# 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?

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: