Dual of a finite dimensional algebra is a coalgebra (ex. from Sweedler) Let $(A, M, u)$ be a finite dimensional algebra where $M: A\otimes A \rightarrow A$ denotes multiplication and $u: k \rightarrow A$ denotes unit. 
I want to prove that $(A^*, \Delta, \varepsilon) $ is a colagebra where
$\Delta: A^*\rightarrow A^* \otimes A^*$ is a composition: 
$$A^* \overset{M^*}{\rightarrow}(A\otimes A)^* \overset{\rho^{-1}}{\rightarrow}A^*\otimes A^*$$ 
And $\rho: V^*\otimes W^* \rightarrow (V\otimes W)^*$ is given by $<\rho(v^*, w^*), v\otimes w>=<v^*, v><w^*,w>$.
I have proven that $\rho$ is injective and since $A$ is finite dimensional $\rho$ is also bijective and we can take the inverse $\rho^{-1}$.
But I have problems understanding how does $\Delta$ work.
By definition we have $<M^*(c^*), a\otimes b>=<c^*, M(a\otimes b)>=c^*(ab)$. But I can't understand what is $\rho^{-1}(M^*(c^*))$, or in other words which element of $A^*\otimes A^*$ can act like $M^*(c^*)$ via $\rho$?
P.S. Please correct me if I have grammar mistakes. Thanks!

Given $M^*c^*=:d^* \in (A \otimes A)^*$, $\rho^{-1}(d^*)=d_1^* \otimes d_2^* \in A^* \otimes A^*$, where $d_1^*(a)=d^*(a \otimes 1)$ and $d_2^*(a)=d^*(1 \otimes a)$.  Notice that $\rho (d_1^* \otimes d_2^*)=d^*$.
