Universal property of the tensor product for balanced maps into abelian groups

Regard $A$ as a $\mathbb{Z}$-module by integer multiplication in its additive group. A group homomorphism between abelian groups is automatically $\mathbb{Z}$-linear: additivity gives $f(na)=nf(a)$ for $n\ge 0$, and $f(-a)=-f(a)$ gives the formula for negative integers. Thus $\mathbb{Z}$-module homomorphisms and group homomorphisms between these underlying additive groups are the same maps.

If $h:M{\otimes }_{R}N\to A$ is a group homomorphism, then $(m,n)\mapsto h(m\otimes n)$ is balanced because the elementary tensors satisfy all three relations in [L1].

Apply [L4] at $S=\mathbb{Z}$ and $X=M\times N$ to extend the set map $b$ uniquely to a $\mathbb{Z}$-linear map $\tilde{b}:F\to A$ satisfying $\tilde{b}(e_{(m,n)})=b(m,n)$.

By [L2], $\tilde{b}$ sends each generator of $H$ to zero: the two additivity generators map respectively to $b(m+m^{\mathrm{\prime }},n)-b(m,n)-b(m^{\mathrm{\prime }},n)$ and $b(m,n+n^{\mathrm{\prime }})-b(m,n)-b(m,n^{\mathrm{\prime }})$, while the balance generator maps to $b(mr,n)-b(m,rn)$. Hence $H\subseteq \ker \tilde{b}$.

By [L5], $\tilde{b}$ factors uniquely through a group homomorphism $\bar{b}:F\mathrm{/}H=M{\otimes }_{R}N\to A$, and $\bar{b}(m\otimes n)=\tilde{b}(e_{(m,n)})=b(m,n)$.

The operations in steps 4.1 and 1.2 are inverse: starting from $b$ recovers its values on every pair, while starting from $h$ produces a homomorphism agreeing with $h$ on every elementary tensor, and those tensors generate $M{\otimes }_{R}N$. This proves both the asserted uniqueness and the displayed bijection.

No selection is made in the construction. If $M=0$ or $N=0$, every balanced map out of $M\times N$ is zero and the relations make every elementary tensor zero, so the same proof gives $M{\otimes }_{R}N=0$; the zero ring is covered as well. Since a module contains its zero element, $M\times N$ is never empty, so there is no separate empty-domain case.