Tensor products of maps Let $V, W, U, X$ be $R$-modules where R is a ring. At what level of generality, if any is it true that the maps (I always mean linear) from $V \otimes W$ to $U \otimes X$ can be identified with $L(V, U)\otimes L(W, X)$ where $L(., .)$ is the space of maps, via the mapping that takes a tensor product of maps to the map that acts on elementary tensors "componentwise?"  I can see that this natural map that might establish the identification is a homomorphism from $L(V, U)\otimes L(W, X)$ to $V \otimes W$ to $U \otimes X$.  When it makes sense to speak of dimensions, I can also see that the dimension is suggestive that perhaps it is an isomorphism.  But is it one at any level of generality of $R$?  This is to justify the usual notation of $f \otimes g$ to refer to the map between tensor products, when the same symbol already refers to the element in the tensor product of $L(. , .)$ spaces.  I suppose even without the isomorphism, and only a homomorphism, the notation is already well-defined, but I'd like to know anyway.
Edit: To clarify for the reader, the universal property has been used twice.  Once to establish that $f \otimes g$ defines a map, and a second time to show that the map taking (f, g) to $f \otimes g$ the map defines yet another map, which is the homomorphism in question.
 A: Ah there is a "bare hands" proof as I desired in the field case.  You just say: suppose $f \otimes g$ considered as a map is 0.  Then it's 0 applied to the fundamental tensors.  Suppose f is not 0.  I will even prove that g is 0.  Let $v$ be a member of $B$ which is a basis for $V$ such that $f(v)$ doesn't vanish. Now for any $w$ in $W$ we find that $f(v)\otimes g(w)=0$.  Suppose toward a contradiction that $g(w)\neq0$.  Then extend $g(w)$ to a basis of X.  Extend $f(v)$ to a basis of U. (I am critically using here that $R$ is a field.  Not even a free module would suffice because I need to know I can extend to a basis, not just that I can find one.)  Then define the bilinear map that takes $(f(v), g(w))$ to 1 and pairs all other basis vector combinations together to 0. (It maps into $R$ the field.)  Then by page 10 characterization (2) here we are done: http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf
That shows injectivity of my natural map.  Surjectivity follows because the dimensions are right. (Again this is a field argument, or at least module over a PID or something like that, but definitely for fields.)
This works when the vector spaces are finite dimensional.
