TITLE: Proof: $C(X×Y)=C(X)⊗C(Y)$
QUESTION [5 upvotes]: Where I can find the proof of the following theorem:
Let $X$ and $Y$ be compact Hausdorff spaces, $C(X)$ and $C(Y)$ the space of continuous functions on $X$ and $Y$ respectively, then we have $C(X×Y)=C(X)⊗C(Y)$.

REPLY [2 votes]: Martin Brandenburg's answers gives a categorical proof using the full strength of Gelfand--Naimark.
One could also try to prove this more directly (using some of the ingredients that come into the proof of Gelfand--Naimark).
There is an obvious morphism $C(X)\otimes C(Y) \to C(X\times Y).$  (Here I just mean the usual tensor product.)  Now the Stone--Weierstrass approximation theorem
will show that this has dense image, and hence that the map from the completed tensor product $C(X) \widehat{\otimes} C(Y) \to C(X\times Y)$ is surjective.
[Added: actually, even this surjectivity isn't clear, as Martin Brandenburg
points out in a comment.]
To show that you get an isomorphism will take a little more work.

Another approach would be to show that the natural map $C(X) \widehat{\otimes} V \to C(X,V)$ (continuous $V$-valued functions) is an isomorphism for some class
of Banach spaces $V$, and to then apply this with $V = C(Y)$, and then use
that $C(X,C(Y)) = C(X\times Y).$  Off the top of my head, though, I'm not
sure how generally this formula holds.  (I don't see why it wouldn't hold for
arbitrary $V$, but I've not thought much about it, and my experience with this
kind of question comes from the $p$-adic context, where things are normally easier.)