The regular module is a tensor unit: $R{\otimes }_{R}N\cong N$ and $M{\otimes }_{R}R\cong M$

The pairing $(r,n)\mapsto rn$ is additive in each variable and satisfies $(rs)n=r(sn)$, so it is balanced; [L1] and [L2] give ${\lambda }_N$.

The pairing $(m,r)\mapsto mr$ is balanced, so [L1] and [L2] give ${\rho }_M$. Define the additive map ${\eta }_M(m)=m\otimes 1_R$. Then ${\rho }_M{\eta }_M(m)=m$, while ${\eta }_M{\rho }_M(m\otimes r)=mr\otimes 1_R=m\otimes r$ by balance; uniqueness in [L1] makes the second composite the identity. Thus ${\rho }_M$ and ${\eta }_M$ are inverse.

The additive map ${\eta }_N:N\to R{\otimes }_{R}N$, $n\mapsto 1_R\otimes n$, satisfies ${\lambda }_N{\eta }_N(n)=n$, while ${\eta }_N{\lambda }_N(r\otimes n)=1_R\otimes rn=r\otimes n$ by balance. Uniqueness in [L1] makes the second composite the identity, so ${\lambda }_N$ and ${\eta }_N$ are inverse.

Each map commutes with any outer scalar action by associativity of that action, checked on elementary tensors. The calculations remain valid for the zero ring and for zero modules, where all displayed maps are the unique maps between zero groups.

Therefore the regular module is a left and right tensor unit with the stated natural formulas.