$E'\otimes_kF'=E'\otimes_k F\cap E\otimes_kF'$? Let $E,F$ be two vector spaces over a field $k$. Let $E'$ be a subspace of $E$ and $F'$ a subspace of $F$. Let $i:E'\rightarrow E$ and $j:F'\rightarrow F$ be the canonical injections. Is it true that
$$\text{Im}(i\otimes j)=\text{Im}(i\otimes 1)\cap\text{Im}(1\otimes j)?$$
The direction $\text{Im}(i\otimes j)\subset\text{Im}(i\otimes 1)\cap\text{Im}(1\otimes j)$, since $i(x)\otimes j(y)=i(x)\otimes y=x\otimes j(y)$ for all $x\in E'$ and $y\in F'$. I'm not sure how to prove the other direction.
Any suggestions?
 A: Let $E''$ be a linear complement of $E'$ in $E$ and $F''$ be a linear complement of $F'$ in $F$, so $E=E'\oplus E''$ and $F=E'\oplus E''$.  Since tensor products distribute over direct sums, we have $$E\otimes F=(E'\otimes F')\oplus (E'\otimes F'')\oplus (E''\otimes F')\oplus (E''\otimes F'')$$ (where I am identifying all the tensor products on the right with their images in $E\otimes F$ under the inclusion maps).  With respect to this direct sum decomposition, $E'\otimes F$ consists of the elements whose last two coordinates are $0$, and $E\otimes F'$ consists of the elements whose second and fourth coordinates are $0$.  So the intersection is the elements whose last three coordinates are all $0$, which is just $E'\otimes F'$.
(Note that the use of something like linear complements here is crucial.  In particular, this result is not true in more general contexts where such complements might not exist.  For instance, if you were taking a tensor product of $\mathbb{Z}$-modules, the intersection of the images of $2\mathbb{Z}\otimes\mathbb{Z}$ and $\mathbb{Z}\otimes 2\mathbb{Z}$ in $\mathbb{Z}\otimes\mathbb{Z}$ is strictly larger than the image of $2\mathbb{Z}\otimes 2\mathbb{Z}$.)
