Susan Montgomery gave the Lemma 1.7.2 without proof in her book 'Hopf algebras and their actions on rings' which states that $M^{H^*}= M^{coH}$ for a right $H$-comodule $M$ with left $H^*$-module structure and $M^H=M^{coH^o}$ for a left $H$-module $M$ with right $H^o$-comodule structure.
Here, $H$ is a Hopf algebra. $M^H=\{m \in M | h \cdot m= \epsilon(h)m, \forall h \in H \}$ and $M^{coH}=\{m\in M | \rho(m)= m \otimes 1\}$ are the sets of invariants and coinvariants respectively.
I'd like to complete the proof. My intuition said that $$ \begin{aligned} \rho(m) = m\otimes 1 &\Leftrightarrow h^* \cdot m =<h^*,1>m\\ &\Leftrightarrow h^*\cdot m= \epsilon(h^*)m. \end{aligned} $$ But I am stuck at $<h^*,1>=\epsilon(h^*)$ and $\Leftarrow$ in the top row. By definition, $h^* \cdot m= \sum <h^*, m_{(1)}> m_{(0)}$, but how to show $\rho(m) = m\otimes 1$ from this?
Thanks in advance.
I have solved this question on my own. For the one hand, $$ <h^*,1_H>=<h^*,u(1_k)>=<u^*(h^*),1_k>=u^*(h^*)=\epsilon_{H^*}(h^*). $$ For the other hand, we can choose a basis $\{x_i\}$ for $M$ since $M$ is a vector space. Assume $\rho(m)= \sum m_i\otimes h_i$, $m=\sum a_jx_j$ and $m_i=\sum a_{ik}x_k$. Because $h^*\cdot m=<h^*,1_H>m$, the left $H^*$-module structure of right $H$-comodule states that
$$ \sum<h^*,h_i>m_i= <h^*,1_H>m, $$ which means $$ \sum<h^*,h_i> \sum a_{ik}x_k= <h^*,1_H> \sum a_jx_j. $$ Consider the coefficient of $x_k$ for both sides, then $$ \sum_i <h^*,a_{ik}h_i>= <h^*,a_k \cdot 1_H>. $$ This means $$ <h^*,a_k \cdot 1_H-\sum_i a_{ik}h_i>=0,\ \forall h^*\in H^*. $$ Note that $<,>$ is nondegenerate, this means $$ a_k \cdot 1_H-\sum_i a_{ik}h_i=0. $$ So $$ \begin{aligned} \rho(m) &= \sum a_{ik}x_k \otimes h_i\\ &=\sum_k x_k \otimes \sum_i a_{ik} h_i\\ &=\sum_k x_k \otimes a_k \cdot 1_H\\ &=\sum_k a_k x_k \otimes 1_H\\ &=m \otimes 1_H. \end{aligned} $$ I also have another proof to avoid considering the basis. Assume again $\rho(m)= \sum m_i\otimes h_i$ with $h_i$ linearly independent. Because $$ \sum<h^*,h_i>m_i= <h^*,1_H>m, \forall h^*, $$ Let $h^*=h_i^*$, then $m_i=<h_i^*,1_H>m$. Thus $$ \begin{aligned} <h^*,1_H>m &= \sum<h^*,h_i>m_i\\ &= \sum<h^*,h_i><h_i^*,1_H>m\\ &= <h^*,\sum<h_i^*,1_H>h_i>m, \forall h^*\in H^*. \end{aligned} $$ Therefore, $$ \sum <h_i^*,1_H>h_i=1_H, $$ which indicates $$ \begin{aligned} \rho(m) &=\sum m_i \otimes h_i\\ &= \sum <h_i^*,1_H>m \otimes h_i\\ &=m \otimes \sum <h_i^*,1_H>h_i\\ &=m \otimes 1_H. \end{aligned} $$
As to $M^H=M^{coH^o}$, one need to consider the right $H^o$-comodule structure of a left $H$-module which was introduced at Lemma 1.6.4 in Susan Montgomery's book i.e. $$ \rho(m)=\sum m_i\otimes f_i, $$ where $a\cdot m= \sum f_i(a)m_i$.