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Let $(T, \mu , \eta )$ a monad on the category $\mathscr{C}$ , with the usual EM (Eilenberg-Moore) adjunction $\langle F_T, U_T, \eta_Y, \epsilon_T \rangle: \mathscr{C}^T \to \mathscr{C}$ where $\mathscr{C}^T$ is the category of $T$-agebras, objects are couples $(A, a)$ where $a: T(A) \to A$ such that $a \circ \eta_A=1$, $a \circ \mu= a \circ T(a) $.

The above adjunction induce on $\mathscr{C}$ the some triple $T$, but on $\mathscr{C}^T$ there is a cotriple $(cT, \delta, \epsilon )$ with $cT(A, a)= (T(A), \mu_A)$, $\epsilon_{(A, a)}: cT(A, a) \to (A, a)$ is give by $a: T(A)\to A$, $\delta: cT(A, a) \to cT\circ cT(A, a) $ i.e. $\delta: (T(A), \mu_A) \to (TT(A), \mu_{TA})) $ is give by $T\eta_A: T(A)\to TTA$. Now a coalgebra of $cT$ is a couple $((A, a), a^1)$ where $(A, a)$ is a $T$-algebra and $a'$ is a $T$-algebra morphism $a': (A, a) \to (T(A), \mu_A)$ such that $a'\circ \epsilon_{A, a}=1 $ and $T(\eta_A)\circ a'= T(a')\circ a' $.

If $T$ come from the free groups adjunction $\langle F, U, \epsilon , \eta \rangle: Gr \to Set$ the condition $a'\circ \epsilon_{A, a}=1 $ makes me suspect that the only cotriples $((A, a), a^1)$ are free-triples (i.e. free groups).

But in the work of J. De Vries (Lecture Notes Math N.540 p.654) in Th 2.3 (p. 659) he show a monadic functor $TTG \to TopGr \times Top$

(where $TTG$ is the topological transformation groups category, $TopGr$ the topological groups category, $Top$ topological spaces category)

where the algebras of the associate monads are the topological actions $\pi: G\times X \to X$.

In Th.2.6 he considered the comonads on the algebras category as in mine initial presentation, and prove that the associated coalgebras are topological actions $\pi: G\times X \to X$ with a continuous cross section $(S, u)$.

So it seems to me that the problem is not trivial, although it come from very simple considerations, however I have not found anywhere.

We can iterate this process of iteration, what we get? What about (co)triple (co)homology? ecc.ecc.

I ask: Do you know where (if exist) I can find a study of this issue?

If not, I hope it can be useful inspiration for those who have shoulders most powerful than mine.

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Such monads-comonads structures were studied (if i recall correctly) by Michael Barr and recently by Bart Jacobs. In some cases it is known that the process stabilizes after a finite number of iterations. I may write more on the subject later. See also the following answer:… – Michal R. Przybylek Jul 16 '13 at 10:58
There are interesting adjunctions $\mathcal{A} \leftrightarrows \mathcal{B}$ that are both monadic and comonadic. For instance, take the tensor–hom adjunction between module categories induced by a faithfully flat extension of commutative rings. – Zhen Lin Jul 16 '13 at 11:20
Have you looked at Applegate-Tierney, Iterated Cotriples: ? – Todd Trimble Jul 16 '13 at 13:19
Todd Trimble: on the iteration of triple or cotriple there is a wide spectrum of work (60 and 70), especially on the cohomology associated (J. Duskin has done a nice book on Mem.AMS). – Buschi Sergio Jul 16 '13 at 17:39
(Trimble2): But what about the EM category of algebras (or coalgebras) I have not found anything, I believe this because the EM adjunction induced by a triple always induces T (the cycle closes repetitive), but if we alternate (as I described), we obtain a non-trivial iteration (I think) of structures of algebras, coalgebras of algebras, algebras of coalgebras of algebras, etc.. – Buschi Sergio Jul 16 '13 at 17:39

To answer one question (implicitly) raised in the post, the free group functor $F: \mathbf{Set} \to \mathbf{Grp}$ is comonadic, so here the monad-comonad tower (I do not know the standard name for this, if one exists) goes from $\mathbf{Set}$ to $\mathbf{Grp}$ and then straight back to $\mathbf{Set}$.

Here is a more general theorem of this type:

  • For $T$ a monad on $\mathbf{Set}$, the free $T$-algebra functor $\mathbf{Set} \to \mathbf{Set}^T$ to the category of $T$-algebras is comonadic provided that the canonical map $T(!): T(0) \to T(1)$ in $\mathbf{Set}^T$ is a regular monomorphism and not an isomorphism.

Edit: Here are alternate hypotheses that often obtain in practice: provided that $T(1)$ is not initial in $\mathbf{Set}^T$ and the underlying set of the initial object $T(0)$ has at least one element, the algebra map $T(!)$ is a split monomorphism and therefore a regular monomorphism, whence $F: \mathbf{Set} \to \mathbf{Set}^T$ is comonadic.

This can be extracted as a consequence of proposition 6.13 (page 24) of this paper of Mesablishvili, for example.

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Todd Trimble: Very interesting, thank you. – Buschi Sergio Jul 17 '13 at 8:28

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