I am trying to understand some aspect of Tensor product.
In this note of Tim Gowers, it says
for every other function of the form, $a_1[v_1,w_1]+ a_2[v_2,w_2]+...+ a_n[v_n,w_n]$, we can find some bilinear map $f$ such that $a_1f(v_1,w_1)+ a_2f(v_2,w_2)+...+ a_nf(v_n,w_n)$ does not equal $0$.
Here, every 'other function of the form' means one that isn't the linear combination of the following four linear combinations.
(1) $ [v,w+w']-[v,w]-[v,w']$
(2) $ [v+v',w]-[v,w]-[v',w]$
(3) $[av,w]-a[v,w]$
(4) $[v,aw]-a[v,w]$
To show the existence of a bilinear form $f$, he creates some formal vector space, uses the quotient spaces, and finally he says it is
To complete the proof, it is enough to find a bilinear map $f$ from $V \times W \to Z\setminus E$ such that $a_1f(v_1,w_1)+ a_2f(v_2,w_2)+...+ a_nf(v_n,w_n)=z+E$ and in particular, is non zero.
Here, $Z$ is the vector space spanned by the formal symbol of $[[v,w]]$ and $E$ is the subspace which consists of the linear combination of the following four linear combinations.
(1) $ [[v,w+w']]-[[v,w]]-[[v,w']]$
(2) $ [[v+v',w]]-[[v,w]]-[[v',w]]$
(3) $ [[av,w]]-a[[v,w]]$
(4) $ [[v,aw]]-a[[v,w]]$
However, I don't know why he says "it is enough to find a bilinear map $f$ from $V \times W \to Z\setminus E\,$" instead of a bilinear map from $V \times W \to \mathbb{R}$. Could you help me to understand this? I couldn't bring all the contents because it is quite lengthy.
If $V$ is any vector space and $x\in V\setminus \{0\}$, then there is $f:V\to K$ with $f(x) = 1$ (where $K$ is the base field).
So to get a linear form $f$ with $f(x)\neq 0$ it suffices to get a linear map with $f(x) \neq 0$