Twist map as a solution of the Quantum Yang-Baxter Equation (QYBE) I am a beginner learning Quantum Groups, I have a question of how to show that that twist map $\tau_{M,M}:M\bigotimes M \rightarrow M\bigotimes M$ is a solution to the QYBE.
I tried to prove it by definition: When $R=\tau$,
$R_{(1,2)}$$R_{(1,3)}$$R_{(2,3)}$=
$(\tau \bigotimes 1_M)(1_M\bigotimes \tau)(\tau \bigotimes 1_M)(1_M \bigotimes \tau)(1_M \bigotimes \tau)$=
$(1_M \bigotimes \tau)(1_M\bigotimes \tau)(\tau \bigotimes 1_M)(1_M \bigotimes \tau)(\tau \bigotimes 1_M)$=
$R_{(2,3)}$$R_{(1,3)}$$R_{(1,2)}$
I have some questions:
1) Is the above method correct?
2) What does $\bigotimes$ mean? I only know vaguely that it is "tensor product".
Sincere thanks for help. 
 A: Let $m_1\otimes m_2\otimes m_3\in M\otimes M\otimes M$, then
$$
\begin{align}
R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1\otimes m_2\otimes m_3)&=
(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(1_M\otimes\tau)(m_1\otimes m_2\otimes m_3)\\
&=(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(m_1\otimes m_3\otimes m_2)\\
&=(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(m_1\otimes m_2\otimes m_3)\\
&=(\tau\otimes 1_M)(1_M\otimes\tau)(m_2\otimes m_1\otimes m_3)\\
&=(\tau\otimes 1_M)(m_2\otimes m_3\otimes m_1)\\
&=m_3\otimes m_2\otimes m_1\\
\end{align}
$$
$$
\begin{align}
R_{(2,3)}R_{(1,3)}R_{(1,2)}(m_1\otimes m_2\otimes m_3)&=
(1_M \otimes \tau)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(m_1\otimes m_2\otimes m_3)\\
&=(1_M \otimes \tau)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(m_2\otimes m_1\otimes m_3)\\
&=(1_M \otimes \tau)(1_M\otimes\tau)(\tau\otimes 1_M)(m_2\otimes m_3\otimes m_1)\\
&=(1_M \otimes \tau)(1_M\otimes\tau)(m_3\otimes m_2\otimes m_1)\\
&=(1_M \otimes \tau)(m_3\otimes m_1\otimes m_2)\\
&=m_3\otimes m_2\otimes m_1\\
\end{align}
$$
so we conclude
$$
R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1\otimes m_2\otimes m_3)=R_{(2,3)}R_{(1,3)}R_{(1,2)}(m_1\otimes m_2\otimes m_3)\tag{1}
$$
Now take arbitrary $u\in M\otimes M\otimes M$, then we have representation 
$$
u=\sum\limits_{i=1}^n m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)}
$$
Hence using $(1)$ we get
$$
\begin{align}
R_{(1,2)}R_{(1,3)}R_{(2,3)}(u)
&=R_{(1,2)}R_{(1,3)}R_{(2,3)}\left(\sum\limits_{i=1}^n m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)}\right)\\
&=\sum\limits_{i=1}^n R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)})\\
&=\sum\limits_{i=1}^n R_{(2,3)}R_{(1,3)}R_{(1,2)}(m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)})\\
&=R_{(2,3)}R_{(1,3)}R_{(1,2)}\left(\sum\limits_{i=1}^n m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)}\right)\\
&=R_{(2,3)}R_{(1,3)}R_{(1,2)}(u)\\
\end{align}
$$
Since $u\in M\otimes M\otimes M$ is arbitrary we conclude
$$
R_{(1,2)}R_{(1,3)}R_{(2,3)}=R_{(2,3)}R_{(1,3)}R_{(1,2)}
$$
