Definition
The tensor product of and is a kind of multiplicative version of the direct sum. We are now going to construct a vector space with the property that .
Theorem Let and be vector spaces over a field . There exists a vector space over , unique up to isomorphism, and a map: , with the following properties (1) The map is bilinear. (2) If is a basis of and a basis of , then is a basis of . Proof. Let be the space of functions from , considered as a set, to . Let be the delta function supported at , that is
0, & \text{otherwise}. \end{cases} $$ Let $K$ be the $F$-subspace generated by the elements $$ \begin{align*} & \delta_{(v_1+v_2,w)} - \delta_{v_1,w} -\delta_{v_2,w}, v_1,v_2 \in V, w \in W,\\ & \delta_{(v,w_1+w_2)} - \delta_{v,w_1} -\delta_{v,w_2}, v \in V, w_1,w_2 \in W,\\ & \delta_{(av,w)} - \delta_{v,aw}, v \in V, w \in W, a \in F,\\ & \delta_{(v,w)} - \delta_{av,w}, v \in V, w \in W, a \in F. \end{align*} $$ We claim that the quotient $F(V \times W)/K$ together with the map: $$ \begin{align*} \beta: & V \times W \to F(V \times W)/K\\ & (v,w) \to \delta_{v,w} + K \end{align*}$$ Remark: We denote the space $F(V \times W) /K$ by $V \otimes W$ and denote $v \otimes w= \delta_{v, w} +K$. Careful: unlike the direct sum, arbitrary elements of $V \otimes W$ are not only of the form $v \otimes w$, but rather finite sums of them: $\sum_{i}v_i \otimes w_i.$ **Definition (The universal property of tensor product)** Let $V$ and $W$ be vector spaces over a field $F$. Let: $\varphi:V \times W \to V \otimes W$ be the bilinear map$(v,w) \to v \otimes w = \delta_{(v,w)} + K$. Then, for every vector space $U$ and every bilinear map: $\theta : V \times W \to U$ there exists a unique linear map $\beta: V \otimes W \to U$ such that the following diagram commutes.\begin{CD} V \times W @>{\varphi}>> V \otimes W \ @V{\theta}VV @VV{ \exists! , \beta}V \ U @= U \end{CD}
**** **Proposition** $$V^*\otimes W \cong Hom(V^* W)$$ ****