$X_i$ be finite set and $F(X_i):=\{f:X_i\to\Bbb{C}\text{ functions}\}$. Then show that $F(X_1\times X_2)=\{(a,b)\mapsto f_1(a)f_2(b)|f_i\in F(X_i)\}$ Let $X_1=\{a_1,\ldots,a_r\}$ and $X_2=\{b_1,\ldots,b_s\}$.
Let $g\in F(X_1\times X_2)$. I need to find $f_i\in F(X_i)$ such that $g(a,b)=f_1(a)f_2(b)$.
I first define $f_2(b_1):=1$. Then I define $f_1(a)=F(a,b_1)$. Then the natural definition of $f_2$
should be $\displaystyle{f_2(b)=\frac{g(a,b)}{f_1(a)}=\frac{g(a,b)}{g(a,b_1)}}$.
But the problem here is that the definition of $f_2(b)$ depends upon $a\in X_1$. So the definition is not defined.
Now, I doubt is the statement actually true? But I think somehow finiteness of $X_1, X_2$ are playing an important role here.
Can anyone help me to figure that out? Thanks for help in advance.
 A: Heuristically, there are good reasons why this has to be false.
To define a function on $X_i$, we need to specify $|X_i|$ values, so the "degree of freedom" is $|X_i|$.
If what you want is true, then to define a function on $X_1\times X_2$, we may only specify values on $X_i$ to get $f_i$, so the degree of freedom is $|X_1|+|X_2|$. But the degree of freedom for $X_1\times X_2$ is $|X_1||X_2|$.
This is related to the fact that if we have two vector spaces $V_1, V_2$, then the tensor product $\dim (V_1\otimes V_2)=\dim(V_1)\dim(V_2)$, and there almost always exist non-deomposable tensors, which has deep consequences in computational complexity theory and quantum mechnics.
For example let $X_1=X_2=\{0,1\}$, and $f(x,y) = 1$ if $x=y$ else $0$. You can show there are no $f_1, f_2$ such that $f(x,y) = f_1(x)f_2(y)$. The idea is clear: $f_1(x)$ doesn't know if it should be $0$ because it cannot communicate with $f_2(y)$ to know whether $x=y$. Congratulations, you just showed that the Bell's state in quantum mechanics cannot be described by the individual systems.
A: The statement is false. Consider $X_1=\{0,1\} =X_2$ and $f(a,b)=\max\{a,b\}$. We have $f(0,0)=0$, thus if we had $f(a,b)=f_1(a) f_1(b)$, then either $f_1(0)=0$ or $f_2(0)=0$. However, then in the first case $1=f(0,1)=f_1(0) f_1(1)=0$ and in the second case $1=f_1(1) f_2(0)=0$, both leading to a contradiction.
