I am working on the following problem:
Let $A \in Mod-R$ and $B \in R-Mod$. Prove that $A \otimes_R B \cong (A \otimes_{\mathbb{Z}} B)/ H$ where $H=\langle ar\otimes_{\mathbb{Z}}b - a\otimes_{\mathbb{Z}}rb \rangle$ is the subgroup generated by those elements.
I don't really know how to do this. I tried to prove that $(A \otimes_R B)/ H$ satisfies the universal property of the tensor product but I couldn't do anything with that.
By the universal property of $A \otimes_R B$, there is a natural homomorphism $A \otimes_R B \to (A \otimes B)/H$, $a \otimes_R b \mapsto a \otimes b + H$. To go the other way around, note that the universal property of $A \otimes B$ gives us a homomorphism $A \otimes_R B \leftarrow A \otimes B$; this map descends to a homomorphism $A \otimes_R B \leftarrow (A \otimes B)/H$, since $H$ belongs to its kernel, and it is clear that this map is an inverse to the first one.