Coalgebra after counit for a monad given by adjunction

Question

Let $L\dashv R$ be an adjunction and $LR$ the associated comonad, with comultiplication $L\eta R\colon LR\to LRLR$ and counit $\varepsilon\colon\mathrm{id}\to LR$. A coalgebra for this comonad is a map $a\colon X\to LRX$ such that

$$\require{AMScd} \begin{CD} X @>{a}>> LRX;\\ @V{a}VV @VV{LRa}V \\ LRX @>{L\eta R_X}>> LRLRX; \end{CD}$$

commutes and such that

$$X\overset{a}{\longrightarrow} LRX\overset{\varepsilon_X}{\longrightarrow} X$$

is the identity of $X$.

Maybe the question is super naive, but if we consider the morphism $a\circ\varepsilon_X \colon LRX\to LRX$, does it commutatively fit the above square? For sure it makes the upper triangle to commute, but what about $L\eta R_X\circ a\circ \varepsilon_X$, is this equal to $LRa$? Could you give a proof or a counterexample?

Answer 1

Ok I think it's not true. Just take the free/forgetful adjunction $L\dashv R$ between groups and sets, and the easiest possible coalgebra $a\colon G\to LRG$ mapping $g\mapsto[g]$. Then $L\eta R_G\circ a\circ \varepsilon_X$ maps $[g_1,\dots,g_n]\mapsto [[g_1\cdot...\cdot g_n]]$ while $LRa$ maps $[g_1,...,g_n]\mapsto [[g_1],\dots,[g_n]]$.

Answer 2

Naturality of the counit implies that $a\circ\epsilon_X\colon LRX\to X\to LRX$ is the same as $\epsilon_{LRX}\circ LRa\colon LRX\to LRLRX\to LRX$. This is one of the equations necessary for the fork $L\eta_{RX}\circ a,LRa\circ a\colon X\to LRX\rightrightarrows LRLRX$ to be split by $\epsilon_X\colon LRX\to X$ and $\epsilon_{LRX}\colon LRXLRX\to LRX$.

Split forks are in particular absolute equalizers, i.e. applying any functor to them makes them an equalizer. In particular, applying the identity functor results in an equalizer. This is usually developed in dually for monads, in which case monad algebras are in particular split coequaliers and so absolute colimits. These are signifincant for the (co)monadicity theorem, whose key hypothesis is that if the result of applying the would-be (co)monadic functor to a parallel pair has a split (co)equalizer, then the original pair had a coequalizer preserved by the functor.

In any case, the other two equations required for splitting the equalizer are the (left) counit axioms (for the comonad and the coalgebra) that $\epsilon_X$ is a retraction of $a\colon X\to LRX$ and that $\epsilon_{LRX}\colon LRLRX\to LRX$ is a retraction of the comlultiplication $L\eta_{RX}\colon LRX\to LRLRX$, i.e., that $\epsilon_X\circ a=\mathrm{id}_X\colon X\to LRX\to X$ and $\epsilon_{LRX}\circ L\eta_{RX}=\mathrm{id}_{LRX}\colon LRX\to LRXLRX\to LRX$.

In particular, $a\circ\epsilon_X\colon LRX\to X\to LRX$ and $L\eta_{RX}\circ\epsilon_{LRX}\colon LRLRX\to LRX\to LRLRX$ are split idempotents. Thus naturality of the counit thus implies we have two factorizations $L\eta_{RX}\circ a\circ\epsilon_X=L\eta_{RX}\circ \epsilon_{LRX}\circ LRa\colon LRX\to LRLRX$ through the two idempotents.

I suspect the above are basically all the relations between morphisms you can get, but a precise proof will require a precise description of all the possible morphisms constructed from a unit and counit of adjunction (or maybe just from a (co)monad and a (co)algebra).