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I am reading this book "a guide to quantum groups" written by V.C. and A.P. But the proof of propersition 4.2.3 on page 121 confused me.enter image description here

Just this place " Applying $id \bigotimes S \bigotimes S^2$ to both sides and reversing the order of the factors ,we obtain ...." My question is why the equation we obtain doesn't have the symbol $\bigotimes$? Why LHS equals to RHS after multiplication?

Xiaosong Peng
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    An observation for future questions: the whole surname(s) of the author(s) often provide a better reference to a book than the title itself, so concealing them kind-of goes in the opposite direction of quoting the precise page and number of proposition. –  May 27 '16 at 12:38

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Applying $\operatorname{id}\otimes S \otimes S^2$ to obtain $$\sum r_ja_i\otimes S(r^ja'_i)\otimes S^2(a''_i)=\sum a'_ir_j\otimes S(a_ir^j)\otimes S^2(a''_i)$$ Next we apply the permutation map $P$ given by $P(x\otimes y\otimes z)=z\otimes y \otimes x$, and the result follows after applying the multiplication map $m$, which is omitted by the author. By doing so, we could use the fact that $S$ is an anti-homomorphism to rewrite the result and gives $(8)$.

Paradox
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