Question about the Sweedler notation and the proof of a property of the antipode Let $(H,m, \eta, \Delta, \epsilon)$ be a Hopf algebra with antipode $s$. A basic property states that $s$ is an antimorphism of algebras. That is, $s(xy)= s(y)s(x)$. The usual proof involves showing that the morphisms $v$ and $w$ defined as
 $\begin{align}v: H \otimes H \rightarrow H  \\ x \otimes y \rightarrow s(xy)\end{align}$
$\begin{align}w: H \otimes H \rightarrow H  \\ x \otimes y \rightarrow s(y)s(x)\end{align}$
are a left and right inverse (with respect to the convolution) of the morphism $m : H \otimes H \rightarrow H$ respectively. By unicity, $v= w$. The problem is in the usual proof of showing that $v * m = \eta \epsilon$ (for example in the book by Sorin Dascalescu, Constantin Nastasescu, Serban Raianu "Hopf algebras" but also for example on the notes here ) they always use the step (in sweedler notation)
$ s(h_{(1)}g_{(1)})h_{(2)}g_{(2)} = s((hg)_{(1)})(hg)_{(2)} $
which they justify as $\Delta$ is a morphism of algebras. But my understanding of sweedler notation gives that $\Delta$ a morphism of algebras implies $h_{1}.g_{1}.h_{2}.g_{2} = (hg)_1. (hg)_2, h, g \in H$, not that  $h_{1}.g_{1} = (hg)_1$ and $h_{2}.g_{2} =  (hg)_2$
So, I don't really know if it's true that 
$ s(h_{(1)}g_{(1)})h_{(2)}g_{(2)} = s((hg)_{(1)})(hg)_{(2)} $. For example I know that if $f$ is a morphism of coalgebras, sweedler notation means that $f(c)_1 \otimes f(c)_2 = f(c_1) \otimes f(c_2)$, not that $f(c)_1 = f(c_1)$. Could this be a mistake in the proof?
 A: Let us assume $\Bbbk$ is the base field and that $\otimes:=\otimes_\Bbbk$. To say that $\Delta$ is a morphism of algebras means that the following diagrams involving $\Bbbk$-vector spaces and $\Bbbk$-linear maps commutes 
$\require{AMScd}$
\begin{CD}
H\otimes H @>{m_H}>> H @.@.@. \Bbbk @= \Bbbk\\
@V{\Delta_{H\otimes H}}VV  @VV{\Delta_H}V @.@. @Vu_{H}VV @VVu_{H\otimes H}V \\
(H \otimes H)\otimes (H\otimes H) @>>{m_{H\otimes H}}> H\otimes H @.@.@. H @>>\Delta_H> H\otimes H
\end{CD}
where $u_H(1_\Bbbk)=1_H$, $u_{H\otimes H}(1_{\Bbbk})=1_H\otimes 1_H$ and (on generators)
\begin{gather*}
\Delta_{H\otimes H}(x\otimes y)=\sum_{(x),(y)}\left(x_{(1)}\otimes y_{(1)}\right)\otimes \left(x_{(2)}\otimes y_{(2)}\right), \\
m_{H\otimes H}((x\otimes y)\otimes (z\otimes t))=(xz)\otimes (yt),
\end{gather*}
for all $x,y,z,t\in H$. In particular
$$\sum_{(xy)}(xy)_{(1)}\otimes (xy)_{(2)}=\sum_{(x),(y)}x_{(1)}x_{(2)}\otimes y_{(1)}y_{(2)}.$$
If you apply now $m_{H}\circ\left(s\otimes H\right)$ to both sides you get the relation you are looking for.
