# Does the embedding of the Kleisli category of a monad into its Eilenberg-Moore category have a right adjoint?

Let $$(T,m,e)$$ be a monad on a category $$\mathcal{A}$$. There is a full and faithful functor $$J_T$$ from the Kleisli category $$\operatorname{Kl}(T)$$ of $$T$$ to its Eilenberg-Moore category $$\operatorname{EM}(T)$$ defined by $$J_T(X)=(T(X),m_X)$$ on objects and sending a morphism $$s:X \rightarrow Y$$ in $$\operatorname{Kl}(T)$$ (i.e. $$s$$ goes from $$X$$ to $$T(Y)$$ as a morphism in $$\mathcal{A}$$) to $$m_Y\circ T(s)$$ (here the composition is in $$\mathcal{A}$$).

My question is whether such functor $$J_T$$ has a right adjoint.

• It feels like this should only be true when the inclusion is an equivalence (like for vector spaces for instance). – Captain Lama Apr 23 at 14:38
• @Captain Lama - Yes. In the case you mention the right adjoint is just the normal right adjoint of the "canonical functor" that goes from $\mathcal{A}$ to $\operatorname{KL}(T)$. But what happens in general? – 3 A's Apr 23 at 14:46

Since every algebra is a coequalizer of free algebras, if $$J$$ is a left adjoint and the Eilenberg-Moore category is cocomplete, then $$J$$ is an equivalence. This follows from the fact that coreflective subcategories of cocomplete categories are closed under colimits. It is very common for the Eilenberg-Moore category to be cocomplete-for instance it suffices that $$\mathcal A$$ be locally presentable and $$T$$ preserve sufficiently filtered colimits, or that every epimorphism splits in $$\mathcal A$$, such as for sets.
EDIT: As Arnaud points out, cocompleteness is not necessary here-the canonical coequalizer in the EM category will be reflected back into the Kleisli category whether or not other colimits exist. So it’s necessary and sufficient that $$J$$ be an equivalence.