# Does the right adjoint of a comonad induce the following comodule map?

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? $$\require{AMScd}$$ $$\begin{CD} GF(M) @>\mu_M>> M\\ @V\delta_{F(M)} VV @VV\xi V\\ G^2F(M) @>G(\mu_M)>> GM \end{CD}$$

• 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. – Tim Campion Apr 5 at 18:00
• Also note that what you call a comodule is often called a coalgebra (though I believe I've seen your terminology before too). – 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.