I'm reading Radford's text on Hopf Algebras right now and I'm a bit confused about the definition of a left A-Hopf algebra. The definition given in the book is:

Let $A$ be a $\Bbbk$-bialgebra. A $\underline{\text{left $A$-Hopf module}}$ is a triple ($M$, $\mu$, $\rho$) where ($M, \mu$) is a left $A$-module, ($M, \rho$) is a left $A$-comodule, and $$ \rho(a m) = a_{(1)}m_{(-1)} \otimes a_{(2)}m_{(0)}.$$

The book uses the convention of Sumless Sweedler notation that

1. $\Delta (a) = a_{(1)} \otimes a_{(2)}$
2. $\rho (m) = m_{(-1)} \otimes m_{(0)}$

I'm just looking to gain some intuition about what the axiom for the $\rho$ map in the definition is trying to convey. It's not clear to me what the meaning of this axiom is.

I'm reading Radford's text on Hopf Algebras right now and I'm a bit confused about the definition of a left $A$-Hopf algebra. The definition given in the book is:

Let $A$ be a $\Bbbk$-bialgebra. A left $A$-Hopf module is a triple $(M, \mu, \rho)$ where $(M, \mu)$ is a left $A$-module, $(M, \rho)$ is a left $A$-comodule, and $$ \rho(a m) = a_{(1)}m_{(-1)} \otimes a_{(2)}m_{(0)}.$$

The book uses the convention of sumless Sweedler notation that

1. $\Delta (a) = a_{(1)} \otimes a_{(2)}$
2. $\rho (m) = m_{(-1)} \otimes m_{(0)}$.

I'm just looking to gain some intuition about what the axiom for the $\rho$ map in the definition is trying to convey. It's not clear to me what the meaning of this axiom is.