If $W \cong U \otimes V$, then what is $W$'s bilinear map?

Question

Suppose that $U, V, W$ are vector spaces over a field $F$. Let $B : U \times V \to U \otimes V$ be the universal bilinear map (any bilinear map exiting $U \times V$ factors through $U \otimes V$ as $B$ followed by a unique linear map). If there is a bilinear map $B' : U \times V \to W$ that behaves just like $B$, then it's fun to prove that $W \cong U \otimes V$. But what about the converse?

If $W \cong U \otimes V$, then there is a bilinear map $B' : U \times V \to W$ with the universal property.

I believe this is true. The tools we have are $B : U \times V \to U \otimes V$ and its universal property, an isomorphism $\phi : U \otimes V \to W$, and general facts about vector spaces (although I suspect the statement will remain true for modules, so maybe only general facts about modules). Well, the bilinear map $\phi \circ B : U \times V \to W$ looks exactly like the sort of thing that could be $B'$. Indeed, if $Z$ is another $F$-vector space and $\widehat{B} : U \times V \to Z$ is bilinear, then there exists a unique linear map $h : U \otimes V \to Z$ such that $h \circ B = \widehat{B}$. Thus $$h \circ \phi^{-1} \circ B' = h \circ \phi^{-1} \circ \phi \circ B = h \circ \mathrm{Id}_{U \otimes V} \circ B = \widehat{B}$$ (with $B' := \phi \circ B$), so there exists a linear map $h \circ \phi^{-1} : W \to Z$ that does what we want. But how do we prove uniqueness of $h \circ \phi^{-1}$?

Answer 1

The point is that just as you composed with $\phi^{-1}$ to turn a map from $U\otimes V$ into a map from $W$, you can compose with $\phi$ to go the other way. Suppose $f:W\to Z$ is a linear map such that $fB'=\widehat{B}$. Then $f\phi$ satisfies $(f\phi)B=fB'=\widehat{B}$, so by uniqueness of $h$, we must have $f\phi=h$. Thus $f=(f\phi)\phi^{-1}=h\phi^{-1}$.

A bit more abstractly, you want to prove there is a unique $f\in\operatorname{Hom}(W,Z)$ such that $fB'=\widehat{B}$. Since $\phi$ is an isomorphism, composition with $\phi$ gives a bijection $\phi^*:\operatorname{Hom}(W,Z)\to\operatorname{Hom}(U\otimes V,Z)$ (whose inverse is given by composition with $\phi^{-1}$). Moreover, since $fB'=f\phi B=\phi^*(f)B$, we have $fB'=\widehat{B}$ iff $\phi^*(f)B=\widehat{B}$. Since $\phi^*$ is a bijection, this means the existence of a unique solution $h$ to $hB=\widehat{B}$ is equivalent to the existence of a unique solution $f$ to $fB'=\widehat{B}$.

Answer 2

Since $\operatorname{im} B$ spans$\color{blue}{^*}$ $U \otimes V$, it follows that $\operatorname{im} B'$ spans $W$, and then $f \circ B' = g \circ B' \implies f=g$ for any pair of linear maps $f$ and $g$ out of $W$.

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$\color{blue}{^*}$If $T$ is the span of $\operatorname{im} B$, we can decompose $B$ as $i \circ \tilde B$, where $\tilde B : U \times V \to T$ is given by $\tilde B(u,v) = B(u,v)$ and $i$ is the inclusion $T \hookrightarrow U \otimes V$. Since $\tilde B$ is clearly bilinear, the universal property of $B$ says that there exists a linear map $j : U \otimes V \to T$ such that $\tilde B = j \circ B$. Now note that, since $B$ itself is bilinear, $\operatorname{id}_{U \otimes V}$ is the unique linear map such that $B = \operatorname{id}_{U \otimes V} \circ\, B$. Thus, as $i \circ j$ is a linear map with the property that $$B = i \circ \tilde B = (i \circ j) \circ B$$ by uniqueness, $i \circ j = \operatorname{id}_{U \otimes V}$, meaning that $i$ is surjective and $T = U \otimes V$.