Let $R$ be a $k$-algebra (R commutative) and $M$, $T$ be $R$-bimodules. $R^e := R \otimes_k R^{op}$
I am stuck on this computation: $T^e \otimes_{R^e}M = T \otimes_RM\otimes_RT$.
By using associativity, $(T \otimes_{k}T^{op}) \otimes_{R^e}M = T \otimes_{k}(T^{op} \otimes_{R^e}M)$. Also, $T^{op}$ (resp. $M$) is a right (resp. left) ${R^e}$-module, hence both $R$-bimodules. Is it true that $T^{op} \otimes_{R^e}M \cong M \otimes_{R^e}T$?