# Why is Hom(V,W)\text{Hom}(V,W) the same thing as V∗⊗WV^* \otimes W?

I have a couple of questions about tensor products:

Why is $\text{Hom}(V,W)$ the same thing as $V^* \otimes W$?

Why is an element of $V^{*\otimes m}\otimes V^{\otimes n}$ the same thing as a multilinear map $V^m \to V^{\otimes n}$?

What is the general formulation of this principle?

The result is generally wrong for infinite-dimensional spaces: see this question.

For finite dimensional space $V$, let’s build an isomorphism $f : V^* \otimes W \to \hom(V,W)$ by defining

This clearly defines a linear map $V^* \otimes W \to \hom(V,W)$ (it’s bilinear in $V^* \times W$). Reciprocally, take a basis $(e_i)$ of $V$, then define $g : \hom(V,W) \to V^* \otimes W$ by:

Where $(e_i^*)$ is the dual basis to $(e_i)$ (I will use a few of its properties in $\color{red}{red}$ below). This is well-defined because $V$ is finite-dimensional (the sum is finite). Let’s check that $f$ and $g$ are inverse to each other:

• For $u : V \to W$, and so $f(g(u)) = u$.

• For $\phi \otimes w \in V^* \otimes W$,

And so $f$ and $g$ are isomorphisms, inverse to each other.

It is known that for finite dimensional $V$, then $(V^*)^{\otimes m} = (V^{\otimes m})^*$. Then an element of $V^{* \otimes m} \otimes V^{\otimes n}$ is an element of $(V^{\otimes m})^* \otimes V^{\otimes n} = \hom(V^{\otimes m}, V^{\otimes n})$. So by definition / universal property of the tensor product, it’s a multilinear map $V^m \to V^{\otimes n}$.