I am trying to prove the following.
Let $V_1, \ldots, V_k$ be finite dimensional vector spaces over a field $F$. There is a natural isomorphism between $V_1^*\otimes\cdots\otimes V_k^*$ and $\mathcal L^k(V_1, \ldots, V_k;\ F)$.
Define a map $A:V_1^*\times\cdots\times V_k^*\to \mathcal L^k(V_1, \ldots, V_k;\ F)$ as \begin{equation*} A(\omega_1, \ldots, \omega_k)(v_1, \ldots, v_k)=\omega(v_1)\cdots\omega_k(v_k) \end{equation*} for all $(\omega_1, \ldots, \omega_k)\in V_1^*\times\cdots\times V_k^*$. It can be seen that $A$ is a multilinear map. By the universal property of tensor product, there exists a unique linear map $\tilde A: V_1^*\otimes\cdots\otimes V_k^*\to \mathcal L^{k}(V_1, \ldots, V_k; \ F)$ such that $\tilde A\circ \pi=A$.
We also know that \begin{equation*} \dim V_1^*\otimes\cdots\otimes V_k^*=\dim \mathcal L^k(V_1, \ldots, V_k; \ F) \end{equation*} Thus we just need to show that $\ker \tilde A=0$ to show that $V_1^*\otimes\cdots\otimes V_k^*$ and $\mathcal L^{k}(V_1, \ldots, V_k; \ F)$ are isomorphic.
My Problem: I want to show the triviality of the kernel in a basis free manner. But here I am stuck.
Can anybody help?
