Let $V,W$ be a vector space over $K$ with basis $\{v_i\}(i\in I), \{w_j\}(w\in J)$
$\begin{eqnarray} S_1 &=& \{(v_1 + v_2,w)-(v_1,w)-(v_2,w)|v_1,v_2\in V,w\in W\}\\
S_2&=&\{(v,w_1+w_2)-(v,w_1)-(v,w_2)|v\in V,w_1,w_2\in W\}\\
S_3&=&\{(kv,w)-k(v,w),(v,kw)-k(v,w)|v\in V,w\in W,k\in K\}\\
S&=&S_1\bigcup S_2\bigcup S_3 \end{eqnarray}$
Now we construct a tensor product of $V$ and $W$ as a quotient space $$V\bigotimes W=(\bigoplus_{i\in V\times W}K)/span\{S\}$$
Let $v\otimes w\in V\bigotimes W$ denote
$(\oplus_{i'\in V\times W} k_{i'})+span\{S\}$
where $(k_{i'}=1$ for $i'=v\times w$ otherwise $k_{i'}=0$)
Is there any way to prove $\{v_i\otimes w_j\}$ (with$(i,j)\in I\times J)$ are basis for the tensor product without using an $\mathbf {UNIVERSAL\quad PROPERTY?}$
It is easy to show $span\{v_i\otimes w_j|i\in I,j\in J\}=\bigoplus_{i\in V\times W}k_i/span\{S\}\quad(=V\bigotimes W)$ since $v\otimes w=(\sum_i k^i_v v_i\otimes \sum_j k^j_w w_j)=\sum_{i,j} k^i_v k^j_w v_i\otimes w_j $ . I can prove it by only using elementwise argument in
$(\bigoplus_{i\in V\times W}k_i)/span\{S\}$. In other words, without using a universal property.
However I feel difficult to prove $\{v_i\otimes w_j|i\in I,j\in J\}$ are linearly independent in $\bigoplus_{i\in V\times W}k_i/span\{S\}$ without using an universal property. I even cannot prove $v_1\otimes w_1\neq v_2\otimes w_2$ in $(\bigoplus_{i\in V\times W}k_i)/span\{S\}\quad (=V\bigotimes W)$ without using an universal property too.
I would appreciate for your answer.
Best Answer
I found the answer.
Suppose $$\begin{eqnarray} v&=&\sum_i k^i_v v_i \\ w&=&\sum_j k^j_w w_j \\ f:\bigoplus_{ v\times w \in V\times W} K_i &\longrightarrow & \bigoplus_{l\in I\times J} K_l \\ \oplus_{v\times w \in V\times W} k_{v\times w} &\longmapsto & \sum_{v\times w \in V\times W}\oplus_{i\times j \in I\times J} k_{v\times w}{k^i_v}{k^j_w} \end{eqnarray} $$
Then it follows that $f(\text{span}\{S\})=\{\oplus_{i\times j \in I\times J} 0 \}$.
Now we find $x_1\otimes y_1 \neq x_2 \otimes y_2,$ since $f(x_1\otimes y_1)\neq f(x_2 \otimes y_2)$, and in the same way, we can prove $\{v_i\otimes w_j\mid i\in I,j\in J\}$ are linearly independent.