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Let $X,Y$ be finite sets, $Fun(X),Fun(X),Fun(X\oplus Y)$ be $\Bbb{C}$-modules of functions mapping the corresponding sets to $\Bbb{C}$.

It's obvious that $dim(Fun(X))=|X|$ and ${\{f_{x_0}:X\rightarrow\Bbb{C}, x_0\in\Bbb{C}|f(x_0)=1,f(x)=0, x\neq x_0\}}$ is a basis.

Let consider a homomorphism $\phi:Fun(X)\otimes Fun(Y)\rightarrow Fun(X\oplus Y), f(x)\otimes g(y)\mapsto f(x)\cdot g(y)$. $\phi$ is correctly defined.

To show that $\phi$ is a isomorphism, consider $\phi^{-1}:f(x,y)\mapsto f(x,0)\otimes f(0,y)$.

Is it enough to show that $\phi$ is an isomorphism?

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No you will need to check that $\phi^{-1}$ is well defined and that $\phi$ and $\phi^{-1}$ are mutual inverses. –  user38268 Apr 22 '12 at 14:37
    
Alternatively, use the fact from one of your earlier questions. –  Zhen Lin Apr 22 '12 at 15:06
    
@BenjaminLim If I try to prove $\phi^{-1}\phi=id$, I get $\phi^{-1}\phi=(f(x)\otimes g(y) \mapsto f(x)\cdot g(0)\otimes f(0)\cdot g(y))$. It means that $\phi^{-1}\phi=id$ only if $f(0)=1$ and $g(0)=1$. Does It mean that $\phi$ is not an isomorphism? –  Sergey Filkin Apr 22 '12 at 16:59
    
@ZhenLin do you mean I can use the universal property of tensor product of modules and first construct a homomorphism $Fun(X)\oplus Fun(Y) \rightarrow Fun(X)\otimes Fun(Y)$, then construct a homomorphism $Fun(X)\oplus Fun(Y) \rightarrow Fun(X,Y)$. From the universal property it follows that there exists a homorphism $Fun(X)\otimes Fun(Y) \rightarrow Fun(X,Y)$. And it remains to prove that it's a isomorphism. Am I right here? –  Sergey Filkin Apr 22 '12 at 17:40
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You seem to be confused. If $X$ is a finite set of $n$ elements, then $\mathbb{C}^X$ is an $n$-dimensional complex vector space. So if $X$ and $Y$ are two finite sets, of size $n$ and $m$ respectively, the product $X \times Y$ (don't write $X \oplus Y$!) is of size $n m$, and all you have to prove is that $\mathbb{C}^n \otimes_\mathbb{C} \mathbb{C}^m \cong \mathbb{C}^{n m}$. This is easy by dimension counting, but it's also a special case of your earlier question about $A^{\oplus n} \otimes_A N$. –  Zhen Lin Apr 22 '12 at 18:01
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