A right comodule $N$ over a coalgebra $(C, \Delta, \epsilon)$ is a pair $(N, \rho)$ where $N$ is a linear space and $\rho : N \rightarrow N\otimes C$ is a linear map such that
- The coassociativity condition is satisfied, $$(\rho \otimes id)\circ \rho = (id \otimes \Delta)\circ \rho$$
- The counit condition is satisfied, $$\iota_{N} \circ(id \otimes \epsilon)\circ \rho = id,$$ where $\iota_{N} : N\otimes k \rightarrow N: n\otimes \lambda \mapsto \lambda n$ is the natural linear isomorphism.
Algebra to Coalgebra Construction
Given two arbitrary $k$-vector spaces $X$ and $Y$ define the linear mapping $$\psi_{X,Y} : X^* \otimes Y^* \rightarrow (X\otimes Y)^*$$ by $\psi_{X,Y}(g\otimes h)(x\otimes y) = g(x)h(y)$. This is well-known to be an isomorphism whenever $X$ and $Y$ are finite dimensional. Being a monomorphism trivially, and an epimorphism by counting dimensions.
As mentioned in the original post the whole intuition for my question came from the following well known fact: given an algebra $(A, \mu, 1)$ the finite dimensional vector space $A^*$ has a coalgebra structure $\Delta : A^* \rightarrow A^* \otimes A^*$ defined in Sweedler notation by $\Delta(f) = \sum f_{(1)} \otimes f_{(2)}$ such that $\sum f_{(1)}(x)f_{(2)}(y) = f(xy)$. This is well-defined because $\Delta(f) = \psi_{A,A}^{-1}(f\circ \mu)$. Justification that $(A^*,\Delta, \epsilon)$ is a coalgebra, with counit $\epsilon(f) = f(1)$, can be found in Susan Montgomery's book for example.
Module to Comodule Construction
I have never seen the previous fact generalised to a module to comodule construction. We assume that $(M, \mathfrak{r})$ is a finite dimensional right $A$-module where the module action is explicitly $\mathfrak{r} : M\otimes A \rightarrow M$, and is associative $\mathfrak{r}\circ (\mathfrak{r} \otimes id) = \mathfrak{r} \circ (id \otimes \mu)$, and unital $\mathfrak{r}(m\otimes 1) = m$, $\forall m \in M$. From now on write $\mathfrak{r}(m\otimes a)$ as $m\cdot a$ interchangeably. The candidate for comultiplication is $\rho : M^* \rightarrow M^* \otimes A^*$ written in Sweedler notation as $\rho(f) = \sum f^{(0)}\otimes f^{(1)} \in M^* \otimes A^*$ such that $\sum f^{(0)}(m)f^{(1)}(a) = f(m\cdot a)$, $\forall m \in M, \forall a \in A$. This is well defined by $\rho(f) = \psi_{M,A}^{-1}(f \circ \mathfrak{r})$ because $f \circ \mathfrak{r} \in (M\otimes A)^*$. I shall provide a proof of the following:
The pair $(M^*, \rho)$ is a comodule over the coalgebra $(A^*, \Delta, \epsilon)$.
Showing Coassociativity
We must show that $(\rho \otimes id)\circ \rho = (id \otimes \Delta) \circ \rho$. Notice that the codomain of both left and right hand side is $M^* \otimes A^* \otimes A^*$, isomorphic to $(M\otimes A \otimes A)^*$ by the mapping $\Psi = \psi_{M\otimes A, A} \circ (\psi_{M,A} \otimes id)$, then $\Psi(f\otimes g \otimes h)(m\otimes a \otimes b) = f(m)g(a)h(b)$. In Sweedler notation we have for $f \in M^*$
\begin{align}
[(\rho \otimes id)\circ \rho] (f) = \sum \rho(f^{(0)}) \otimes f^{(1)} = \sum (f^{(0)})^{(0)} \otimes (f^{(0)})^{(1)} \otimes f^{(1)}
\end{align}
for the left hand side, and for the right hand side
\begin{align}
[(id \otimes \Delta) \circ \rho](f) = \sum f^{(0)} \otimes \Delta(f^{(1)}) = \sum f^{(0)} \otimes (f^{(1)})_{(0)} \otimes (f^{(1)})_{(1)}
\end{align}
Now $\Psi$ applied to the first equation gives us
\begin{align*}
\Psi([(\rho \otimes id)\circ \rho] (f))(m\otimes a \otimes b) &= \sum (f^{(0)})^{(0)}(m)(f^{(0)})^{(1)}(a) f^{(1)}(b) \\ &= \sum f^{(0)}(m\cdot a)f^{(1)}(b) \\ &= \sum f((m\cdot a) \cdot b).
\end{align*}
Moreover $\Psi$ applied to the second one gives
\begin{align*}
\Psi([(id \otimes \Delta) \circ \rho](f))(m\otimes a \otimes b) &= \sum f^{(0)}(m) (f^{(1)})_{(0)}(a) (f^{(1)})_{(1)}(b) \\ &= \sum f^{(0)}(m) f^{(1)}(ab) \\ &= \sum f(m\cdot ab)
\end{align*}
Comparing the last two lines of the previous equations and applying $\Psi^{-1}$ we see that, because $M$ is an associative module, both $(\rho \otimes id)\circ \rho$ and $(id \otimes \Delta) \circ \rho$ are equal. We have shown that $\rho$ is coassociative.
Counit Condition
Here we must prove that $\iota_{M}\circ (id\otimes \epsilon)\circ \rho = id$. Since $\rho(f) = \sum f^{(0)} \otimes f^{(1)}$, applying $id \otimes \epsilon$ gives us $\sum f^{(0)}\otimes f^{(1)}(1)$, which when applying $\iota_M$ gives us $\sum f^{(0)}f^{(1)}(1)$. Now $\sum f^{(0)}(m)f^{(1)}(1) = f(m\cdot 1) = f(m)$, $\forall m\in M$ so that $\rho$ satisfies the counit condition.
