## Tensors #3

§4. Uniqueness of the tensor product

Starting with the usual setup, suppose $(\mathsf{T}, \otimes)$ and $(\mathsf{T}', \otimes')$ are two pairs satisfying the universal property. We can give a slick proof of uniqueness as follows: observe that due to the universal property of $(\mathsf{T}', \otimes')$, the bilinear map $\otimes : \mathsf{V} \times \mathsf{W} \to \mathsf{T}$ induces a linear map $f : \mathsf{T}' \to \mathsf{T}$ such that $f(v \otimes' w) = v \otimes w$. Similarly, due to the universal property of $(\mathsf{T}, \otimes)$, the bilinear map $\otimes' : \mathsf{V} \times \mathsf{W} \to \mathsf{T}'$ induces a linear map $g : \mathsf{T} \to \mathsf{T}'$ such that $g(v \otimes w) = v \otimes' w$.

Now $f(g(v \otimes w)) = v \otimes w$, and $g(f(v \otimes' w)) = v \otimes' w$, and by applying (T1) twice, we see that the vectors $v \otimes w$ and $v \otimes' w$ generate $\mathsf{T}$ and $\mathsf{T}'$ respectively, so therefore $f$ and $g$ are inverse linear isomorphisms and we’re done.