Vector space/Tensor product/Dual space/Relation/Fact/Proof

Proof

For fixed linear forms $f_1\in {V_1}^{\ast },\dots ,f_n\in {V_n}^{\ast }$, the mapping

$V_1\times \cdots \times V_n⟶K,(v_1,\dots ,v_n)⟼f_1(v_1)\cdots f_n(v_n),$

is multilinear due to exercise; therefore, it defines a linear form on $V_1{\otimes }_{}\cdots {\otimes }_{}{V}_n$. This yields the mapping

$\Psi :{V_1}^{\ast }\times \cdots \times {V_n}^{\ast }⟶{(V_1{\otimes }_{}\cdots {\otimes }_{}{V}_n)}^{\ast },(f_1,\dots ,f_n)⟼((v_1{\otimes }_{}\cdots {\otimes }_{}{v}_n)\mapsto f_1(v_1)\cdots f_n(v_n)).$

This assignment $\Psi$ is also multilinear, and gives a linear mapping

${V_1}^{\ast }{\otimes }_{}\cdots {\otimes }_{}{V_n}^{\ast }⟶{(V_1{\otimes }_{}\cdots {\otimes }_{}{V}_n)}^{\ast }.$

Due to fact and fact, both spaces have the same dimension. Let $v_{ij}$, $1\le j\le {\dim }_K\left(V_i\right)$, be bases of $V_i$. Then the $v_{1j_1}{\otimes }_{}\cdots {\otimes }_{}{v}_{n{j}_n}$ form, according to fact  (3), a basis of $V_1{\otimes }_{}\cdots {\otimes }_{}{V}_n$, and the dual basis is a basis of the dual space. We claim the equality of the linear mappings

$\Psi (v_{1j_1}^{\ast }{\otimes }_{}\cdots {\otimes }_{}{v}_{n{j}_n}^{\ast })={(v_{1j_1}{\otimes }_{}\cdots {\otimes }_{}{v}_{n{j}_n})}^{\ast }.$

This equality follows from the fact that both mappings give, when applied to the basis elements $v_{1k_1}{\otimes }_{}\cdots {\otimes }_{}{v}_{n{k}_n}$, in case $(k_1,\dots ,k_n)=(j_1,\dots ,j_n)$ the value $1$, and else the value $0$. Therefore, $\Psi$ is surjective, and then also injective.

To fact