On tensors

Let $V_1, V_2, W_1, W_2, U_1, U_2 \in \; \mathbb{K}$ -Vect, $V_1 \xrightarrow{\;\; \alpha_1 \;\; }W_1 \xrightarrow{\;\; \beta_1 \;\; }U_1$ , $V_2 \xrightarrow{\alpha_2}W_2 \xrightarrow{\beta_2}U_2 \;\; \mathbb{K}$ -linear. Prove that

$(\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2)=(\beta_1 \alpha_1)\otimes(\beta_2 \alpha_2)$

Solution

Recall the general definition of the tensor product of linear maps, we have successively:

$\begin{align*} ((\beta_1 \alpha_1) \otimes (\beta_2 \alpha_2))(v_1 \otimes v_2) &= (\beta_1 \alpha_1)(v_1) \otimes (\beta_2 \alpha_2)(v_2) \\ &=\beta_1(\alpha_1(v_1)) \otimes \beta_2(\alpha_2(v_2)) \\ &= (\beta_1 \otimes \beta_2)(\alpha_1(v_1) \otimes \alpha_2(v_2)) \\ &=((\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2))(v_1 \otimes v_2) \end{align*}$

Thus, the two linear maps $V_1 \otimes V_2 \rightarrow U_1 \otimes U_2$ are equal when composed with the canonical bilinear map $V_1 \times V_2 \to V_1 \otimes V_2$ , hence equal (by the universal property).