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.

  • $\begingroup$ It feels like this should only be true when the inclusion is an equivalence (like for vector spaces for instance). $\endgroup$ Apr 23, 2020 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, 2020 at 14:46

1 Answer 1


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.

  • 1
    $\begingroup$ 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, 2020 at 8:16
  • $\begingroup$ @Arnaud D. Of course, thanks. $\endgroup$ Apr 24, 2020 at 8:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.