Existence of a canonical surjective homomorphism on tensor products I got stuck with Liu's "Algebraic Geometry and Aritmetic Curves" exercise 1.1.7.
Let $B$ be an $A$-algebra, and let $M$, $N$ be $B$-modules. Why do there exists a canonical surjective homomorphism $M\otimes_AN\to M\otimes_BN$?
 A: a) One solution, not depending on the construction of the tensor product but on its universal property,  is to realize that the map $f: M\times N\to M\otimes_B N:(m,n)\mapsto m\otimes n$ is $A$-bilinear and thus factors through an $A$-linear morphism $f: M\otimes_A N\to M\otimes_B N:m\otimes n \mapsto m\otimes n$, which is obviously surjective.
Very nice, but how does this mixing of $A$- modules and $B$- modules make sense? 
b) The point is that a $B$-module has an underlying structure of $A$-module thanks to the ring morphism $\phi:A\to B$ obtained from the $A$-algebra structure on $B$.
Explicitly, the product $a\star m$ of the scalar $a\in A$ and the vector $m\in M$ is given by the product $\phi(a)\cdot m$ of the scalar $\phi(a)\in B$ and the vector $m\in M$. 
c) And now  checking that $f$ above is $A$-bilinear amounts to showing identities like $f(a\star m,n)=a\star f(m,n) $  which translates to $(\phi(a)\cdot m)\otimes n=\phi(a)\cdot (m\otimes n) $.
And this last equality is obviously true in the $B$-module  $M\otimes_B N$  
