Inducing a comodule structure on Hom If C is an R-coalgebra and M is an R-module... then is it possible to endow $Hom_{_RMod}(C,M)$ or $Hom_{_RMod}(M,C)$ with the strucutre of a C-comodule?
 A: Hint. We have a canonical span $\hom_R(M,C) \to \hom_R(M,C \otimes_R C) \leftarrow \hom_R(M,C) \otimes_R C$. The right arrow is an isomorphism when $M$ is finitely generated projective (first check $M=R$, then direct sums, then direct summands).
I don't see such a construction for $\hom_R(C,M)$.
A: Maybe this is not what you were asking for, but it should be useful.
Let $k$ be a field, and $C$ be a finite-dimensional $k$-coalgebra. The question is are there any $C$-comodule structures on its linear dual $C^{\ast}\,:=\mathrm{Hom}(C,\,k)$?
Since $C$ is a coalgebra, then $C^{\ast}$ is naturally an algebra. We write $C^{\ast}=:\,A$. Then $A^{\ast}\cong C$ because $C$ is finite dimensional. There is an $A$-bimodule structure on $A^{\ast}$, hence on $C$. Dualizing this bimodule structure, you should get a $C$-bicomodule structure on $C^{\ast}$. For example, the $A$-left module structure on $C$
$$A \otimes C \rightarrow C$$
would induce a $C$-left comodule structure on $A$
$$C \otimes A \leftarrow A\,.$$
Conceptually, the construction above is straightforward. However, you might want to know how to write down the coactions explicitly. Let $\{\xi_{i}\}$ be a basis of $C$ and $\{\xi^{\ast}_{i}\}$ be its dual basis. Given $f\in C^{\ast}$, the right coaction $\Delta_{r}$ is given by 
$$\Delta_{r}(f)\,=\,\sum_{i}\xi^{\ast}_{i}\bigotimes (\mathrm{id}\otimes f)\Delta(\xi_{i})\,,$$
while the left coaction $\Delta_{l}$ is given by
$$\Delta_{l}(f)\,=\,\sum_{i}(f\otimes\mathrm{id})\Delta(\xi_{i})\bigotimes \xi^{\ast}_{i}\,.$$
It would be a routine exercise to check those are indeed induced by the right and left actions of $A$ on $C$.
