# How should I understand “every bilinear function $f$ on $V\times W$”?

Let $U$ and $V$ be vector spaces, and let $b:U\times V\to X$ be a bilinear map from $U\times V$ to a vector space $X$. Suppose that for every bilinear map $f$ defined on $U\times V$ there is a unique linear map $c$ defined on $X$ such that $f=cb$. Then there is an isomorphism $i:X\to U\otimes V$ such that $u\otimes v=ib(u,v)$ for every $(u,v)$ in $U\otimes V$.

There are several other places in this article where the author uses "every bilinear map $f$ defined on $U\times V$ " without specifying the range of the function.

As I understand, it may mean one of the following items:

1. $\forall f\in{\mathcal L}(V,W;Y)$ for some vector space Y;
2. $\forall f\in \bigcup_{Y\text{is a vector space}}{\mathcal L}(V,W;Y).$

Which one is correct and how should I understand it?

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2. is meant, though you probably need only the cases $Y=U\otimes V$ and $Y=X$ in the proof. –  Stefan Walter Jul 13 '11 at 16:27
Sorry, I don't actually remember $\mathcal{L}$. Is this the set of all linear transformations from $V\times W$ to $Y$? –  mathmath8128 Jul 13 '11 at 17:34
@abraham: just read the question. It follows that ${\mathcal L}(V,W;Y)$ denotes the space of bilinear maps on $U\times V$. –  wildildildlife Jul 13 '11 at 19:23

$\forall Y$ a vector space and $\forall f:U\times V \to Y$ $\exists ! c: X\to Y$ s.t. $f = cb$.
The first statement would usually be expressed by something like "Suppose that there is a vector space $Y$ such that for every bilinear map $f$ from $U\times V$ to $Y$..." -- one wouldn't usually imply such a distinguished object without introducing it. In a sense, the target space $Y$ is a free variable of the statement, and free variables usually imply a universal quantifier applied to them (2), not an existential quantifier (1).