= Universal property of a tensor product
{title2=$B=L\circ\tau$}
For a <bilinear map> $B:V\times W\to U$, there is exactly one <linear map> $L:V\otimes_F W\to U$ satisfying $L(v\otimes w)=B(v,w)$. Equivalently, $B=L\circ\tau$, where $\tau(v,w)=v\otimes w$. The <universal property of a tensor product> concerns the <vector space> together with this distinguished <bilinear map>.<Linear Algebra Done Right by Sheldon Axler (4th edition)>[§9D, Theorem 9.79(a), printed p. 375: unique linear factorization]{ref}
Here is a complete calculation over the <real numbers>. Define the <bilinear map>
$$
B(v,w)=\bigl(v_1w_1+2v_1w_2+v_2w_2,\ v_2w_1-v_1w_2\bigr).
$$
Its values on pairs of <basis vectors> determine $L$ on the four <tensor-product basis> elements:
$$
\begin{aligned}
L(e_1\otimes f_1)&=(1,0),&L(e_1\otimes f_2)&=(2,-1),\\
L(e_2\otimes f_1)&=(0,1),&L(e_2\otimes f_2)&=(1,0).
\end{aligned}
$$
Consequently, for $t=\sum_{i,j=1}^2c_{ij}e_i\otimes f_j$,
$$
L(t)=(c_{11}+2c_{12}+c_{22},\ c_{21}-c_{12}).
$$
For $v=(2,1)$ and $w=(-1,3)$, the direct calculation gives $B(v,w)=(-2+12+3,-1-6)=(13,-7)$. Using the expansion of $v\otimes w$ instead gives $L(v\otimes w)=-2(1,0)+6(2,-1)-(0,1)+3(1,0)=(13,-7)$. The <linear map> also handles a nonseparable <tensor> such as $t=e_1\otimes f_1+e_2\otimes f_2$: $L(t)=(2,0)$, even though no single pair has $v\otimes w=t$.
\Image[/tensor-product-bilinear-factorization.png]
{title=A bilinear calculation factors through one linear map on four tensor coefficients}
A short proof uses a <tensor-product basis>: assign $L(e_i\otimes f_j)=B(e_i,f_j)$ and extend by <linearity>. Expanding both inputs verifies $L(v\otimes w)=B(v,w)$ by <bilinearity>; the assigned values on a <basis> force uniqueness. More generally, every <multilinear map> $F:V_1\times\cdots\times V_r\to U$ factors uniquely through a <linear map> from $V_1\otimes\cdots\otimes V_r$. This lets formulas involving several inputs define <linear maps> without choosing <coordinates>; <tensor contraction> uses the evaluation pairing between a <vector> and a <covector>.
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