The whole (intuitive) idea of the necessity of tensor product vector space is almost understood, I mean: given our prior experience with the concept of multiplication (basic elementary one like in fields, scalar multiplication and so on) and various products on vector calculus (like scalar product, vector product, kronecker product, etc…) we would like to give sense and meaning to a element like:
$$ t = v\otimes w $$
Which is a candidate of a heuristic concept of a what means a product of vectors. Also, based on our knowledge of the concept of "product", it seems that the whole construct is based on what a product "should look like", i.e., this product (or this combination of vectors/this well defined one element) must satisfy some basic rules of bilinearity (a.k.a. distributive laws):
$$ (v+u)\otimes w = v \otimes w +u \otimes w $$
$$ v\otimes (w+u) = v \otimes w +v \otimes u $$
$$ (\lambda v) \otimes w = \lambda (v \otimes w) \tag{1}$$
$$ v \otimes (\lambda w) = \lambda (v \otimes w) $$
Again, we want all those things. But some fundamental problems arises when we ask for too much:
$1)$ We do not know which set contains the element $v\otimes w$; therefore we must to find (construct) such algebraic structure.
$2)$ We do not know if such a "product" $\otimes$, really exist satisfying bilinearity; therefore we must to find (define) such operation.
The point is that, from linear algebra we do have one space which contains more or less the fundamental structure for all that we required above; this space is called Quotient Vector Space.
Now begins my doubt. So, in order to construct the vector space which contains the symbols $v\otimes w$ (The Tensor Product $\mathfrak{V}\bigotimes \mathfrak{W}$) and also encodes the well property of a product defined via relations $(1)$, we must to consider other prior algebraic structures and objects:
$1)$ Cartesian Products
$2)$ Bilinear Maps
$3)$ Quotient Vector Spaces (and isomorphism theorems and universal property)
$4)$ Free Vector Spaces
The "building blocks" $1)$ $2)$ $4)$ together with the notion of equivalence classes given by $3)$, gives to you the coordinate free construction of $\mathfrak{V}\bigotimes \mathfrak{W}$.
Question:
So, my problem is actually with the preference for a particular vector space called Free Vector Space. First of all, why we need them for tensor product construction? In other words, why this particular vector space and not a cartesian product with vector space structure ? I know that the tensor product is not the free vector space, but rather a quotient vector space which use free vector spaces!
Best Answer
You need something bigger than $V\times W$ because, for instance, we have that $$(u+v,w) = (u,w)+(v,0),$$while the condition $$(u+v)\otimes w=(u\otimes w)+(v\otimes w)$$should correspond (under your idea) to $$(u+v,w)=(u,w)+(v,w),$$which is not true! The underlying issue here is that being bilinear in the variables $u$ and $v$ is not the same as being linear in the pair-variable $(u,v)$, which is also reflected in the fact that the dimension of a cartesian product is the sum of dimensions, while the dimension of a tensor product is the product of the dimensions. If you really want to feel the difference first-hand, exercise: show that if $B:V\times W \to Z$ is simultaneously bilinear and linear, then $B=0$.
And there's more to it. Assume that all spaces have finite dimension. Then $V\otimes W$ and $V\times W$ are isomorphic (perhaps non-naturally) if and only if $\dim V\dim W = \dim V+\dim W$. The solutions are given by $\dim V = \dim W/(\dim W -1)$ for $\dim W \neq 1$, but this has no non-trivial integer solutions other than $\dim V = \dim W =2$. Meaning that $V\otimes W$ and $V\times W$ are never isomorphic unless $V=W=\{0\}$ or both have dimension $2$.
The natural guess for something bigger than $V\times W$ which gives you enough room to impose the distributive property of $\otimes$ (via the quotient) is ${\rm Free}(V\times W)$. Now, the thing is that you don't need to define $V\otimes W = {\rm Free}(V\times W)/_\sim$. By categorical arguments, all the "tensor products" of $V$ and $W$ are isomorphic, and while this settles uniqueness, it does not say anything about existence of $V\otimes W$. The construction with the quotient of the free vector space is one possible construction.
Once you have proven the existence of $V\otimes W$ (which is really all for which this construction is good for), all the algebraic properties can be deduced via universal properties, and you never need to think about this construction again.