Vector space/Tensor product/Universal property/Fact/Proof

Proof

1. follows immediately from the definition of the tensor product.
2. Since the ${}v_{1}\otimes _{}\cdots \otimes _{}v_{n}$ are a ${}K$-generating system of ${}V_{1}\otimes _{K}\cdots \otimes _{K}V_{n}$, and because

   ${}{\bar {\psi }}(v_{1}\otimes _{}\cdots \otimes _{}v_{n})=\psi (v_{1},\ldots ,v_{n})\,$

   must hold, there can exist at most one such a linear mapping. To show existence, we consider the ${}K$-vector space ${}F$ from the construction of the tensor product. The ${}e_{(v_{1},\ldots ,v_{n})}$ form a basis of ${}F$; therefore, the assignment

   ${}{\tilde {\psi }}{\left(e_{(v_{1},\ldots ,v_{n})}\right)}:=\psi (v_{1},\ldots ,v_{n})\,$

   defines a linear mapping

   ${\tilde {\psi }}\colon F\longrightarrow W.$

   Because of the multilinearity of ${}\psi$, the linear subspace ${}U$ is mapped by ${}{\tilde {\psi }}$ to ${}0$. Hence, according to the factorization theorem, this mapping induces a ${}K$-linear mapping

   ${\overline {\psi }}\colon F/U\cong V_{1}\otimes _{K}\cdots \otimes _{K}V_{n}\longrightarrow W.$

To fact