Isomorphism of a torsion product and a quotioten of torsion product.

I have the following problem:

If $A'$ a submodule of the right $R$-module $A$ and $B'$ a submodule of the left $R$-module $B$, then $A/A' \otimes B/B' \cong (A\otimes B)/C$ where $C$ is the subgroup of all elements of the form $a'\otimes b$ and $a \otimes b'$ with $a\in A$, $a'\in A$, $b\in B$, $b'\in B'$.

Let $\pi: A\times B\rightarrow A/A'\times B/B'$ be the canonical map. By the universal propierty of tensor product there exist and $\phi:A\otimes_R B\rightarrow A/A'\times B/B'$ such that $\phi\circ\iota=\pi$. Then we can define $\bar\phi=\iota'\circ \phi$. I want to prove $\ker \bar\phi=C$. How can I do that? Or anyone know another aproximation?

Note: $\iota:A\times B \rightarrow A\otimes B$ and $\iota:A/A'\times B/B' \rightarrow A/A'\otimes B/B'$ are canonical.

Thanks a lot!

Let us informally write $$A/A'=\{a+a',a\in A,a'\in A'\},B/B'=\{b+b',b\in B,b'\in B'\}$$ where $a,b$ are class representatives and $a',b'$ can be any elements in $A', B'$. We now see the tensor product $$(a+a')\otimes (b+b')=a\otimes b+a'\otimes b+a\otimes b'+a'\otimes b'$$ But $a'\otimes b, a\otimes b',a'\otimes b'\equiv 0$ in $A/A'\otimes B/B'$. Therefore the whole product is just $a\otimes b$. It is easy to see this construction is the same as giving a decomposition of $A\otimes B$, quotient out submodule $C$ generated by elements like $a\otimes b'$ and $a'\otimes b$. So the two modules must be isomorphic.