Checking the antipode for the dual of the Group Hopf Algebra Consider the vector space $\mathbb{C}G\,=\left\{ f:\,G\longrightarrow\mathbb{C}\right\}$ , and define the following  


*

*multiplication: $\mu\left(f\otimes
   g\right)\left(x\right):=f\left(x\right)g\left(x\right)$,

*co-multiplication: $\triangle\left(f\right)\left(x\otimes   
   y\right)\,:=f\left(xy\right)$,   

*unit: $\eta\left(x\right):=1_{G}$,

*co-unit: $\epsilon\left(f\right)\,:=f\left(1_{G}\right)$,

*antipode: $S\left(f\right)\left(x\right):=f\left(x^{-1}\right)$.


It is well known it is a Hopf Algebra (look at the second example on Wikipedia article). Anyway I'm trying to demonstrate that is valid the relation for the antipode 
$$\mu\circ\left(S\otimes id\right)\circ\triangle=\eta\circ\varepsilon=\mu\circ\left(id\otimes S\right)\circ\triangle,$$
and I'm getting quite confused on how should I proceed since it result to me that
$$\left(\mu\circ\left(S\otimes id\right)\circ\triangle\left(f\right)\right)\left(x\otimes y\right)=f\left(x^{-1}y\right)\neq f\left(xy^{-1}\right)$$ 
Can anybody write che proof explicitly or at least give me an hint on how to proceed?
 A: Write $A$ for your Hopf algebra, and $k={\mathbb C}$, and $e$ for the identity element of $G$. 
For clarity, let's view $A$ only using the RHS of your definition - namely, as $k$-valued functions on $G$, and denote by  $\delta\in A$ for the function defined by 
$$\delta (g ) = 1\in k\text{, for all $g\in G$.}$$
We have 


*

*$\varepsilon \colon A \to k$ is evaluation at $e$: $$\varepsilon a  = a( e).$$ 

*$\eta \colon k \to A$, with $\eta (z ) = z\delta$, for $z\in k$.

*Therefore, one has $\eta \circ \varepsilon \colon A \to A$, with 
$$\left[(\eta\circ \varepsilon) (a )\right] (g) = a(e) \delta (g) =a(e).$$


On the other hand,


*

*$\Delta\colon A \to A\otimes A$. The algebra $A\otimes A$ is the set of $k$-valued functions on $G\times G$, and  $(\Delta a) (g, h) = a ( gh ).$
(Your evaluation of the tensor product $x\otimes y$ didn't make sense, and I think that is part of the confusion in your attempt to prove the equality.)


*

*$\mu (a \otimes b ) (g) = a ( g ) b ( g)$. In other words, $\mu$ is the dual to the diagonal map $G \to G \times G$.



Your equality (to be proved)
$$ \mu\circ\left(S\otimes id\right)\circ\triangle=\eta\circ\varepsilon=\mu\circ\left(id\otimes S\right)\circ\triangle$$
is an equality of $A\to A$ maps - i.e., we should evaluate both sides on an arbitrary element $a \in A$, and see that the result is the same function on $G$.
So, if $a\in A$, and $g\in G$, you want to test the value  (in $k$) of 
$$ \left[(\mu\circ\left(S\otimes id\right)\circ\triangle) (a) \right] (g) $$
Running through the definitions, we get that that is equal to
$$ \left[(( S \otimes id )\circ \triangle ) (a )\right] ( g, g ) =\triangle (a) ( g^{-1},g ) = a ( g^{-1} g) = a(e).$$
Edit/Addendum
As requested, here is a clarification (I hope!) of 'Running through the definitions':
First of all, it might be worthwhile to repeat the above 'dual to diagonal' (maybe 'transpose' would have been a better word than 'dual') statement for $\mu$. Identify $A\otimes A$ with the $k$-valued functions on $G \times G$. Then, if $c\in A\otimes A$, we have, by duality/transpose, i.e., by definition after the identification of $A\otimes A$ with functions on $G \times G$, that
 $$\mu (c ) (g) = c (g,g ).$$
Next, set  $\phi= \left(S\otimes id\right)\circ\triangle $.  
Then 
$$ (\mu\circ\left(S\otimes id\right)\circ\triangle) (a) = (\mu \circ \phi ) ( a )  = \mu ( \phi ( a ) ).$$
Therefore, by the previous observation, $$ \mu (\phi ( a ) ) ( g ) = \phi (a ) ( g, g).$$
(Note that $\phi : A \to A\otimes A$, so the evaluation of $\phi(a)$ on $(g,g)$ makes sense.)
Similarly, $$\phi ( a ) = (\  \left(S\otimes id\right)\circ\triangle\  ) (a ) = \left(S\otimes id\right) ( \triangle (a ) ).$$
Therefore 
$$ \phi(a) (g,g) = (\triangle (a ) ) (g^{-1}, g).$$
This last is equal to 
$ a ( g^{-1}g ) = a (e ) $.
A: I write $f = \sum \lambda_g \delta_g$, and compute $\eta \circ \varepsilon (f) = \eta(\lambda_1) = \lambda_1 (\sum_g \delta_g)$.
Now $\Delta(f) = \sum_{g,h} \lambda_g \delta_h \otimes \delta_{h^{-1}g}$ so $((S \otimes id) \circ \Delta )(f) = \sum_{g,h} \lambda_g \delta_{h^{-1}} \otimes \delta_{h^{-1}g}$. 
Finally $\mu \circ (S \otimes id) \circ \Delta(f) = \sum_{g,h} \lambda_g \delta_{h^{-1}}\delta_{h^{-1}g} = \lambda_1 (\sum_g \delta_g)$
