Tensor Product Definition

Question

NOTATION: $S$ denotes the set of all vector spaces over a commutative ring $K$. $ A \times B$ denotes the cartesian product between the two sets $A$ and $B$. $f \circ g$ denotes the composition of the maps $f$ and $g$.

Consider the definition of the tensor product space:

Let $V, W \in S$.

Then $ V \otimes W $ is a new vector space equipped with an additional map $\sigma: V \times W \rightarrow V \otimes W$ such that $\forall \; U \in S$ and homeomorphisms $f : V \times W \rightarrow U $, $ \; \exists !$ $\psi : V \otimes W \rightarrow U$ such that $ f = \psi \circ \sigma$.

Here is a link to the associated commutative diagram, though the symbols I chose are different than those shown in the diagram. https://i.stack.imgur.com/SWNnE.png

My question with this definition has to do with the uniqueness of the vector space homeomorphism $\psi$. What does this stipulation say about $ V \otimes W $? More specifically, what does this requirement say about the relationship between the basis of $ V \otimes W $ and the basis of $ V \times W $?

Also, if my definition of the tensor product space is incorrect or incomplete in some way (which is highly likely!), I would appreciate any suggestions in improving it.

Thanks in advance!

Answer 1

This is a definition of the tensor product with a universal property.

• The existence of $\psi$ for any $f$ grants that $V\otimes W$ has enough power of expression to describe any bilinear map $f : V\times W \mapsto U$, in the sense that $f$ is described by the corresponding $\psi$.
• The uniqueness of $\psi$ grants uniformity of this description process : indeed, we now have a functional $f \mapsto \psi$, which can be seen as a parametrisation of the class of bilinear maps with domain $V\times W$. It is also useful as it makes $V\otimes W$ unique up to isomorphism.