TITLE: Why is the tensor product constructed in this way?
QUESTION [19 upvotes]: I've already asked about the definition of tensor product here and now I understand the steps of the construction. I'm just in doubt about the motivation to construct it in that way. Well, if all that we want is to have tuples of vectors that behave linearly on addition and multiplication by scalar, couldn't we just take all vector spaces $L_1, L_2,\dots,L_p$, form their cartesian product $L_1\times L_2\times \cdots \times L_p$ and simply introduce operations analogous to that of $\mathbb{R}^n$ ? 
We would get a space of tuples of vectors on wich all those linear properties are obeyed. What's the reason/motivation to define the tensor product using the free vector space and that quotient to impose linearity ? Can someone point me out the motivation for that definition ?
Thanks very much in advance.

REPLY [5 votes]: The tensor product IS an analogue of the Cartesian product, but it is AN analogue of the Cartesian product. What I mean by this is the following.
When working with sets, the Cartesian product happens to satisfy TWO properties. 

The first property is that if you have two functions $f: A\to X$ and $g\colon A\to Y$, you can "pair them up" to get a unique function $(f,g)=h\colon A\to X\times Y$, and conversely any function $h\colon A\to X\times Y$ there are unique functions $f\colon A\to X$ and $g\colon A\to Y$ so that $h=(f,g)$.
The second property is that any $X$-parametrized family of functions $\{f_x\colon Y\to B\}_{x\in X}$ corresponds to a function $f\colon X\times Y\to B$ under the identification $f(x,y)=f_x(y)$.

When working with vector spaces (or $R$-modules), the direct product satisfies the first property, while the tensor product satisfies the second property. In fact, the first property is the definition of product, the second property is the definition of tensor product, and is a theorem that these definitions can be satisfied by sets and by vector spaces, and in fact by the same sets, but by different vector spaces.
The proof of the theorem for sets is just checking that the Cartesian product (which we have constructed) satisfies the properties. To prove the theorem for vector spaces, we need to construct both the direct product and the tensor product. (Of course, the two properties for vector spaces will talk about linear functions and linear parametrizations rather than just functions and parametrizations.)
In order to perform these constructions, we have to take note of certain (easy to check, but conceptually new) relationships between vector spaces and sets. The keyword here is adjoint functors, though I will not use it, merely illustrate it.
Properties of free vector spaces
First, for any vector space $V$ we have the underlying set $\mathcal SV$ consisting of the vectors in $V$. Additionally, for any linear map $\phi\colon V\to W$ we have the underlying set-map $\mathcal S\phi\colon\mathcal SV\to\mathcal SW$ given by $(\mathcal S\phi)(v)=w$.
Second, for any set $S$ we have the free vector space $\mathcal FV$ consisting of finite linear combinations of elements of $S$ (this is the vector space with basis $S$). Free vector spaces satisfy the following property:

Any function $f\colon S\to \mathcal SV$ corresponds to a unique linear function $\phi\colon\mathcal FS\to V$, and conversely. In other words, linear functions are uniquely determined by their values on bases.

In particular, note that the identity function $\DeclareMathOperator{\id}{id}\id\colon\mathcal SV\to\mathcal SV$ has to correspond to a unique linear map $\xi\colon \mathcal F\mathcal SV\to V$. This map is significant because its kernel consists precisely of the linearity relations of vectors on $V$. It follows that $f\colon\mathcal SV\to\mathcal SW$ is the underlying set-map of a linear map $\phi\colon V\to W$ if and only if the induced map $\psi\colon \mathcal F\mathcal S V\to V$ vanishes on $\ker\xi$. Furthermore, one can show that when that is the case, $\psi$ factors though $\xi$ as $\psi=\phi\circ\xi$ where $f=\mathcal S\phi$.
Even better, if we have a map $f\colon\mathcal SV\to X$, this gives an induced map
$\psi\colon\mathcal F\mathcal S V\to\mathcal FX$, If $\psi$ vanishes on $\ker\xi$, then it factors as $\psi=\phi\circ\xi$ with $\phi\colon V\to FX$, and the image has to have underlying set $X$ in the sense that $\mathcal S(\operatorname{im}(\phi))=X$. Now, the condition that $\psi$ vanishes on $\ker\xi$ depends only on $f$, and corresponds exactly to the case when $f\colon SV\to X$ induces on $X$ the structure of a vector space $X_f$ so that $f=\mathcal S\phi$ for $\phi\colon V\to X_f$.
Direct product
Consider two linear functions $\phi\colon U\to V_1$ and $\psi\colon U\to V_2$, and let $f\colon \mathcal SU\to\mathcal SV_1$ and $g\colon\mathcal SU\to\mathcal SV_2$ be the underlying set-maps $f=\mathcal S\phi$, $g=\mathcal S\psi$. Then we definitely have a set-map $(f,g)\colon\mathcal SU\to\mathcal SV_1\times\mathcal SV_2$. It is not difficult to check that $(f,g)$ induces a vector-space structure on $\mathcal SV_1\times\mathcal SV_2$, though it is slightly more work to show that this is the same vector space structure for any pair of maps $f$ $g$. Nevertheless, this is the gives us the direct product for vector spaces is $V_1\oplus V_2$. 
Tensor product
Consider a $V_1$-parametrized family of (linear) functions $\{\phi_v\colon V_2\to W\}_{v\in V_1}$. Applying the forgetful functor, we get an $\mathcal SV_1$-parametrized family of maps $\{f_{v_1}\colon\mathcal SV_2\to\mathcal SW\}_{v\in \mathcal SV_1}$. By the second property of the Cartesian product, this gives a map $f\colon \mathcal SV_1\times\mathcal S V_2\to\mathcal SW$. Applying the free functor gives us a function $\phi\colon\mathcal F(\mathcal SV_1\times\mathcal SV_2)\to W$ which is linear. Thus, $V_1$-parametrized families $\{\phi_v\colon V_2\to W\}_{v\in V_1}$ of linear maps embed as linear maps $\phi\colon\mathcal F(\mathcal SV_1\times\mathcal SV_2)\to W$. Taking some care, we can show that the necessary and sufficient condition for $f\colon\mathcal SV_1\times\mathcal SV_2\to\mathcal SW$ to come from a linear parametrization is that the induced $\phi$ vanishes on $\ker\xi_1\oplus\mathcal\ker\xi_2$ where $\xi_i\colon\mathcal F\mathcal S V_i\to V_i$ are the linear maps induced by $\id\colon\mathcal SV_i\to\mathcal SV_i$. Hence taking the quotient with respect to that subspace (which is just the one spanned by the multilinearity relations) gives a new vector space $V_1\otimes V_2$ with the property that $V_1$-linearly parametrized families of maps $\{\phi_v\colon V_2\to W\}_{v\in V_1}$ naturally correspond to linear maps $\bar\phi\colon V_1\otimes V_2\to W$.