Linear maps on tensor products Short question. Suppose we have vector spaces $V_1,V_2,V_3,V_4$ and a linear map
$f: V_1\otimes V_2 \to V_3 \otimes V_4$.
Are there always linear maps $f_1: V_1 \to V_3$ and
$f_2: V_2 \to V_4$, such that
$f\simeq f_1 \otimes f_2$?
If yes, why is it?
 A: Even a linear map $V\otimes V\to V\otimes V$ needn't be of the form $A\otimes B$, e.g. $v\otimes w\mapsto w\otimes v$; this example indicates the possibility of linear maps "mixing things" across the $\otimes$ symbol.
The embedding ${\rm hom}(V_1,V_3)\otimes\hom(V_2,V_4)\hookrightarrow\hom(V_1\otimes V_2,V_3\otimes V_4)$ is easily seen to be a surjection if $V_1,V_2,V_3,V_4$ are all finite-dimensional, simply by comparing dimensions, but there will necessarily be elements in $\hom\otimes\hom$ that are not expressible as pure tensors, except in the very special circumstances of $\dim V_1=\dim V_3=1$ or $\dim V_2=\dim V_4=1$.
A: No, this is not always the case, for example if $k$ is a field then consider the map $k \otimes k \to k^2 \otimes k^2$ defined by $1 \otimes 1 \mapsto (1, 0) \otimes (1, 0) + (0, 1) \otimes (0, 1)$.
Another way to see this is that if these are spaces over the finite field $\mathbb F_p$ and $V_1, V_2, V_3, V_4$ have dimensions $a, b, c, d$ respectively then $\hom(V_1, V_3)$ has $p^{ac}$ elements, $\hom(V_2, V_4)$ has $p^{bd}$ elements, and $\hom(V_1 \otimes V_2, V_3 \otimes V_4)$ has $p^{abcd}$ elements but there are $p^{ac + bd}$ maps in $\hom(V_1, V_3) \times \hom(V_2, V_4)$ and hence $p^{ac + bd - 1}$ maps of the form $f_1 \otimes f_2$.
