Name of theorem for the factorisability of pure-vector preserving linear maps? 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.
 A: 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 $S_0$.
