$\text{Hom}_k(M,N)\cong M^*\otimes_k N$ as Hopf-algebra modules. I'm reading Representations and Cohomology by D.J. Benson. At the beginning of the third chapter the following is explained:
Let $\Lambda$ be a bialgebra over $R$ and $M,N$ left $\Lambda$-modules. We make $M\otimes_R N$ into a $\Lambda$-module as follows: If
$$\Delta(\lambda)=\sum_i \mu_i\otimes \nu_i$$ then 
$$\lambda(m\otimes n)=\sum_i \mu_i(m)\otimes \nu_i(n).$$
We can also make $R$ into a $\Lambda$-module via $\lambda(r)=\varepsilon(\lambda)r$.
If $\Lambda$ is a Hopf-algebra over $R$ and $M,N$ are left $\Lambda$-modules, then we make $\text{Hom}_R(M,N)$ into a $\Lambda$-module as follows: If $$\Delta(\lambda)=\sum_i \mu_i\otimes \nu_i$$ and $\phi\in \text{Hom}_R(M,N)$, then 
$$\lambda(\phi)(m)=\sum_i\mu_i(\phi(\eta(\nu_i)(m))),$$
where $\eta$ means the antipode of $\Lambda$.
We write $M^*=\text{Hom}_R(M,R)$. Note that we are viewing $M^*$ as a left $\Lambda$-module. Because of the antipode $\eta$, we can regard right $\Lambda$-modules as left $\Lambda$-modules via $\lambda m = m\eta(\lambda)$ and vice-versa.
Now, suppose that $R=k$ is a field and $\Lambda$ is a cocommutative Hopf-algebra. Suppose that $M,N$ are two left $\Lambda$-modules that are finite dimensional as $k$-vector spaces, then the natural vector space isomorphism
$$\text{Hom}_k(M,N)\cong M^*\otimes_kN$$ is a $\Lambda$-module isomorphism.
I don't understand why this last statement is true. 
I know that the natural bijection is given by 
$$f:M^*\otimes_k N\rightarrow \text{Hom}_k(M,N):\phi\otimes w\mapsto \phi(\cdot)w.$$ I only have to show that this map is a $\Lambda$-module morphism. So let $\lambda\in \Lambda$ such that $\Delta(\lambda)=\sum_i\mu_i\otimes \nu_i=\sum_i\nu_i\otimes \mu_i$ (cocommutativity), and we calculate
\begin{eqnarray*}
 (\lambda\cdot f(\phi\otimes w))(v) &=& \sum_i \mu_i(f(\phi\otimes w)(\eta(\nu_i)(v)))\\
&=& \sum_i \mu_i(\phi(\eta(\nu_i)(v))w),\\
f(\lambda\cdot (\phi\otimes w))(v) &=& f(\sum_i (\mu_i\phi)\otimes (\nu_i w))(v)\\
&=& \sum_i (\mu_i\phi)(v)\nu_i w.
\end{eqnarray*}
By cocommutativity $\mu_i$ and $\nu_i$ are interchangeable in the calculation, however if I want to proceed, we need to consider $\Delta(\mu_i)$ to calculate $\mu_i\phi$ since $\mu_i\in \Lambda$ and $\phi\in \text{Hom}_k(M,k)$ (which we gave a $\Lambda$-module structure.) Now this leads to nowhere, so how do I see that this really is a $\Lambda$-module morphism?
Thank you in advance!
 A: I hope I solved my question, though it would be nice if anyone was willing to check this answer. Using the notation above, we get the following:
Suppose that $\Delta(\lambda)=\sum_i\mu_i\otimes\nu_i$ and for any $i$, $\Delta(\mu_i)=\sum_j \alpha_{ij}\otimes \beta_{ij}$. Then 
\begin{eqnarray*}
 (\lambda\cdot f(\phi\otimes w))(v) &=& \sum_i\mu_i(f(\phi\otimes w)(\eta(\nu_i)(v)))\\
&=& \sum_i \mu_i(\phi(\eta(\nu_i)(v))w)\\
&=& \sum_i \phi(\eta(\nu_i)(v))\mu_iw,\\
f(\lambda\cdot (\phi\otimes w))(v) &=& f(\sum_i(\mu_i\cdot\phi)\otimes (\nu_iw))(v)\\
&=& \sum_i f ((\mu_i\cdot\phi)\otimes (\nu_iw))(v)\\
&=& \sum_i (\mu_i\cdot \phi)(v)\nu_iw\\
&=& \sum_i \left[ \sum_j \alpha_{ij}(\phi(\eta(\beta_{ij})(v))) \right]\nu_iw\\
&=& \sum_i \left[\sum_j \varepsilon(\alpha_{ij})\phi(\eta(\beta_{ij})(v))\right]\nu_iw\\
&=& \sum_i \left[ \sum_j \phi(\varepsilon(\alpha_{ij})\eta(\beta_{ij})(v))\right]\nu_iw\\
&=& \sum_i \left[\phi( \eta(\sum_j\varepsilon(\alpha_{ij})\beta_{ij}(v))\right]\nu_iw\\
&=& \sum_i \phi(\eta(\mu_i)(v))\nu_iw\\
&=& \sum_i \phi(\eta(\nu_i)(v))\mu_iw.
\end{eqnarray*}
Hence $f$ is a $\Lambda$-module morphism map, I hope.
