Why is the coproduct $\Delta$ of the quantum double $D(H)$ an algebra homomorphism? If $H$ is a a finite dimensional Hopf algebra, $H^\ast\otimes H$ has a Hopf algebra structure with multiplication
$$
(\phi\otimes h)(\psi\otimes g)=\sum \psi_2\phi\otimes h_2g\langle Sh_1,\psi_1\rangle\langle h_3,\psi_3\rangle
$$
and coproduct
$$
\Delta(\phi\otimes h)=\sum (\phi_1\otimes h_1)\otimes (\phi_2\otimes h_2).
$$ 
The proof that $\Delta$ is an algebra morphism goes as
$$
\begin{align*}
\Delta(\phi\otimes h)\Delta(\psi\otimes g)&= (\phi_1\otimes h_1)(\psi_1\otimes g_1)\otimes(\phi_2\otimes h_2)(\psi_2\otimes g_2)\\
&= \psi_2\phi_2\otimes h_2g_1\otimes \psi_5\phi_2\otimes h_5g_2\langle Sh_1,\psi_1\rangle\langle h_3,\psi_3\rangle\langle Sh_4,\psi_4\rangle\langle h_6,\psi_6\rangle\\
&= \psi_2\phi_1\otimes h_2g_1\otimes\psi_3\phi_2\otimes h_3g_2\langle Sh_1,\psi_1\rangle\langle h_4,\psi_4\rangle\\
&=\Delta((\phi\otimes h)(\psi\otimes g))
\end{align*}
$$
I don't get how the third equality follows. Does $\psi_5\phi_2\otimes h_5g_2\langle h_3,\psi_3\rangle\langle Sh_4,\psi_4\rangle$ shrink back to $\psi_3\phi_2\otimes h_3g_2$ or something? 
Maybe something like
$$
\langle h_3,\psi_3\rangle\langle Sh_4,\psi_4\rangle=\langle h_3\otimes Sh_4,\psi_3\otimes \psi_4\rangle=\langle h_3\otimes Sh_4,\Delta\psi_3\rangle=\langle h_3Sh_4,\psi_3\rangle=\epsilon(h_3)
$$
and then relabeling the neutral notation changes $\psi_5\phi_2$ to $\psi_3\phi_2$ and the $h_5$ in $h_5g_2$ becomes $h_3$ after multiplying by $\epsilon(h_3)$ after relabeling indices?
 A: Well, loosely speaking that's the idea, but it's formally invalid.  When using Sweedler notation you can't just siphon off indices you don't like and "isolate" the parts you do want: If you have indices $3,4,5$ then you have to have $1,2$ somewhere.  Indices only make sense in complete sets.  But if you take your idea and put it into the full expression then it works out:
\begin{align*}
\psi_2\phi_2\otimes & h_2g_1\otimes \psi_5\phi_2\otimes h_5g_2\langle Sh_1,\psi_1\rangle\langle h_3,\psi_3\rangle\langle Sh_4,\psi_4\rangle\langle h_6,\psi_6\rangle\\ &= \psi_2\phi_2\otimes h_2g_1\otimes \psi_5\phi_2\otimes h_5g_2\langle Sh_1,\psi_1\rangle\langle h_3\otimes Sh_4,\psi_3\otimes \psi_4\rangle\langle h_6,\psi_6\rangle\\
&=\psi_2\phi_2\otimes h_2g_1\otimes \psi_4\phi_2\otimes h_5g_2\langle Sh_1,\psi_1\rangle\langle h_3\otimes Sh_4,\Delta \psi_3\rangle\langle h_6,\psi_5\rangle\\
&= \psi_2\phi_2\otimes h_2g_1\otimes \psi_4\phi_2\otimes h_5g_2\langle Sh_1,\psi_1\rangle\langle h_3 Sh_4, \psi_3\rangle\langle h_6,\psi_5\rangle\\
&= \psi_2\phi_2\otimes h_2g_1\otimes \psi_4\phi_2\otimes h_4 g_2\langle Sh_1,\psi_1\rangle\langle \epsilon(h_3),\psi_3\rangle\langle h_5,\psi_5\rangle\\
&= \psi_2\phi_2\otimes h_2g_1\otimes \psi_3\phi_2\otimes h_3 g_2\langle Sh_1,\psi_1\rangle\langle h_4,\psi_4\rangle\\
\end{align*}
as desired.
