Direction of map between tensor products 
Suppose $A$ and $B$ are commutative rings, $A\to B$ is a ring map, and $M, N$ are $B$-modules. Is there a map $M\otimes_A N \to M\otimes_B N$, or in the other direction?

This should be very simple but for some reason I'm confused.
Thanks for your help.
 A: There is a natural map $M\otimes_AN\to M\otimes_BN$, which maps $m\otimes n$ to $m\otimes n$.
A: Here is another way to see why a "natural" map $M \otimes_A N \rightarrow M \otimes_B N$ exists: we have that $M \otimes_B N \cong (M \otimes_A N) / E$ where $E$ is the $A$-submodule of $M \otimes_A N$ that is generated by all elements of the form $x \otimes_A by - bx \otimes_A y$ for all $x \in M, y \in N, b \in B$.
Hence the map $M \otimes_A N \rightarrow M \otimes_B N$ is nothing but the natural projection$M \otimes_A N \rightarrow (M \otimes_A N)/E$.
A: Just an elaboration on Mariano's answer. There's a map $M \otimes_A N \to M \otimes_B N$ induced by the $A$-bilinear map $(m, n) \mapsto m \otimes n$. This definitely works.
Why not swap $A$ and $B$ here? My reasoning is that for this to be natural I would want a natural $B$-module structure on $M \otimes_A N$ and I don't see one -- it seems I'd have to pick one of the factors. Even if I did this, I would need to have $(bm) \otimes n = m \otimes (bn)$ inside of $M \otimes_A N$ and there's no reason for that to happen: take $A = k$ and $B = M = N = k[x]$.
Maybe this example could help us remember the direction: does it make more sense to have a surjective map $k[x, y] \to k[x]$ or the other way around?
A: Some thoughts: Perhaps part of the confusion is that you can think of these objects as being $A$ or $B$-modules in many different wayts. We can think of $M,N$ as being left and right $B$ modules, which I'll denote as  $\sideset{_{B\ } }{_B} M$ and $\sideset{_{B\ } }{_B} N$. A ring homomorphism $r$ lets us recast these modules on either the right or the left. So suppose we want to tensor the modules over $A$. We recast $M$ as a left-B, right-A module using $r$, and I'll denote the resulting object by $\sideset{_{B\ } }{_r} M$. $N$ is recast as $\sideset{_{r\ } }{_B}N$. We now take the tensor product: $\sideset{_{B\ } }{_r} M \otimes_A \sideset{_{r\ } }{_B}N$ and see a $B$-module. The inside stuff is "hidden" from us and we can only interact with the object from the left and right, using the multiplication from $B$. 
But there is also a $B$-module $\sideset{_{B\ }}{_B} M \otimes_B \sideset{_{B\ } }{_B}N,$ so I count four different left-modules defined using the tensor product: the two $B$-modules just defined, and the result from changing their scalars using $r$. So there are lots of options for what you want your maps to do, and you have to carefully think about what kind of map you have (module homomorphism? abelian group homomorphism?)
I'm not sure what kind of map you want in the original post, but perhaps this is it: there are two ways to obtain an $A$-module. We can 1) restrict $M,N$ on the inside, tensor over $A$, and then restrict the resulting $B$-module OR 2) tensor over $B$ and restrict the resulting $B$ module. With my notation, the first is written $\sideset{_{r\ }}{_r} M \otimes_A \sideset{_{r\ } }{_r}N$, while the second is $\sideset{_{r\ }}{_B} M \otimes_B \sideset{_{B\ } }{_r}N$. These live in $A$-module world. Now, if we send the pair $(m,n)$ to $m\otimes_B n$, then this is obviously $A$-bilinear, and so we do see an $A$-module homomorphism $$\sideset{_{r\ }}{_r} M \otimes_A \sideset{_{r\ } }{_r}N \rightarrow \sideset{_{r\ }}{_B} M \otimes_B \sideset{_{B\ } }{_r}N.$$ The point is that the stuff on the left is pairs modulo moving $r(a)$ from one side to the other, and the right side is pairs modulo moving anything in $B$ from one side to the other, so is "smaller", and we get a surjective map. 
Note that you might be tempted to try and define the same map $$\sideset{_{B\ }}{_r} M \otimes_A \sideset{_{r\ } }{_B}N \rightarrow \sideset{_{B\ }}{_B} M \otimes_B \sideset{_{B\ } }{_B}N$$ up on the level of $B$-modules using the same universal property. This won't quite work, because the tensor product is between $A$-modules, not $B$-modules! So the LHS no longer possesses the universal property. But I wouldn't be surprised if there was some sort of universal property for $r(A)$-bilinear maps in $B$-modules enjoyed by the LHS that would allow this map to exist. Perhaps someone will prove me wrong.
A: Every $B$-bilinear map is also $A$-bilinear (when we restrict scalars via $A \to B$, which I will denote by $|_A$). Thus, there is a natural transformation $\mathrm{Bilin}_B(M,N,-) \hookrightarrow \mathrm{Bilin}_A(M|_A,N|_A,(-)|_A)$ of functors $\mathsf{Mod}(B) \to \mathsf{Set}$. The Yoneda Lemma tells us that this corresponds to a homomorphism of $A$-modules $M|_A \otimes_A N|_A \to (M \otimes_B N)|_A$. People usually just forget the forgetful functor ;) and write $M \otimes_A B \to M \otimes_B N$ for this.
