Bialgebra and Hopf algebra I'm reading ''HOPF ALGEBRAS'' by S. Dascalescu, C. Nastasescu and S. Raianu. At the beginning of the four chapter the following is explained:
''Proposition 4.1.6. Let H be a finite dimensional bigalgebra. Then $H^*$ together with algebra structure which is dual to the coalgebra structure of H, and with the coalgebra structure which is dual to the algebra structure of H is a bialgebra, which is called the dual bialgebra of H''.
Proof. We denote by $\Delta $ and $\epsilon$ the comultiplication and the counit of $H,$ and by the $\delta$ and $E$ the comultiplication and the counit of $H^*.$ We racall that for $h^*\in H^*,$ we have $E(h^*) = h^*(1)$ and $\delta(h^*) = \sum h_1^*\otimes h_2^*,$ where $h^*(hg) =\sum h_1^*(h)h_2^*(g) $ for any $h, g\in H.$ Let us show that $\delta$ is a morphism of algebra. Indeed, if $h^*, g^*\in H^*$ and $\delta(h^*) = \sum h_1^*\otimes h_2^*,$ $\delta(g^*) = \sum g_1^* \otimes g_2^*,$ then for any $h, g\in H,$ we have
$$ \begin{array}{lll}
(h^*g^*)(hg) &=& \sum h^*(h_1g_1)g ^*(h_2g_2)\\
&=&\sum h_1^*(h_1)h_2^*(g_1)g_1^*(h_2)g_2^*(g_2)\\
&=& \sum (h_1^*g_1^*)(h) (h_2^*g_2^*)(g).
\end{array}$$
My question is why  $$(h^*g^*)(hg) = \sum h^*(h_1g_1)g ^*(h_2g_2) ?$$ $$ (h_1^*g_1^*)(h)  = h_1^*(h_1)g_1^*(h_2)  ?$$
Thank you advance!
 A: The multiplication of $H^*$ comes from dualizing the comultiplication of $H$ as follows:
Since $\Delta \colon H \to H \otimes H$ is linear it induces a linear map $\Delta \colon (H \otimes H)^* \to H^*$, $f \mapsto f \circ \Delta$.
We have a canonical inclusion $\varphi \colon H^* \otimes H^* \to (H \otimes H)^*$ which is given on simple tensors by
$$
    \varphi(h_1^* \otimes h_2^*)(g_1 \otimes g_2)
  = h^*_1(g_1) h^*_2(g_2).
$$
(Since $H$ is finite dimensional this is actually an isomorphism.)
The multiplication of $H^*$ is now given by the composition $\Delta^* \circ \varphi \colon H^* \otimes H^* \to H^*$.
This multiplication can also be described explicitely on elements:
For $h^*_1, h^*_2 \in H^*$ and $g \in H$ with $\Delta(g) = \sum_{(g)} g_1 \otimes g_2$ we have
$$
  (h^*_1 h^*_2)(g) = \sum_{(g)} h^*_1(g_1) h^*_2(g_2).
$$
This does in particular explain the second formula.
For the first formula we also have to use that $\Delta \colon H \to H \otimes H$ is multiplicative:
Given that $\Delta(h) = \sum_{(h)} h_1 \otimes h_2$ and $\Delta(g) = \sum_{(g)} g_1 \otimes g_2$ we find that
$$
    \Delta(hg)
  = \Delta(h) \Delta(g)
  = \left( \sum_{(h)} h_1 \otimes h_2 \right)
    \left( \sum_{(g)} g_1 \otimes g_2 \right)
  = \sum_{(h),(g)} (h_1 g_1) \otimes (h_2 g_2).
$$
By using the previous explicit formula for the multiplication of $H^*$ we find that
\begin{align*}
    (h^* g^*)(h g)
 &= \sum_{(h),(g)} h^*(h_1 g_1) g^*(h_2 g_2)
 \\
 &= \sum_{(h),(g),(h^*),(g^*)} h^*_1(h_1) h^*_2(g_1) g^*_1(h_2) g^*_2(g_2)
 \\
 &= \sum_{(h^*), (g^*)}
    \left(
      \sum_{(h)} h^*_1(h_1) g^*_1(h_2)
    \right)
    \left(
      \sum_{(g)} h^*_2(g_1) g^*_2(g_2)
    \right)
  \\
  &= \sum_{(h^*), (g^*)} (h^*_1 g^*_1)(h) (h^*_2 g^*_2)(g).
\end{align*}
