---
tag: Y8824EH5DH
kind: proof
name: Proof of the universal property of the tensor product from its construction
msc: [15A69]
proves: 0YDRV01R8F
uses: [0G31VJ5YS5, 0RJF668P1T, 0R8X9XA83A, 0ZERJX6PH6, 02YFME3RGW, 0EPTJA5QV4, 0BJKVF90Z6, 0SVR29SV1V]
sources:
  - url: https://kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod.pdf
    locator: Theorem 3.2
---

## Proof

We use the construction $V \otimes W = L / N$ of the [[tensor product|0G31VJ5YS5]]. Here $L$ has basis the vectors $[v, w]$ for $(v, w) \in V \times W$.

1. 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 $w \in W$.
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|02YFME3RGW]] of $T$ is a [[linear subspace|0EPTJA5QV4]] that contains all generators of $N$. So it contains their span $N$.
5. By the [[universal property of the quotient space|0ZERJX6PH6]], there is a linear map $\tilde h \colon L/N \to U$ with $\tilde h([v, w] + N) = T([v, w])$.
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|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.
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.

## Idea

A 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.