So I was trying to prove that if $(H,m, \eta, \Delta, \epsilon)$ is a Hopf algebra with antipode $s$, then $s$ is an antimorphism of co-algebras, that is
$\Delta \ s = (s \otimes s) \ \Delta^{op}$
The proof involes showing the functions $F = \Delta \ s$ and $G = (s \otimes s) \ \Delta^{op} $, $ F, G : H \rightarrow H \otimes H$ are right and left inverses of $\Delta$ with respect to the convolution product. The part of showing $G * \Delta = \eta \epsilon$ is what gets me because lines 1 and 2 are simply nonsenical to me. In line 0 I think the author is defining $h_1 = (h_1)_1, h_2 = (h_1)_2$ and so on, which seems fine. But then in line 1 the autor just swaps around $h_4$ with $h_3$ and suddenly $h_2$ becomes $(h_2)_1$ when it was $(h_1)_2$. I just don't understand how he swaps around the $h_i$. Same in line 2,
$ S(h_1) \epsilon (h_2) h_3$ doesn't seem to be the same as $S(h_1) \epsilon((h_2)_1) (h_2)_2$
Any input on this magic will be greatly appreciated.
After thinking a little I realized the reason to this is using the sweedler notation for co associavity. For example:
$ h_1 \otimes h_2 \otimes h_3 \otimes h_4 = h_1 \otimes (h_2)_1 \otimes (h_2)_2 \otimes h_3$, swaping first and third coordinate and then second and first gives
$h_2 \otimes h_3 \otimes h_1 \otimes h_4 = (h_2)_1 \otimes (h_2)_2 \otimes h_1 \otimes h_3$. Applying the function $S \otimes Id \otimes S \otimes Id$ on both sides gives
$S(h_2) \otimes h_3 \otimes S(h_1) \otimes h_4= S((h_2)_1) \otimes (h_2)_2 \otimes S(h_1) \otimes h_3$, and then the function $m\otimes m$ on both sides
$S(h_2).h_3 \otimes S(h_1). h_4 = S((h_2)_1).(h_2)_2 \otimes S(h_1).h_3$ which was the desired result.
$\\$
Part 2 starts similarly,
$h_1 \otimes h_2 \otimes h_3 =h_1 \otimes(h_2)_1 \otimes (h_2)_2$, swaping first and second gives
$h_2 \otimes h_1 \otimes h_3= (h_2)_1 \otimes h_1 \otimes (h_2)_2$, and applying the function $ \epsilon \otimes S \otimes Id$ on both sides results in
$\epsilon (h_2) \otimes S(h_1) \otimes h_3 = \epsilon ((h_2)_1) \otimes S(h_1) \otimes (h_2)_2$ and then applying $Id \otimes m$
$\epsilon (h_2) \otimes S(h_1) . h_3 = \epsilon ((h_2)_1) \otimes S(h_1) . (h_2)_2$
$ \epsilon ((h_2)_1) \otimes S(h_1) . (h_2)_2 = 1 \otimes S(h_1) .\epsilon ((h_2)_1) . (h_2)_2 = 1 \otimes S(h_1) .h_2 $
and that's it