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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 $\pi(v,w) (f:V \times W \rightarrow \mathbb{R}) = f(v,w).$

$\mathbf{CORRECTION:}$ $B$ should be the space of bilinear maps $V \times W \to \mathbb{R}$, not $V \times W \to U$ as previously stated.

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.

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Ideally I'd like to show that $\widetilde{\Theta \circ f \circ \pi} = f$ but I can't seem to derive that from the definition. – Paul Slevin Apr 30 '12 at 3:09
It's the map $\pi: V \times W \to B^*$. Given an element $f \in B$, $\pi(v,w)$ sends $f$ to $f(v,w) \in \mathbb{R}$. I defined this map as part of the construction, so I'm not sure it's correct. However it seems like the only reasonable choice. Apologies if this wasn't clear in my original post. – Paul Slevin Apr 30 '12 at 12:12
You're right. I guess there is a fundamental problem there! Perhaps we need to define $$\pi(v,w) (f) = \sum_{i=1}^n u_i^* (f(v,w))?$$ where $\{ u_1, \ldots, u_n\}$ is a basis for $U$? – Paul Slevin Apr 30 '12 at 12:24
But surely it won't be surjective? Because in general the map $\large\otimes: V \times W \to V \otimes W$ is not surjective? I'm not sure, maybe it can be. – Paul Slevin Apr 30 '12 at 12:48
no it can't be surjective! It only maps into elementary tensors, for example there's no element which maps to $m \otimes n + m' \otimes n'$ under the tensor map. I think. – Paul Slevin Apr 30 '12 at 22:45
up vote 2 down vote accepted

After a great deal of soul searching and chats with fellow mathematicians, here is the solution.

Suppose that $V$ has basis $\{e_1, \ldots e_n \}$ and $W$ has basis $\{f_1, \ldots f_m \}$. Then it is quite easy to show that the vector space $B$ has basis $$\{ (e_if_j)^* \mid i=1,\ldots, n \ ; \ j = 1,\ldots m \} $$ where $(e_if_j)^* (e_{k},f_{l}):=0 \iff (k,l) = (i,j)$, and $1$ otherwise. Hence $\dim B = \dim V \dim W$. It follows that the dual space $B^*$ has $\dim B^* = \dim V \dim W$. I claim that $$\{ \pi(e_i,f_j) \mid i = 1, \ldots, n \ ; \ j=1,\ldots, m \}$$ is a basis for $B^*$. This is easy to see since we need only check that this set is linearly independent, which is clearly is. Hence $B^*$ is generated by the image of $\pi$.

Suppose for $g: B^* \to U$, we have that $g\pi = 0$. Suppose that $g \not= 0$. Then there is some basis element $ \pi(e_i, f_j)$ such that $g\pi(e_i,f_j) \not= 0$. This is a contradiction, so $g=0$. This of course implies that the map above is unique, so $B^*$ is the tensor product of $V$ and $W$.

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