Showing $\lambda: A \otimes C^* \rightarrow \text{Hom}(C,A)$ is a morphism of algebras 
Show that $\lambda: A \otimes C^* \rightarrow \text{Hom}(C,A)$ is a morphism of algebras.

Let either $C^*$ or $A$ be finite dimensional, and let $\lambda$ be the isomorphism $\lambda: A \otimes C^* \rightarrow \text{Hom}(C,A)$ defined by $$\lambda(a,\gamma)(x)=a\gamma(x)$$ for $a \in A, \gamma \in C^*$ and $x \in C$.
Let $(C,\Delta,\epsilon)$ be a colalgebra and $(A, \mu, \nu)$ be an algebra where $\Delta, \mu$ are the coproduct and product whilst $\epsilon, \nu$ are the counit and unit.
Define the convolution $@$ for $f,g \in \text{Hom}(C,A)$ by $$(f @ g)(x) = \mu (f \otimes g) \Delta(x)$$
Then we have the following string of equalities for $a,b \in A$ and $\alpha,\beta \in C^*$
$$\lambda(a \otimes \alpha) @ \lambda(b \otimes \beta)(x)=\Sigma_{(x)}\alpha(x')\beta(x'')ab= (\alpha \beta)(x)ab=(\lambda(ab \otimes \alpha \beta))(x)$$
Can somebody explain to me the equailty $\Sigma_{(x)}\alpha(x')\beta(x'')ab= (\alpha \beta)(x)ab$ please?

 A: The dual space $C^*$ is an algebra because $C$ is a coalgebra.
The multiplication of $C^*$ comes from “dualizing” the comultiplication of $C$.
Let us be more precise.
The comultiplication
$$
  \Delta
  \colon
  C
  \to
  C \otimes C
$$
induces a linear map
$$
  \Delta^*
  \colon
  (C \otimes C)^*
  \to
  C^* \,,
  \quad
  \Delta^*(\varphi)(x)
  =
  \varphi( \Delta(x) ) \,.
$$
We also have a linear map
$$
  \Phi
  \colon
  C^* \otimes C^*
  \to
  (C \otimes C)^*
$$
given by
$$
  \Phi(\alpha \otimes \beta)(x \otimes y)
  =
  \alpha(x) \beta(y)
$$
for all $\alpha, \beta \in C^*$ and $x, y \in C$.
The composite
$$
  \Delta^* \circ \Phi
  \colon
  C^* \otimes C^*
  \to
  C^*
$$
is the induced multiplication on $C^*$, which makes $C^*$ into an algebra.
This composite is given by
\begin{align*}
  (\Delta^* \circ \Phi)(\alpha \otimes \beta)(x)
  &=
  \Delta^*( \Phi( \alpha \otimes \beta) )(x)
  \\
  &=
  \Phi( \alpha \otimes \beta )( \Delta(x) )
  \\
  &=
  \Phi( \alpha \otimes \beta )\left( \sum_{(x)} x' \otimes X'' \right)
  \\
  &=
  \sum_{(x)} \Phi( \alpha \otimes \beta )( x' \otimes x'' )
  \\
  &=
  \sum_{(x)} \alpha(x') \beta(x'')
\end{align*}
for all $\alpha, \beta \in C^*$ and $x \in C$.
The above calculation explains that the multiplication on the algebra $C^*$ is given by
$$
  (\alpha \beta)(x)
  =
  \sum_{(x)} \alpha(x') \beta(x'')
$$
for all $\alpha, \beta \in C^*$ and $x \in C$.
The identity in question,
$$
  (\alpha \beta)(x) ab
  =
  \sum_{(x)} \alpha(x') \beta(x'') ab \,,
$$
follows directly from this.
