I was wondering if anyone could explain why the following proof of linear independence is not valid.
Choose the identity function, $i: V \times W \to V \times W$. This function is clearly bilinear so by definition of the tensor product it induces a linear function $\bar{i}:V \otimes W \to V \times W$ defined by $\bar{i}(v \otimes w) = (v,w).$ Next we observe that if the bases are given as above, then $\{(v_i,w_j)\}_{i \in I, j \in J}$ forms a basis for $V \times W$. Now we may use the result that if the image of a set of elements under a linear function is a basis, then those elements are linearly independent, along with $\{v_i \otimes w_j\}_{i \in I, j \in J} \subset V \otimes W$ as the set with the observation that $\bar{i}(v_i \otimes w_j) = (v_i,w_j)$ to conclude that $\{v_i \otimes w_j\}_{i \in I, j \in J}$ is linearly independent.
The identity function is not bilinear. $(a+b,c)=(a,c)+(b,0)\neq (a,c)+(b,c)$.