In the article. It is said that the inverse of the map $$ {\displaystyle c_{V,W}:V\otimes W\to W\otimes V}, \\ {\displaystyle c(v\otimes w):=v_{(-1)}{\boldsymbol {.}}w\otimes v_{(0)},} $$ is $$ {\displaystyle c_{V,W}^{-1}(w\otimes v):=v_{(0)}\otimes S(v_{(-1)}){\boldsymbol {.}}w.} $$ I am trying to verify this fact as follows. \begin{align} & c^{-1}(c(v \otimes w)) \\ & = c^{-1}(v_{(-1)}.w \otimes v_{(0)}) \\ & = (v_{(0)})_{(0)} \otimes S( (v_{(0)})_{(-1)} ).(v_{(-1)}.w) \\ & = (v_{(0)})_{(0)} \otimes (S((v_{(0)})_{(-1)}) v_{(-1)}).w \\ & = v_{(0)} \otimes (S(v_{(-1)})v_{(-2)}).w \\ & = v_{(0)} \otimes v_{(-1)}.w. \end{align} But this is not $v \otimes w$. How to show that $c^{-1}$ defined above is the inverse of $c$? Thank you very much.
Edit: here $V$ is a Yetter-Drinfeld module over $H$.
I think the idea is to write $$ c^{-1}(v_{(-1)}\cdot w \otimes v_{(0)}) = v_{(0)(0)} \otimes S(v_{(0)(-1)})\cdot(v_{(-1)}\cdot w) $$ then use the comodule condition $$ v_{(-1)}\otimes v_{(0)(-1)}\otimes v_{(0)(0)} = v_{(-1)(1)}\otimes v_{(-1)(2)}\otimes v_{(0)} $$ to get $$= v_{(0)} \otimes S(v_{(-1)(2)})\cdot (v_{(-1)(1)}\cdot w) = v_{(0)}\otimes (S(v_{(-1)(2)})v_{(-1)(1)})w = v_{(0)}\otimes \epsilon(v_{(-1)})w$$ using an antipode axiom, and finally $$ = v\otimes w$$ as $v=v_{(0)}\epsilon(v_{(-1)})$ by one of the coalgebra axioms.