TITLE: Tensor product equivalent definitions?
QUESTION [5 upvotes]: Are these two definitions of a tensor product equivalent?
A tensor product of two vector spaces $V$ and $W$ over field $K$ is a pair $(T,\otimes)$ where $T$ a vector space over $K$ and $\otimes$ a bilinear map with the property that ...
Def 1.(ref) ... for every bilinear map $B_{L}\colon V\times W\to X$ where $X$ any vector space over $K$, there exists a unique linear map $F_{\otimes}\colon T\to X$ so that $B_{L}=F_{\otimes}\circ\otimes$.
Def 2.(ref) ... if $\{\bar{e}_{i}\}$ is a basis of $V$ and if $\{\bar{f}_{j}\}$ is a basis of $W$ then $\{\bar{e}_{i}\otimes\bar{f}_{j}\}$ is a basis of $T$ (used notation $\bar{e}_{i}\otimes\bar{f}_{j}\equiv \otimes(\bar{e}_{i},\bar{f}_{j})$)

REPLY [2 votes]: Recall that for linear maps $f : V \to W$, if we have a basis $\{ v_i \}$ for V and know all of the values $w_i = f(v_i)$, then that is enough to determine the value of $f(v)$ for every vector $v \in V$. Conversely, any choice we make for the $w_i$ gives us a linear map.
The main thing to note here is that we have a similar fact for bilinear maps! If $f : U \times V \to W$ is bilinear, we have bases $\{ u_i \}$ and $\{ v_j \}$, and we know the values $w_{ij} = f(u_i, v_j)$, that is enough to determine every value $f(u,v)$. Conversely, any choice for the $w_{ij}$'s gives us a bilinear map.
But now notice the similarity of the previous two paragraphs. If we had a new based* vector space $T$ whose basis vectors were labelled with pairs $(i,j)$, then picking a bilinear map from $U \times V \to W$ is the same thing as picking a linear map $T \to W$!
Once you recognize this, all of the rest is just grinding out the details.
*: a based vector space is just a vector space for which we've chosen a particular basis.