Basic Confusion About Tensor Products Let $A$ and $B$ be subspaces of vector spaces $V$ and $W$ respectively.
Given $a\in A$ and $b\in B$, there are two possible interpretations of $a\otimes b$: we can think of it as a member of $A\otimes B$ or as a member of $V\otimes W$.
So I tried to free myself of this apparent ambiguity by defining $f:A\times B\to V\otimes W$ as $f(x, y)=x\otimes y$ for all $(x, y)\in A\times B$, where, of course, $x\otimes y$ is thought of as a member of $V\otimes W$ and not $A\otimes B$.
Since $f$ is a bilinear map, it induces a unique linear map $\bar f:A\otimes B\to V\otimes W$ which send $x\otimes y\in A\otimes B$ to $x\otimes y\in V\otimes W$ for all $x\in A$ and $y\in B$.

If $\bar f$ were injective, then the ambiguity could be dissolved. But I am unable to show it.

What am I missing?

Choose bases $(E_i)$ and $(F_j)$ for $A$ and $B$ and extend them to bases for $V$ and $W$, respectively. We get a basis $(E_i\otimes F_j)$ for $A\otimes B$, and in the same way the extensions give a basis for $V\otimes W$. Then, we get that $\bar{f}$ maps a basis of $A\otimes B$ to a linearly independent subset of $V\otimes W$ (a subset of a basis). This shows that $\bar{f}$ is injective.
