Can we recover the bases of two infinite-dimensional vector spaces into a tensor product? I know that in general, if V,W are infinite-dimensional vector spaces with basis {${v_i}$} and {${w_j}$} respectively it holds that V⊗W has as basis {${v_i⊗w_j}$}.
My question is: what about the reciprocal? That is: if {${v_i}$} and {${w_j}$} are families of vectors in V and W respectively such that the family {${v_i⊗w_j}$} is a basis of V⊗W, are {${v_i}$} and {${w_j}$} bases of V and W respectively?
 A: The answer is yes: if $\{v_i \otimes w_j\}$ is a basis of $V \otimes W$ then $\{v_i\}$ and $\{w_j\}$ are bases of $V$ and $W$, respectively. Actually, we need to account for the following slight technicality: we must assume that $V, W \neq 0$.
Proof. Since $V, W \neq 0$, $\{w_j\} \neq \varnothing$; fix some vector $w$ from the indexed set $\{w_j\}$. Let $g : V \otimes W \to V$ be the linear function defined by $v_i \otimes w_j \mapsto v_i$. Let $f : V \to V \otimes W$ be the linear function $f(v) = v \otimes w$. By inspection, $g \circ f = \operatorname{id}_{V}$, so $g$ is surjective, which means that $\{v_i\}$ spans $V$. To show that $\{v_i\}$ is linearly independent, suppose $\alpha_1 v_{i_1} + \dots + \alpha_k v_{i_k} = 0$ for some scalars $\alpha_1, \dots, \alpha_k$ and some indices $i_1, \dots, i_k$. Then
$$0 = f(\alpha_1 v_{i_1} + \dots + \alpha_k v_{i_k}) = \alpha_1 (v_{i_1} \otimes w) + \dots + \alpha_k (v_{i_k} \otimes w).$$
Since $\{v_i \otimes w_j\}$ is linearly independent, we conclude that $\alpha_1 = \cdots = \alpha_k = 0$, as desired. This shows that $\{v_i\}$ is a basis of $V$; symmetrically we have that $\{w_j\}$ is a basis of $W$.
Edit: posted a little too soon, the proof is complete now!
