Proof of the universal property of the tensor product from its construction

Proof

We use the construction $V \otimes W = L / N$ of the tensor product. Here $L$ has basis the vectors $[v, w]$ for $(v, w) \in V \times W$.

1. By the extension from a basis, there is exactly one linear map $T \colon L \to U$ with $T([v, w]) = h(v, w)$ for all $v \in V$ and $w \in W$.
2. Let $v, v' \in V$ and $w \in W$. Then $T([v + v', w] - [v, w] - [v', w]) = h(v + v', w) - h(v, w) - h(v', w) = 0$, since $h$ is linear in its first argument.
3. In the same way, $T$ maps the three other kinds of generators of $N$ to $0$. This uses the additivity of $h$ in $w$, and $h(av, w) = a h(v, w) = h(v, aw)$.
4. The kernel of $T$ is a linear subspace that contains all generators of $N$. So it contains their span $N$.
5. By the universal property of the quotient space, there is a linear map $\tilde h \colon L/N \to U$ with $\tilde h([v, w] + N) = T([v, w])$.
6. Since $v \otimes w$ is the class $[v, w] + N$, step 5 says $\tilde h(v \otimes w) = h(v, w)$. This proves existence.
7. For uniqueness, the vectors $v \otimes w$ span $V \otimes W$. Indeed each element of the quotient space is $x + N$ with $x \in L$, and $x = \sum_k c_k [v_k, w_k]$ is a finite combination.
8. So $x + N = \sum_k c_k\, (v_k \otimes w_k)$, since the quotient map is linear.
9. Let $g \colon V \otimes W \to U$ be linear with $g(v \otimes w) = h(v, w)$ for all $v, w$. By step 8 and linearity, $g(x + N) = \sum_k c_k h(v_k, w_k) = \tilde h(x + N)$.
10. So $g = \tilde h$, and $\tilde h$ is the only linear map with the required property.