every bilinear map is a tensor product if the dimensions matche

Question

In the following every vector spaces are real and finite dimensional.

The following statement seems too good to be true:

Let $V,W$ be vector spaces. If $Z$ is a vector space of dimension $dim(V)dim(W)$ and $f:V\times W\to Z$ is bilinear, then $f$ is a tensor product.

My proof. (heavily adapted from the last proof here)

Let $h:V\times W\to T$ be some tensor product (universal property). There exists a linear map $\tilde f:T\to Z$ such that $f=\tilde f\circ h$.

Since the dimensions of $T$ and $Z$ match, the map $\tilde f$ is a vector space isomorphism.

It is now easy to see that $f$ satisfy the universal property. If $b:V\times W\to Y$ is bilinear, by universal property of $T$, there exists $\tilde b:T\to Y$ such that $b=\tilde b\circ h$.

Now the map $b'=\tilde b\circ \tilde f^{-1}$ is a lift of $b$, so that the map $f$ satisfy the universal property.

QUESTIONS :

• Is the statement true ?
• Is my proof correct ?

Answer 1

As shown by the other answers, it is false. But at least the following is true:

If $f : V \times W \to Z$ is a bilinear map, $\dim(Z) = \dim(V) \dim(W) < \infty$, and $Z$ is generated by the image of $f$, then $f$ is a tensor product of $V$ and $W$.

Namely, the induced linear map $\overline{f} : V \otimes W \to Z$ is surjective by assumption, but also $\dim(V \otimes W) = \dim(Z) < \infty$, so it must be an isomorphism.

Answer 2

No, this is quite false, e.g. we could have $f = 0$. This line of your proof:

Since the dimensions of $T$ and $Z$ match, the map $\tilde f$ is a vector space isomorphism.

is nonsense. You need to know that $\tilde{f}$ is either injective or surjective, and neither of those necessarily holds here since if $f = 0$ then $\tilde{f} = 0$.