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.

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About mlbaker

just another guy trying to make the diagrams commute.
This entry was posted in abstract algebra, linear algebra and tagged , , , . Bookmark the permalink.

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