(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?


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.

  • $\begingroup$ This clears a lot of things. Thanks. $\endgroup$ Feb 25 '15 at 15:01
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    $\begingroup$ Maybe I should add that we know $\pi( V \times W) $ spans $ V \otimes W $ since it contains a basis namely $\{ e_i \otimes f_j : 1 \le i \le m, 1 \le j \le n \}$. Using this we know that $V \otimes W$ has dimension $mn$, so we know that $\pi$ can not be surjective because $V \otimes W$ has dimension $m+n$. $\endgroup$
    – user60589
    Feb 25 '15 at 15:10
  • $\begingroup$ @user60589 You wrote $\otimes$ instead of $\times$. $\endgroup$ Feb 26 '15 at 8:21

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.

  • $\begingroup$ @caffeinemachine You're welcome. $\endgroup$ Feb 25 '15 at 15:04

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.

  • $\begingroup$ Am I wrong in thinking that $\pi(V\times W)$ satisfies the universal property? $\endgroup$ Feb 25 '15 at 14:26
  • $\begingroup$ I think if $\pi$ is surjective then it does. But we don't know necessarily know that $\pi$ is surjective, right? $\endgroup$ Feb 25 '15 at 14:29
  • $\begingroup$ $\pi$ is not surjective (usually) indeed (nor is it injective). But what do you mean by $\pi$ mapping the image of $V \times W $ into $V \otimes W$ injectively? The image of $V \times W$ is already a subset of $V \otimes W$. $\endgroup$ Feb 25 '15 at 15:02
  • $\begingroup$ I meant that every element in the image of $\pi$ has a unique preimage. That is $\forall\,y\in\pi(V\times W)\,\exists!\,x\inV\times X$ such that $\pi(x)=y$. $\endgroup$ Feb 25 '15 at 15:09

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