Let $\mathcal{C}$ be a category and $\mathcal{G}=(G,\delta, \epsilon)$ be a comonad on $\mathcal{C}$. Here $G: \mathcal{C}\to \mathcal{C}$ is a functor, $\delta: G\to G^2$ and $\epsilon: G\to id_{\mathcal{C}}$ are natural transformations satisfying $G(\delta)\circ \delta=\delta G\circ \delta$ and $G(\epsilon)\circ \delta=id_G=\epsilon G\circ \delta$.

A $\mathcal{G}$-comodule is a pair $(M,\xi)$ where $M$ is an object in $\mathcal{C}$ and $\xi: M\to G(M)$ is a morphism such that $\delta_M\circ \xi=G(\xi)\circ \xi$ and $\epsilon_M\circ \xi=id_M$. Moreover a $\mathcal{G}$-comodule map $\Phi:(M,\xi)\to (N,\eta)$ is a morphism $\Phi:M\to N$ such that $\eta\circ \Phi=G(\Phi)\circ \xi$.

Now suppose that the functor $G$ has a right adjoint $F$ with $\mu:GF\to id_{\mathcal{C}}$ and $\iota: id_{\mathcal{C}}\to FG$ be the adjunctions. For a $\mathcal{G}$-comodule $(M,\xi)$, $GF(M)$ has a $\mathcal{G}$-comodule structure with coaction $\delta_{F(M)}:GF(M)\to G^2F(M)$. On the other hand, we have $\mu_M:GF(M)\to M$, a morphism in $\mathcal{C}$.

My question is: does $\mu_M$ induce a $\mathcal{G}$-comodule map? More precisely, does the following diagram commute? $\require{AMScd}$ \begin{CD} GF(M) @>\mu_M>> M\\ @V\delta_{F(M)} VV @VV\xi V\\ G^2F(M) @>G(\mu_M)>> GM \end{CD}

  • $\begingroup$ Just a comment on notation: when talking generically about an adjunction, one of the most common default notations is that $F$ is the left adjoint, $G$ is the right adjoint, $\eta: 1 \Rightarrow GF$ is the unit, and $\varepsilon: FG \Rightarrow 1$ is the counit. This is not a universal convention, but it's common enough that I find it clashes with your notation and makes your post a bit confusing to read. I'd suggest using some letter other than $F$ for the right adjoint of $G$ (maybe $H$ or something) and reserving $\eta, \varepsilon$ to be the unit and counit of this adjunction. $\endgroup$ – Tim Campion Apr 5 at 18:00
  • $\begingroup$ Also note that what you call a comodule is often called a coalgebra (though I believe I've seen your terminology before too). $\endgroup$ – Tim Campion Apr 5 at 18:02

$\DeclareMathOperator{\id}{id}$Note that the $\mathcal G$-comodule structure on $GF(M)$ is the co-free one on $F(M)$; in particular, it does not depend on the map $\xi$ defining the comodule structure on $M$. For this reason, I would not expect this diagram to commute, and it is indeed easy to find a counterexample:

Let $\mathcal C = \operatorname{Set}$ (which can be replaced by any cartesian closed category) and $A\in\operatorname{Set}$ be arbitrary. Set $G(M) = A\times M$, with $\delta_M = \Delta_A\times\id_M:A\times M\to A\times A\times M$ and $\epsilon_M = \operatorname{pr}_2:A\times M\to M$. Then the functor $\big(M,\delta_M:M\to A\times M\big)\mapsto \big(M\xrightarrow{\operatorname{pr}_1\circ\delta_M}A\big)$ defines an equivalence of categories between $\mathcal G$-comodules and the over-category $\operatorname{Set}_{/A}$. The adjoint $F$ is given by $M\mapsto M^A$.

Given a $\mathcal G$-comodule $\big(M\xrightarrow{g\times\id_M}A\times M\big)$, your commutative diagram defines two maps $A\times M^A\to A\times M$. The top-right-corner composition sends a pair $(a,f)$ to $(g(f(a)),f(a))$, while the bottom-left-corner composition sends it to $(a,f(a))$. Taking $g$ the constant map with value $a_0\in A$ and $a\in A\smallsetminus\{a_0\}$ gives the required counterexample.


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