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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.

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  • $\begingroup$ It feels like this should only be true when the inclusion is an equivalence (like for vector spaces for instance). $\endgroup$ – Captain Lama Apr 23 at 14:38
  • $\begingroup$ @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? $\endgroup$ – 3 A's Apr 23 at 14:46
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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.

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    $\begingroup$ Why do you even need the Eilenberg-Moore category to be cocomplete ? A fully faithful left adjoint creates colimits that exist in its codomain, and here we know that the coequalizer exists in the Eilenberg-Moore category. $\endgroup$ – Arnaud D. Apr 24 at 8:16
  • $\begingroup$ @Arnaud D. Of course, thanks. $\endgroup$ – Kevin Arlin Apr 24 at 8:24

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