# bijection between bilinear maps and linear map from tensor product

Let $V$, $W$ and $U$ be finite dimensional vector spaces over a field $R$, and let $b:V \times W \to U$ be a bilinear map, and let $\hat b:V \otimes W \to U$ be the linear map defined by $\hat b(v \otimes w)=b(v,w)$, and extended linearly to all elements of $V\otimes W$. The tensor product $V\otimes W$ is defined as a quotient of the vector space whose basis is $V\times W$ by the subspace $\bar S$ whose basis is the set $S$, where $S$ is all terms of the form (1) $(v_1+v_2,w)-(v_1,w)-(v_2,w)$, (2) $(v,w_1+w_2)-(v,w_1)-(v,w_2)$, (3) $r(v,w)-(rv,w)$ and (4) $r(v,w)-(v,rw)$. I don't know how to prove that $\hat b$ is well defined. Namely, if $(v_1, w_1)-(v_2,w_2)\in \bar S$, how do I prove that the bilinearity of $b$ implies that $b(v_1,w_1)=b(v_2,w_2)$. Can somebody please explain in detail how does it follow?

• Try using the fact that $(v,w) \mapsto v \otimes w$ is bilinear. Jun 29, 2018 at 14:53

$\overline{S}$ is generated by $S$, but $S$ is in general no basis. Let $F(V \times W)$ denote the vector space with basis $V \times W$. Define $b' : F(V \times W) \to U$ as the unique linear map such that $b'(v,w) = b(v,w)$ on the basis $V \times W$. Now verify that $b'(s) = 0$ for all $s \in S$. This implies $\overline{S} \subset ker(b')$ and gives you a unique linear $\hat{b} : V \otimes W = F(V \times W)/\overline{S} \to U$ such that $\hat{b} \circ p = b'$, where $p : F(V \times W) \to V \otimes W$ denotes the quotient map.