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.