Identifying tensors with functions in $C^*$ -algebras We know the result $C(Y,C(X)) \cong C(X)\otimes C(Y)$, I don't able to construct the isomorphism mapping that by starting  an arbitrary function $f$ from $C(X,Y)$, how to get tensor element in $C(X\otimes Y)$, Please help if anybody knows this.
 A: The equality as stated makes no sense: first, because $X\otimes Y$ is not defined for topological spaces (the tensor product is defined on vector spaces); second, because if we replace it with $C(X\times Y)$, it is not true: you would have $C([0,1])=C([0,1],\mathbb C)\simeq C([0,1]\times \mathbb C)$, which is false ($C(X)\simeq C(Z)$ implies $X$ and $Z$ are homeomorphic). 
The equality that holds is $$ C(X\times Y)\simeq C(X)\otimes C(Y).$$ To write the isomorphism, denote by $f\times g$ the function $(x,y)\longmapsto f(x)g(y)$. The set 
$$
E=\operatorname{span}\{f\times g:\ f\in C(X),\ g\in C(Y)\}
$$
is dense in $C(X\times Y)$; this is seen by Stone-Weierstrass, since $E$ is a $*$-algebra, contains the identity, and separates points. So we may define a linear map $\alpha:E\to C(X)\otimes C(Y)$ by $$ \alpha(f\times g)=f\otimes g.$$It is not hard to see that $\alpha$ is a $*$-homomorphism. 
Next, one proves that $\alpha$ is isometric. This requires one to look at the underlying Hilbert space, since it is needed to express the norm in $C(X)\otimes C(Y)$ explicitly. See this answer for details. 
Once you know that $\alpha$ is isometric, it is defined in $C(X\times Y)$, and being an isometry with dense range, it is onto. Thus $\alpha$ is an isomorphism. 
