Let $\displaystyle U,V$ and $\displaystyle W$ be finite dimensional vector spaces, and define $\displaystyle B$ to be the vector space of all bilinear maps $\displaystyle V \times W \to \mathbb{R}$. Given a bilinear map $\displaystyle \alpha : V \times W \rightarrow U$, define $\displaystyle \tilde{\alpha}: B^* \rightarrow U^{**}$ by $\displaystyle \alpha(\psi)(\sigma) = \psi (\sigma \circ \alpha)$. Define a map $\displaystyle \pi : V \times W \rightarrow B^*$ by $\displaystyle \pi(v,w) (f:V \times W \rightarrow \mathbb{R}) = f(v,w).$

In order to show that $\displaystyle B^*$ satifies the universal property of the tensor product, I have to show that given a map $\displaystyle \alpha : V \times W \rightarrow U$, then there is a unique $\displaystyle \tilde{\alpha} : B^* \rightarrow U^{**}$ such that $\displaystyle \Theta \circ \tilde{\alpha} \circ \pi = \alpha$, where $\displaystyle \Theta:U^{**} \to U$ is the canonical isomorphism.


It is quite clear that $\displaystyle \tilde{\alpha}$ defined above satisfies this property, but I am having trouble proving uniqueness. I would like to show that given $\displaystyle f:B^* \to U^{**}$ such that $\displaystyle \Theta\circ f \circ \pi = \alpha$, then $\displaystyle f= \tilde{\alpha}$, however I am getting nowhere. Any help would be appreciated, thank you. Ideally I'd like to show that $\displaystyle \widetilde{\Theta \circ f \circ \pi} = f$.

Note that I don't want to invoke the existence of a map $\displaystyle \operatorname{Hom} ( V \otimes W, U ) \cong \operatorname{Bilinear}(V \times W, U)$, I want to verify directly that $\displaystyle B^*$ is in fact the tensor product.