Do we have $\epsilon(h)=\epsilon(h_{(1)})\epsilon(h_{(2)})$?

Question

Let $H$ be a Hopf algebra and $\epsilon: H \to \mathbb{C}$ the counit. Do we have $\epsilon(h)=\epsilon(h_{(1)})\epsilon(h_{(2)})$?

I think that $h = \epsilon(h_{(1)})h_{(2)}$. Therefore $\epsilon(h) = \epsilon(\epsilon(h_{(1)})h_{(2)}) = \epsilon(h_{(1)})\epsilon(h_{(2)})$. Is this correct? Here $h_{(1)}\otimes h_{(2)} = \Delta(h)$. Thank you very much.

Answer 1

Your argument is correct: $$\epsilon(h) = \epsilon(\epsilon(h_{(1)})h_{(2)}) = \epsilon(h_{(1)})\epsilon(h_{(2)})$$ due to the facts that:

• the compatibility condition between the comultiplication and the counity is: $$ (Id\otimes\epsilon)\circ\Delta=(\epsilon\otimes Id)\circ\Delta= Id \Leftrightarrow \\ \Leftrightarrow h_{(1)}\epsilon(h_{(2)}))=\epsilon(h_{(1)})h_{(2)}=h $$
• and the counity map $\epsilon: H \to \mathbb{C}$, is by definition, $\mathbb{C}$-linear.