Unit commutes with $H$-action.

Question

Let $H$ be a Hopf algebra. Let $A$ be an $H$-module algebra. Then the unit map $\eta: k \to A$ commutes with the $H$-action. It is said that "$\eta: k \to A$ commutes with the $H$-action" is equivalent to the condition $h.1=\epsilon(h)1$, where $\epsilon$ is the counit of $H$. Why "$\eta: k \to A$ commutes with the $H$-action" is equivalent to the condition $h.1=\epsilon(h)1$? Thank you very much.

Answer 1

The $H$-module structure on $k$ is defined by $h\lambda = \epsilon\left(h\right)\lambda$ for every $\lambda \in k$ and $h \in H$. Thus, for every $\lambda \in k$ and $h \in H$, we have $\eta\left(h\lambda\right) = \eta\left(\epsilon\left(h\right)\lambda\right) = \epsilon\left(h\right)\lambda 1_A$ (by the definition of $\eta$).

Now, we have the following chain of logical equivalences:

$\left(\eta : k \to A \text{ commutes with the } H\text{-action}\right)$

$\Longleftrightarrow \left(h\cdot \underbrace{\eta\left(\lambda\right)}_{= \lambda 1_A} = \underbrace{\eta\left(h\lambda\right)}_{=\epsilon\left(h\right)\lambda 1_A} \text{ for all } \lambda \in k\right)$

$\Longleftrightarrow \left(h\cdot \lambda 1_A = \epsilon\left(h\right)\lambda 1_A \text{ for all } \lambda \in k\right)$

$\Longleftrightarrow \left(h\cdot 1_A = \epsilon\left(h\right)\right)$.

(In the last equivalence, the $\Longrightarrow$ direction follows from setting $\lambda = 1_k$, whereas the $\Longleftarrow$ direction follows from multiplying by $\lambda$.)