Stuck showing that the group ring $\mathbb{C}[G]$ is a bi-algebra

I am learning Hopf algebras via the following online notes on geometric representation theory,

https://www.maths.ed.ac.uk/~djordan/QGpublic.pdf

Exercise 1.7 of the Hopf Algebra section asks to show that $\mathbb{C}[G]$ with coproduct $\Delta(g) = g \otimes g$ and counit $\epsilon(g) = \delta_{e,g}$ gives a bi-algebra structure, where $G$ is a finite group. Using definition 1.6 this amounts to showing that both $\Delta$ and $\epsilon$ are algebra morphisms.

I am ok with everything except showing that $\epsilon$ is an algebra morphism, that is, showing that the following square commutes

$$\require{AMScd} \begin{CD} \mathbb{C}[G] @<{\mu}<< \mathbb{C}[G] \otimes \mathbb{C}[G];\\ @V{\epsilon}VV @VV{\epsilon \otimes \epsilon}V \\ \mathbb{C} @<{\cdot}<< \mathbb{C} \otimes \mathbb{C}; \end{CD}$$

as if $r_1 \otimes r_2 \in \mathbb{C}[G] \otimes \mathbb{C}[G]$ then surely $\epsilon \circ \mu(r_1\otimes r_2) = \epsilon(r_1r_2) = {(r_1r_2)}_e$, i.e. the $[e]$ coefficient of $r_1r_2 \in \mathbb{C}[G]$ whereas $(\cdot \circ \epsilon \otimes \epsilon)(r_1 \otimes r_2) = (r_1)_e \cdot (r_2)_e \in \mathbb{C}$, and these are not the same thing.

What am I misunderstanding? Thanks!

Solution 1:

In the usual definition of the coalgebra structure on $\mathbb{C}[G]$, the counit is defined as $\epsilon(g) = 1$ for all $g\in G\subset \mathbb{C}[G]$, i.e. it sends all the group elements to $1$, then extended linearly. Thus $\epsilon(r_1 r_2)$ should sum up the coefficient of all basis elements of $\mathbb{C}[G]$, instead of just taking the $[e]$ coefficient.

See e.g. Example 1.5.3 in Majid's Foundations of Quantum Group Theory, or in Kassel's Quantum Groups, Chapter III's Section 1 Example 3, and Section 2 Example 2, and Section 3 Example 2, taken together.