Showing $V_1^* \otimes \dots \otimes V_k^* \otimes W \cong \mathrm{L}(V_1,\dots,V_k;W)$

Question

This is exercise 12-4 in the book Introduction to Smooth Manifolds by John M. Lee:

Let $V_1,\dots,V_k$ and $W$ be finite-dimensional real vector spaces. Then $$V_1^* \otimes \dots \otimes V_k^* \otimes W \cong \mathrm{L}(V_1,\dots,V_k;W)$$ canonically, where $\mathrm{L}(V_1,\dots,V_k;W)$ denotes the set of multilinear mappings from $V_1\times \dots \times V_k$ to $W$.

It is clear, that we use the universal property of the tensor product space, so we construct a mapping $$\Phi : \begin{cases} V_1^* \times \dots \times V_k^* \times W \to \mathrm{L}(V_1,\dots,V_k;W)\\ (\varphi_1,\dots,\varphi_k,w) \mapsto \left((v_1,\dots,v_k) \mapsto \varphi_1(v_1) \cdots \varphi_k(v_k)w\right) \end{cases}$$

Now $\Phi$ is multilinear and thus we get a linear mapping $$ \widetilde{\Phi} : V_1^* \otimes \dots \otimes V_k^* \otimes W \to \mathrm{L}(V_1,\dots,V_k;W) $$

such that $\Phi = \widetilde{\Phi} \circ \otimes$. It is easily checked that $\ker\widetilde{\Phi} = \{0\}$ and thus $\widetilde{\Phi}$ is injective. How do we show, that $\widetilde{\Phi}$ is surjective not knowing what the dimension of $\mathrm{L}(V_1,\dots,V_k;W)$ is? I want to use this exercise to determine the dimension of $\mathrm{L}(V_1,\dots,V_k;W)$.

Answer 1

Suppose you have $f\in\operatorname{L}(V_1,\dots,V_k;W)$. By the universal property, you get a linear map $\hat{f}\colon V_1\otimes\dots \otimes V_k\to W$ and conversely, so what you really have to show is

1. $L(V,W)\cong V^*\otimes W$
2. $(V_1\otimes\dots \otimes V_k)^*\cong V_1^*\otimes\dots \otimes V_k^*$

Note that 2 is a special case of the theorem, but with $W=\mathbb{R}$ (in general, the base field), where you just need injectivity and the fact that the dimension of the two spaces is clearly the same.

For 1 it's again proving injectivity and equality of dimensions.