TITLE: A diagram for understanding action/coaction compatibility in a Yetter-Drinfeld module
QUESTION [7 upvotes]: For a Hopf algebra $H$ with antipode $S$, let $M$ be a left $H$-module with the action $h \otimes m \mapsto \rho(h,m)$, and also a left $H$-comodule with coaction $\delta \colon m \mapsto m^{(-1)} \otimes m^{(0)}$. For $M$ to be a Yetter-Drinfeld module, it must satisfy the compatibility condition 
$$
  \delta(\rho(h,m)) = h_{(1)}m^{(-1)} S(h_{(3)}) \otimes \rho(h_{(2)},m^{(0)})\,.
$$
Is there a natural commutative diagram to draw that illustrates this compatibility condition? I've included my attempt below in a CW answer. Also, is there any more reason behind this condition besides "it's the condition we need to be true to get the nice braiding to work out in the Yetter-Drinfeld category?".

REPLY [3 votes]: The equivalent way of writing the Yetter-Drinfeld compatibility condition between left action and left coaction described in this post can be easily translated into a commutative diagram. It looks like this
$$
(\nabla\otimes\rho)(1\otimes\tau\otimes 1)(\Delta\otimes\delta)=(\nabla^\mathrm{op}\otimes 1)(1\otimes\delta)(1\otimes\rho)(\Delta^\mathrm{op}\otimes 1)\,.
$$
Similary equations work for ${}_H\mathcal{YD}^H$ instead of ${}_H^H\mathcal{YD}$, see for example Kassel, Definition IX.5.1, page 220. These equations tell us how $\rho$ and $\delta$ "commute". We need precisely this condition to compute the Drinfeld center of ${}_H\mathrm{Mod}$.