In this MSE answer, it is explained maps out of the tensor product of (not necessarily commutative) $R$-algebras classifies pointwise commuting pairs of arrows from the factors in the following sense $$\mathrm{Hom}(A \otimes_R B, C) = \{ (g,h) \in \mathrm{Hom}(A,C) \times \mathrm{Hom}(B,C) : \forall a \in A, b \in B, g(a) \text{ and } h(b) \text{ commute}\}.$$
This notion of "pointwise commutation" is related to a more abstract notion - that of cooperating arrows, which appears in the book Mal'cev, Protomodular, Homological and Semi-Abelian Categories by Bourn and Borceux. The relevant details are below.
Definition. A unital category is a finitely complete pointed category such that the unit injections $\jmath_i:A_i\to A_1\times A_2$ given by $\jmath_1=(1,0),\jmath_2=(0,1)$ where $0,1$ are the zero and identity arrows are jointly extremally epimorphic.
Example. The unital algebraic varieties are unital categories.
Definition. In a unital category, arrows $A\overset{f}{\to}C\overset{g}{\leftarrow}B$ are said to cooperate if there's a unique arrow $\varphi_{f,g} : A\times B\to C$ such that $\varphi_{f,g}\circ\jmath_1 = f$ and $\varphi_{f,g}\circ\jmath_2 = g$.
Example. Monoid homomorphisms cooperate iff their images commute pointwise, i.e $f(x)g(y)=g(y)f(x)$.
Thus, the non-commutative tensor product classifies pairs of ring-homomorphisms which cooperate as multiplicative monoid homomorphisms. On the other hand, the category of unital rings is not unital because it has no zero object, though the presence of a multiplicative unit is of crucial importance in the argument in the linked answer. I am confused.
- What is the right way to organize this information and relate the symmetric monoidal structure with the notion of cooperation?
In this comment to the same answer, it is written:
Actually this is more than an analogy. Take any symmetric monoidal category, then the tensor product of algebra objects makes sense and satisfies the universal property that it classifies "commuting" morphisms of algebras. For $R$-modules this gives the tensor product of $R$-algebras, for sets this gives the (tensor) product of monoids (for example, groups).
- What is the meaning of "tensor product of algebra objects", and what is the notion of "commuting"? Is it related to cooperating?
Added. Just trying to organize the information involved.
- I think we're starting with a (symmetric?) monoidal $(\mathsf C,\otimes)$, e.g $(\mathsf{Set},\times)$ and $(\mathsf{Ab},\otimes,\mathbb Z)$.
- Then, we want a tensor product of internal monoids, call it $\boxtimes$.
- I guess we want $\boxtimes$ to come with injections into it, so that we may analogously define cooperation of internal monoid morphisms: internal monoid morphisms $A\overset{f}{\to}C\overset{g}{\leftarrow}B$ are said to cooperate if there's a unique arrow $A\boxtimes B\to C$ making the obvious diagram with the injections commute.
- Then this $\boxtimes$ is actually characterized by $$\mathsf{Mon}(\mathsf C)(A\boxtimes B,C)\cong \left\{ (g,h)\in \mathsf{Mon}(\mathsf C)(A,C)\times \mathsf{Mon}(\mathsf C)(B,C) \mid g,h\text{ cooperate} \right\},$$ and it is this $\boxtimes$ which retrieves the tensor product of rings. Is that right?