Isomorphism between $\mathbb{C} X \otimes \mathbb{C} X$ and $\mathbb{C} (X \times X)$ Let $G$ be a finite group and $X$ a finite set. Denote by $\mathbb{C}X$ the vector space of functions from $X$ to $\mathbb{C}$. In a book I found the following statement:
$\varphi : \mathbb{C} X \otimes \mathbb{C} X \rightarrow \mathbb{C} (X \times X)$ defined by $\varphi(f_1 \otimes f_2) = h_{f_1, f_2}$ there $h : X \times X \rightarrow \mathbb{C}$ is defined by $h_{f_1, f_2}(x, y) = f_1(x) f_2(y)$ is an isomorphism. How do I show that it is linear and injective (surjectivity is trivial)? 
If we take $f_1 \otimes f_2, f_3 \otimes f_4 \in \mathbb{C} X \otimes \mathbb{C} X$ we need to see that $$\varphi(f_1 \otimes f_2 + f_3 \otimes f_4) = \varphi(f_1 \otimes f_2)  + \varphi(f_3 \otimes f_4)$$ but how can we saw something about $f_1 \otimes f_2 + f_3 \otimes f_4$?
For scalar multiplication it is simple since we know $(\lambda f_1) \otimes f_2 = \lambda (f_1 \otimes f_2)$.
For injectivity, for $\varphi(f_1 \otimes f_2) = \varphi(f_3 \otimes f_4)$ we want to show that $f_1 \otimes f_2 = f_3 \otimes f_4$. That happens if and only if $f_1 = f_3$ and $f_2 = f_4$, right? 
My problem is that we can have $f_1(x) f_2(y) = f_3(x) f_4(y)$ for all $x, y \in X$ and not have $f_1 = f_3$ and $f_2 = f_4$.
 A: As you are learning, proving anything about tensor products can be difficult.  One eventually picks up on the right way of doing these things, and this is a good example to work out in full.  Here are the steps:


*

*An element of $\def\C{\mathbb{C}}\C X \otimes \C X$ can be made into a function on $X \times X$ by writing $(f_1 \otimes f_2)(x,y) = f_1(x) f_2(y)$ for simple tensors and defining the complete map by imposing linearity; i.e.
$$\left(\sum_{i = 1}^n f_{1,i} \otimes f_{2,i}\right)(x,y) = \sum_{i = 1}^n f_{1,i}(x) f_{2,i}(y).$$
The more abstract way of saying this is that in order to define a linear map out of a tensor product such as $\C X \otimes \C X$, it is equivalent to define a bilinear map out of the regular product $\C X \times \C X$.  This bilinear map is given by the formula for simple tensors, just replacing $f_1 \otimes f_2$ by the pair $(f_1, f_2)$.  It automatically extends to a linear map on $\C X \otimes \C X$.  Let's call this map $\phi$, for future reference.

*To check that $\phi$ is an isomorphism we need to use the particulars of the problem.  It's usually difficult to show that maps from a tensor product are injective, so we start with surjectivity.  Suppose we have a function $g(x,y)$ on $X \times X$ (thus, $g \in \C(X \times X)$); we wish to write it as a sum of functions of the form $f_1(x) f_2(y)$.  We do this in the stupidest way imaginable: for each pair $(a,b)$, let
$$g_{a,b}(x,y) = \begin{cases} g(a,b) & (x,y) = (a,b) \\ 0 & \text{otherwise} \end{cases}$$
be the function whose only value is the one at $(a,b)$, so we have
$$g(x,y) = \sum_{(a,b) \in X \times X} g_{a,b}(x,y),$$
which is a finite sum since $X$ is finite.  Now I claim that each $g_{a,b}(x,y)$ is $\phi$ of some simple tensor; i.e. of the form $f_1(x) f_2(y)$.  These functions are basically the same and defined like $g_{a,b}:$
$$f_c(x) = \begin{cases} 1 & x = c \\ 0 & \text{otherwise} \end{cases}$$
so that $g(a,b)f_a(x) f_b(y) = g_{a,b}(x,y)$, as you can check.  Putting it all together, we have
$$g(x,y) = \sum_{(a,b) \in X \times X} g(a,b)f_a(x) f_b(y) = \phi\biggl(\sum_{(a,b) \in X \times X} g(a,b)(f_a \otimes f_b)\biggr)(x,y).$$
This shows that $g = \phi(\sum_{a,b} g(a,b)(f_a \otimes f_b))$, so $\phi$ is surjective.

*Showing that $\phi$ is injective is, as I said, tricky.  In this case, the trick is to use a dimension argument: namely, both $\C X \times \C X$ and $\C (X \times X)$ have the same finite dimension $\def\card#1{\lvert #1 \rvert}\card{X}^2$.  The proof of this is that for any finite set $Y$, the vector space $\C Y$ has as a basis exactly the functions I denoted by $f_c$ above, taken over all $c \in Y$.  Since the tensor product multiplies dimensions, we have
$$\dim{\C X \otimes \C X} = \dim(\C X) \dim(\C X) = \card{X}^2 = \card{X \times X} = \dim \C(X \times X).$$
So we have a surjective linear map $\phi$ between two vector spaces of the same dimension.  It follows that the kernel must have dimension zero, i.e. is the zero vector space, so $\phi$ is injective as well.
This is the only proof I know if this statement; when $X$ is infinite, $\phi$ is not even surjective, because you can't take an infinite sum over the points of $X \times X$.  (However, there are variants in which we impose some kind of topology on the vector spaces that would allow such sums, and this restores the isomorphism when done right.)
