Base of tensor product Given the tensor product $V_1 \otimes V_2$ (so we have a bilinear map $\otimes : V_1 \times V_2 \to V_1 \otimes V_2$) and bases $B_i$ of $V_i$ (not need to be finite) how to prove using only the universal property (u.p.) that $\otimes(B_1 \times B_2)$ is a base of $V_1 \otimes V_2$?
My attempt is to consider the map $f: V_1 \times V_2 \to  \langle \otimes(B_1 \times B_2) \rangle$ with $f(v_1,v_2)=v_1 \otimes v_2$ and apply u. p. in somehow.
 A: Let $B_1:=\{u_1,\dots, u_m\}$ and $B_2:=\{v_1,\dots,v_n\}$ denote bases of $V_1$ and $V_2$ respectively.  From the bilinearity of the tensor product it follows that
$$
u_i\otimes v_j,\hspace{0.2in} i=1,\dots, m,~j=1,\dots, n
$$
spans $V_1\otimes V_2$.  What we need to do is prove that $\{u_i\otimes v_j\}$ is linearly independent.
So suppose then that
$$
\sum_{i,j} c_{ij} u_i\otimes v_j = 0.
$$
Let $f_{ij}: V_1\times V_2\rightarrow \mathbb{R}$ be the bilinear map defined by
$$
f_{ij}(u_k,v_l)=\delta_{ik}\delta_{jl}
$$
In other words, $f_{ij}(u_k,v_l)=0$ whenever $k\neq i$ or $j\neq l$ and $f_{ij}(u_k,v_l)=1$ when $k=i$ and $l=j$.  By the universal property of the tensor product, there exists a linear map $\widehat{f}_{ij}: V_1\otimes V_2\rightarrow \mathbb{R}$ such that
$$
\widehat{f}_{ij}\circ \otimes = f_{ij}
$$
Hence, $\widehat{f}_{ij}(u_k\otimes v_l)=1$ only when $k=i$ and $j=l$.  So
$$
\begin{align*}
0&=\widehat{f}_{ij}\left(\sum_{k,l}c_{kl}u_k\otimes v_l\right)\\
&=\sum_{k,l}c_{kl}\widehat{f}_{ij}(u_k\otimes v_l)\\
&=\sum_{k,l}c_{kl}\delta_{ki}\delta_{lj}\\
&=c_{ij}
\end{align*}
$$
which proves linear independence of $\{u_k\otimes v_l\}$.
