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

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

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

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