Let $k$ be a field, over which we consider algebras and coalgebras. A $k$-coalgebra is a comonoid object in $k$-modules, and a $k$-algebra is a monoid object in $k$-modules. A $k$-bialgebra is equivalently a monoid object in $k$-coalgebras or a comonoid object in $k$-algebras.
Let $A$ be a coalgebra. Do we have an isomorphism $A \cong [A, A]_{A \text{-comod}}$, just like we have an isomorphism $B \cong [B, B]_{B \text{-mod}}$?
Let $A$ be bialgebra. The map $f : [A, A]_{A \text{-comod}} \rightarrow [A^*, A^*]_{A^* \text{-mod}}$ sending $f$ to $f^*$ seems to be a $k$-module map (and it has $(fg)^* = g^* f^*$).
From (1) and (2), it seems we have a map $A \cong [A, A]_{A \text{-comod} } \rightarrow [A^*, A^*]_{A^* \text{-mod}} \cong A^*$ for any bialgebra $A$. In other words, and what the point of this has all been, we have a canonical map $A \rightarrow A^*$, unlike the situation in simple $k$-modules, where we require a bilinear form.
Firstly, is all this correct? Secondly, is there a simpler way to express this canonical map $A \rightarrow A^*$? What is an explicit form? I suppose it would be $a \mapsto (b \mapsto \epsilon(ab))$. I suppose I'm trying to ask an array of similar questions as to what this map is about. I guess I am trying to ask an array of questions - I'm looking for someone to say a bit about whether this all seems correct and whether they have an easier way of expressing it.
It seems this is a nice way of viewing the $k$-vector space isomorphism $k[G] \cong k(G)$ (between the function algebra and the group algebra of a group). These are dual Hopf algebras and the above scenario gives maps between them.