Proving antipode of Hopf algebra property

Question

So I am trying to show $\epsilon(S(a)) = \epsilon(a)$, where $\epsilon$ is the co-unit and $S$ is the antipode. \begin{align} \epsilon(S(a)) &= \epsilon( S( \sum_{(a)} a^{(1)} \epsilon(a^{(2)}) ) ) \\ &= \epsilon( S( \sum_{(a)} \epsilon(a^{(1)}) a^{(2)} ) ) \\ &= \epsilon( \sum_{(a)} \epsilon(a^{(1)}) S(a^{(2)}) ) \\ &= \epsilon( \epsilon(a) I ) \\ &= \epsilon(a) \end{align} However, I am unsure of the last 3 lines. Particularly the case $$ \sum_{(a)} \epsilon(a^{(1)}) S(a^{(2)}) = \epsilon(a)I $$ What is the identity that is used here?

Answer 1

The identity in question $$ \sum_{(a)} ε( a^{(1)} ) S( a^{(2)} ) = ε(a) 1_H \tag{$\ast$} $$ isn’t true. To see this, we consider for a non-trivial group $G$ and its group-algebra $[G]$. The comultiplication, counit and antipode of $[G]$ are given by $$ Δ(g) = g ⊗ g \,, \quad ε(g) = 1 \,, \quad S(g) = g^{-1} $$ for every element $g$ of $G$. It follows that for every non-trivial element $g$ of $G$, $$ \sum_{(g)} ε( g^{(1)} ) S( g^{(2)} ) = ε( g ) S( g ) = 1 ⋅ g^{-1} = g^{-1} ≠ 1 = ε(g) \,. $$

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If we take a step back, then we see that the identity $(\ast)$ can’t make sense. We arived at the left-hand side of $(\ast)$ from the chain of (correct) equalities $$ S(a) = S\biggl( \sum_{(a)} ε( a^{(1)} ) a^{(2)} \biggr) = \sum_{(a)} ε( a^{(1)} ) S( a^{(2)} ) \,. $$ Therefore, the formula $(\ast)$ is just a complicated reformulation of the equivalent formula $$ S(a) = ε(a) 1_H \,. $$ And this is clearly nonsense.