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  "body": "\n## Statement\n\nLet $F$ be a field, and let $V$, $W$ and $U$ be vector spaces over $F$.\nLet $h \\colon V \\times W \\to U$ be a bilinear map.\nThen there is exactly one linear map $\\tilde h \\colon V \\otimes W \\to U$ such that\n$$\\tilde h(v \\otimes w) = h(v, w) \\qquad \\text{for all } v \\in V \\text{ and } w \\in W,$$\nwhere $V \\otimes W$ is the [[tensor product|0BF2DE2SYK]].\n\n## Remarks\n\nSo the bilinear maps $V \\times W \\to U$ correspond one to one with the [[linear maps|0NRYHJPQG2]] $V \\otimes W \\to U$.\nThis property determines $V \\otimes W$ up to a unique isomorphism that fixes the vectors $v \\otimes w$.\nUniqueness holds because the vectors $v \\otimes w$ span $V \\otimes W$.\n\n## Explanations\n\n### L1 human\n\nEvery bilinear map out of $V \\times W$ factors through the tensor product in exactly one way.\nSo the study of bilinear maps reduces to the study of linear maps.\nThis property is the usual definition of the tensor product in more abstract settings.\n",
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