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}