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  "body": "\n## Proof\n\nWe use the construction $V \\otimes W = L / N$ of the [[tensor product|0G31VJ5YS5]]. Here $L$ has basis the vectors $[v, w]$ for $(v, w) \\in V \\times W$.\n\n1. By [[the extension from a basis|0R8X9XA83A]], there is exactly one [[linear map|0RJF668P1T]] $T \\colon L \\to U$ with $T([v, w]) = h(v, w)$ for all $v \\in V$ and $w \\in W$.\n2. 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.\n3. 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 $h(av, w) = a h(v, w) = h(v, aw)$.\n4. The [[kernel|02YFME3RGW]] of $T$ is a [[linear subspace|0EPTJA5QV4]] that contains all generators of $N$. So it contains their span $N$.\n5. By the [[universal property of the quotient space|0ZERJX6PH6]], there is a linear map $\\tilde h \\colon L/N \\to U$ with $\\tilde h([v, w] + N) = T([v, w])$.\n6. Since $v \\otimes w$ is the class $[v, w] + N$, step 5 says $\\tilde h(v \\otimes w) = h(v, w)$. This proves existence.\n7. For uniqueness, the vectors $v \\otimes w$ [[span|0BJKVF90Z6]] $V \\otimes W$. Indeed each element of the [[quotient space|0SVR29SV1V]] is $x + N$ with $x \\in L$, and $x = \\sum_k c_k [v_k, w_k]$ is a finite combination.\n8. So $x + N = \\sum_k c_k\\, (v_k \\otimes w_k)$, since the quotient map is linear.\n9. 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, $g(x + N) = \\sum_k c_k h(v_k, w_k) = \\tilde h(x + N)$.\n10. So $g = \\tilde h$, and $\\tilde h$ is the only linear map with the required property.\n\n## Idea\n\nA linear map out of $L$ may send the basis vectors $[v, w]$ anywhere, so we send them to $h(v, w)$. Bilinearity of $h$ says exactly that the relations in $N$ go to $0$, so the map passes to the quotient. Uniqueness holds because the vectors $v \\otimes w$ span the tensor product.\n",
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