Can you always construct a map $A\otimes B\to A\times B$?

Question

Suppose we have two $R$-mods $A,\,B$. Can we always construct a homomorphism $A\otimes B\to A\times B$?

Answer 1

Yes, the trivial homomorphism. But apart from that, there is no natural homomorphism.

Proof. Suppose that $\eta_{A,B} : A \otimes B \to A \times B$ is a natural homomorphism, meaning that $\eta_{A,B}$ defined for all $R$-modules $A,B$ and $\eta = (\eta_{A,B})_{A,B}$ is a natural transformation. Let $(r_1,r_2) := \eta_{R,R}(1 \otimes 1) \in R \times R$. If $A,B$ are two $R$-modules and $a \in A$, $b \in B$, then $\eta_{A,B}(a \otimes b) = (r_1 a,r_2 b)$ because of naturality of $\eta$ with respect to $a : R \to A$ and $b : R \to B$. It follows $0 = \eta_{R,R}(0 \otimes 1) = (0,r_2)$, hence $r_2=0$, and likewise $r_1=0$. Thus, $\eta_{A,B}=0$ for all $A,B$. $\square$