Basic Question Regarding the Universal Property of the Tensor Product.

Question

(All vector spaces are over a fixed field $F$).

Universal Property of Tensor Product. Given two finite dimensional vector spaces $V$ and $W$, the tensor product of $V$ and $W$ is a vector space $V\otimes W$, along with a multilinear map $\pi:V\times W\to V\otimes W$ such that whenever there is multilinear map $A:V\times W\to X$ to any vector space $X$, there exists a unique linear map $\bar A:V\otimes W\to X$ such that $\bar A\circ \pi = A$.

Notation. We write $\pi(v,w)$ as $v\otimes w$.

First Question: Is the map $\pi :V\times W\to V\otimes W$ sujective?

I think it should be surjective since $\pi(V\times W)$ also satisfies the universal property of tensor product of $V$ and $W$. Since $V\otimes W$ is given by a universal property, it is unique upto a unique isomorphism and thus $\pi$ should be surjective.

Please correct me if I am wrong anywhere in the the above said things.

Main Question: Assuming all is well so far, I am confused in the following:

Let $(e_1,\ldots, e_m)$ and $(f_1,\ldots, f_n)$ be bases for $V$ and $W$ repectively.

I was wondering what member of $V\times W$ maps to $e_1\otimes f_1 + e_2\otimes f_2$.

Since $\pi$ is surjective, some member should map to it.

But I am lost as to how to find it.

Can somebody help?

Answer 1

The map $\pi$ is not linear, so the image of $\pi$ is not a vector space and can not satisfy the universal property.

We have $\pi((e_1,f_1)+(e_2,f_2)) \ne \pi((e_1,f_1))+\pi((e_2,f_2))$.

A multilinear map is just linear in a single component if we fix all other components.

Answer 2

I am new to this stuff, so take my answer with a grain of salt. Your map $\pi$ is usually thought of as the inclusion map, is my understanding. So it need not be surjective. But $\pi$ maps the image of $V\times W$ into $V\otimes W$ injectively.

Answer 3

In general it is not possible for $\pi$ to be surjective because the domain has a smaller dimension than the range provided then dimension of each factor is at least $2$. If $V$ has dimension $m$ and $W$ has dimension $n$, then $V\times W$ has dimension $m+n$ and $V\otimes W$ has dimension $mn$.

$\pi(V\times W)$ is not a subspace of $V\otimes W$, but it contains a basis, hence its span is the whole space.