mathcal{C}}\to T$ and $\mu :T^2\to T$ satisfyingIts Kleisli category has the same objects as $\mathcal{C}$, with ${\mathcal{C}}_T(A,B)=\mathcal{C}(A,TB)$, identity ${\eta }_A$, and compositeThe unit laws follow from naturality of $\eta$ and the two monad unit laws. For $h:C\to TD$, the two triple composites reduce, using naturality of $\mu$, tothey agree by the monad associativity law. Thus this is a category.

Define the free functor into a Kleisli category $J$ and the right-hand functor $U$ byThe identity and composition formulas for $J$ areFor $U$, $U({\eta }_A)={\mu }_{A}T{\eta }_A=1_{TA}$, andwhere the last step is naturality of $\mu$. Hence both mappings are functors. The hom-set identity ${\mathcal{C}}_T(JA,B)=\mathcal{C}(A,UB)$ gives the adjunction $J⊣U$. Its unit is $\eta$, and its counit at $B$ is the Kleisli arrow represented by $1_{TB}:TB\to TB$. Applying $U$ to that counit gives ${\mu }_B$, so the induced monad is the specified one.

For any $L⊣R:\mathcal{D}\to \mathcal{C}$ inducing this same monad, with counit $\epsilon$, define the Kleisli comparison functorThe triangular identities give $K({\eta }_A)=1_{LA}$, and naturality of the counit together with $\mu =R\epsilon L$ gives $K(g\star f)=K(g)K(f)$. Explicitly, naturality moves ${\epsilon }_{LB}$ past $L(g)$, then moves the resulting counit past ${\epsilon }_{LC}$, reducing the composite to ${\epsilon }_{LC}L({\mu }_{C}T(g)f)$. Thus $K$ is a functor, with $KJ=L$ and $RK=U$, and it sends the Kleisli counit to ${\epsilon }_L$.

An adjunction morphism here is required to commute with the left and right adjoints and preserve their adjunction structure. Such a morphism must have the above object map; every Kleisli arrow $f$ factors as its counit after $J(f)$, so its arrow map is forced as well. This proves initiality of the Kleisli adjunction: