# Kerodon

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Example 2.1.0.2. Let $k$ be a field and let $U$, $V$, and $W$ be vector spaces over $k$. Recall that a function $b: U \times V \rightarrow W$ is said to be $k$-bilinear if it satisfies the identities

$b(u+u',v) = b(u,v) + b(u',v) \quad \quad b(u,v + v') = b(u,v) + b(u,v')$
$b( \lambda u, v) = \lambda b(u, v) = b(u, \lambda v) \text{ for \lambda \in k.}$

We say that a $k$-bilinear map $b: U \times V \rightarrow W$ is universal if, for any $k$-vector space $W'$, composition with $b$ induces a bijection

$\{ \text{k-linear maps W \rightarrow W'} \} \simeq \{ \text{k-bilinear maps U \times V \rightarrow W'} \} .$

If this condition is satisfied, then $W$ is determined (up to unique isomorphism) by $U$ and $V$; we refer to $W$ as the tensor product of $U$ and $V$ and denote it by $U \otimes _{k} V$ . The construction $(U,V) \mapsto U \otimes _{k} V$ then determines a functor

$\otimes _{k}: \operatorname{Vect}_{k} \times \operatorname{Vect}_{k} \rightarrow \operatorname{Vect}_{k},$

which we will refer to as the tensor product functor. It is associative in the following sense: for every triple of vector spaces $U,V,W \in \operatorname{Vect}_{k}$, there exists a canonical isomorphism

$U \otimes _{k} (V \otimes _{k} W) \xrightarrow {\sim } (U \otimes _{k} V) \otimes _{k} W \quad \quad u \otimes (v \otimes w) \mapsto (u \otimes v) \otimes w.$