Take a category $\mathcal{C}$ with a monad $T$ and construct the the Eilenberg-Moore category $\mathcal{C}^T$, the adjunction that arises is the terminal splitting of the monad $M$. Denote the resulting comonad on $\mathcal{C}^T$ as $S$, one can now construct its co-Eilenberg-Moore category $(\mathcal{C}^T)^{S}$ and there is a unique functor $L:\mathcal{C} \to (\mathcal{C}^T)^{S}$.

This seems to be a fairly canonical construction that somebody must have looked at before. I'm particularly interested in the case where $\mathcal{C}$ is locally presentable and the monad is accessible - it follows that case that $\mathcal{C}^{TS}$ is presentable and $L$ is accessible.

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    $\begingroup$ Did you have a specific question? One not terribly difficult result is that if one takes a monad $M$ on $Set$ and passes to the co-Eilenberg-Moore category over $Set^M$, then one gets $Set$ back provided that $F(!): F(0) \to F(1)$ is a regular monomorphism in $Set^M$ but not an isomorphism. This condition holds in a great many examples. The result is due to Mesablishvili -- see the nLab: ncatlab.org/nlab/show/comonadic+functor $\endgroup$
    – Todd Trimble
    Sep 19, 2019 at 20:01
  • $\begingroup$ @ToddTrimble Thanks for pointing out that material - I was able to prove a much stronger result than I had expected. $\endgroup$ Sep 20, 2019 at 3:23

1 Answer 1


This situation is studied in detail in Pavlovic–Hughes's The nucleus of an adjunction and the Street monad on monads and applied to study the Dedekind–MacNeille completion in Tight limits and completions from Dedekind-MacNeille to Lambek-Isbell by the same authors.

  • $\begingroup$ Oh wow, this is a deep cut. Now I have to try and remember why I was thinking about that… $\endgroup$ Jul 14, 2022 at 3:22

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