Tensor product of vector spaces: operations on the congruence class After having read the Wiki article on the Tensor Product, I have tried constructing $\mathbb{R}\otimes\mathbb{R}$ (over the field $\mathbb{R}$), to check my understanding.
Step 1: construct $\mathbb{R}\times\mathbb{R} := \{(a,b); a\in \mathbb{R}, b\in \mathbb{R}\}$. So for example, $\{1,2\},\{0,5\}$ and $\{2,7\}$ are all in $\mathbb{R}\times\mathbb{R}$.
Step 2: Construct $F(\mathbb{R}\times\mathbb{R}):= \{g: \mathbb{R}\times\mathbb{R}\to \mathbb{R}; |\mbox{supp}(g)|<\infty\}.$ So for example, the function
$$g(x,y) = \begin{cases} 3 &&\mbox{ if $(x,y)=(1,2)$}\\
6 &&\mbox{ if $(x,y)=(0,5)$}\\
1&&\mbox{ if $(x,y)=(2,7)$}\\
0 &&\mbox{ else }
\end{cases}$$
is in $F(\mathbb{R}\times\mathbb{R})$.
Step 3: Construct congruence classes to ensure bilinearity. So for example, one congruence class could be $\{(1,2),(1,1)+(1,1), 2(0.5,2), 2(1,1),...\}$.
This is where I stop understanding. What do the elements of $F(V\times W)$ (such as the function $g$) have to do with these equivalence classes? How are these relations 'on $F(V\times W)$', given that the elements of $F(V\times W)$ are functions?
 A: One obvious basis for $F(V \times W)$ is $\left\{\delta_{(u,w)}, u \in V, w \in W, \right\}$. In wiki they denote those basis elements by $(u,w)$ and define the quotient relations on the basis, extending them by linearity to arbitrary element of $F(V\times W)$. For instance, a function mapping $(1,2)$ to $1$ is identified with function mapping $(1,1)$ to $2$ (in particular $\delta_{(1,2)}=\delta_{(2,1)}$). That can be confusing a little bit but if you think of those delta's as indicator functions it becomes transparent. I think tensor product construction is covered relatively well in Lang's algebra textbook.
A: While it is of course very useful for proving the existence of tensor products in great generality and independent of, e.g, the choice of a basis, the free vector space approach is, in my opinion, not that useful for understanding what the tensor product of two vector spaces actually is. 
I think it's much easier to (accept the fact that it exists and can be and is a universal object) and to use known properties of the tensor product to figure out what a certain tensor product actually is. 
In your special case ($\mathbb{R}\otimes \mathbb{R}$) one of the most useful facts about tensor spaces is the dimension formula, which tells you that this is a real vector space of dimension $1$, so it's actually $\mathbb{R}$ itself. 
Your mileage may vary, of course.
