As a lemma for a small proof in a paper I'm writing, I need to prove that given some vector spaces $V = \bigotimes_i V^i$, $W = \bigotimes_j W^j$ over the same field $\mathbb{F}$, if a linear function $S: V \to W$ preserves pure vectors (i.e., $\forall \bigotimes_i v^i, \exists \bigotimes w^j$ s.t. $S(\bigotimes_i v^i) = \bigotimes_j w^j$), then it can be factorised into a set of linear maps and constant terms $\{S^j\}_j$ such that $\bigotimes_j S^j = S$. I only need this to hold for the case where $\{V^i\}_i$, $\{W^j\}_j$ contain only finitely many vector spaces each, but generality is appreciated.
I have already managed to find a proof, but it is comically long and tedious. I am convinced that this is such a simple, elementary theorem, that is must have already been discovered and named long ago.
Is this a standard result, and if it is, what is its name? I would really appreciate not having to finish typesetting my long proof just to serve as a lemma in an appendix.
edit: as Omnomnomnom pointed out, the theorem does not quite hold for the problem as stated above, however I meant (but failed to include) that all the vector spaces and their constituent vectors are labelled, and so we can freely permute the order of our terms without ambiguity.
The statement does not generally hold. I suspect that it holds, however, if it is known that each of the spaces $V_i$ are of distinct dimension, as are the spaces $W_j$. Or, we could allow for permutations of the identical spaces.
For a counterexample, take $V = W = \Bbb R^2 \otimes \Bbb R^2$, and consider the map $S$ defined such that $$ S(v \otimes w) = w \otimes v. $$ $S$ preserves pure tensors, but it cannot be factorized in the fashion you suggest.
To see that this is the case, note that if we identify $\Bbb R^2 \otimes \Bbb R^2$ with $\Bbb R^{2 \times 2}$ via $v \otimes w \mapsto wv^T$, then $S$ corresponds to the map $X \mapsto X^T$. On the other hand, the "factorizable" maps $S_1 \otimes S_2$ correspond precisely to maps on $\Bbb R^{2 \times 2}$ of the form $X \mapsto AXB$. Since there exist no $A,B$ for which $AXB = X^T$ for all $X$, we conclude that $S$ cannot be factorized.
Here is an idea for an inductive proof of the correct version of the theorem. Here, we want to show that we can write $S = \pi \circ \bigotimes_j S^j$ for some "permutation" $\pi$.
The statement for $V = V_1 \otimes V_2$ can be made more efficient by exploiting the isomorphism explained above.
Suppose that we have proven the statement for $V = V_1 \otimes V_2$. We have $$ S: (V_1 \otimes \cdots \otimes V_n) \otimes V_{n+1} \to \bigotimes_j W_j, $$ and this map preserves pure tensors. Let $V = V_1 \otimes \cdots \otimes V_n$. Applying the statement for two spaces shows that $S = \pi_1 \circ (S_0 \otimes S_n)$. Now, apply the inductive hypothesis to $S_0$.