I'm reading a proof of the following theorem



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> If $H$ is a Hopf algebra with invertible antipode then Yetter-Drinfeld modules of finite dimension form a rigid category. 
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In this proof we define $V^\*$ in a natural way as $\mathrm{Lin}\_k(V,k)$. We define action and coaction on $V^\*$ by $\langle h\rhd f,v \rangle:=\langle f,Sh\rhd v\rangle$ and $\langle\delta(f),v\rangle:=S^{-1}\left(v\_{(-1)}\right)\langle f,v\_{(0)}\rangle$.


There is one step in this proof which I don't understand, namely :


$S^{-1}\left(\left(Sh\right)\_{(1)}v\_{(-1)}\right)\left\langle f, \left(Sh\right)\_{(2)}\rhd v\_{(0)}\right\rangle=\\=S^{-1}\left(\left(\left(Sh\right)\_{(1)}\rhd v\right)\_{(-1)}\left(Sh\right)\_{(2)}\right)\left\langle f, \left(\left(Sh\right)\_{(1)}\rhd v\right)\_{(0)} \right\rangle$ 


How to obtain this equality ? (Maybe it is obvious, but I don't know how to do it.)



Yetter-Drinfeld modules as rigid category
Not sure if a reply is still needed to this.


This is simply the YD-compatibility condition on V (in the form of e.g. Kassel, IX.5, equation (5.2)) applied to $Sh \otimes v$, and afterwards the output in $V$ is evaluated against the element $f$ of $V^\*$, while $S^{-1}$ is applied to the output in the Hopf algebras $H$.
The stated duality theorem is correct, but not used in this equation at all.


The kind of question may also be more suitable for math stackexchange as it is all textbook level material.


