A Basic Question Regarding the Universal Property of the Tensor Product.

[caffeinemachine]

(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?

 

[Opalg]

(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 $\pi :V\times W\to V\otimes W$ surjective?

No, that map is not usually surjective. In fact, if $\dim V = m$ and $\dim W = n$, then $\dim V\times W = m+n$, but $\dim V \otimes W = mn$.

One way to see this is that if $\{v_1,\ldots,v_m\}$ is a basis for $V$ and $\{w_1,\ldots,w_n\}$ is a basis for $W$, then $\{v_1,\ldots,v_m,w_1,\ldots,w_n\}$ is a basis for $V\times W$, and $\{v_i\otimes w_j : 1\leqslant i \leqslant m,1\leqslant j\leqslant n\}$ is a basis for $V\otimes W$.

 

[caffeinemachine]

No, that map is not usually surjective. In fact, if $\dim V = m$ and $\dim W = n$, then $\dim V\times W = m+n$, but $\dim V \otimes W = mn$.

One way to see this is that if $\{v_1,\ldots,v_m\}$ is a basis for $V$ and $\{w_1,\ldots,w_n\}$ is a basis for $W$, then $\{v_1,\ldots,v_m,w_1,\ldots,w_n\}$ is a basis for $V\times W$, and $\{v_i\otimes w_j : 1\leqslant i \leqslant m,1\leqslant j\leqslant n\}$ is a basis for $V\otimes W$.

That clears things up. Thanks.