# Ring of continuous functions, reference request.

I am looking for a reference for the following facts in functional analysis and topology. (If these "facts" are not true, I suppose I'm looking for the closest approximation which is true.)

Let $X$ be a locally compact Hausdorff topological space. Let $C(X)$ denote the ring of continuous complex-valued functions on $X$, endowed with the compact-open topology. Then $C(X)$ is a complete locally convex topological (complex) vector space (this can be found in Kothe, vol. 2, I think).

Now let $Y$ be another locally compact Hausdorff topological space. From Kothe, vol. 2, I know that $C(X \times Y)$ is naturally isomorphic to $C(X) \hat \otimes C(Y)$, where $\hat \otimes$ denotes the completion with respect to the injective tensor product topology.

I believe that pointwise multiplication $C(X) \times C(X) \rightarrow C(X)$ extends (uniquely) to a continuous linear map from $C(X) \hat \otimes C(X)$ to $C(X)$.

If $f: X \rightarrow Y$ is a continuous function, then precomposition with $f$ yields a continuous $C$-algebra homomorphism $f^\ast: C(Y) \rightarrow C(X)$.

I believe the following to be true:

Theorem: For every continuous algebra homomorphism $\phi: C(Y) \rightarrow C(X)$, there exists a unique continuous map $f: X \rightarrow Y$ such that $\phi = f^\ast$.

In other words, I wish that $C( \bullet )$ is a faithful functor from the category of locally compact Hausdorff spaces and continuous maps to the category of rings in the (symmetric monoidal under $\hat \otimes$) category of complete locally convex topological vector spaces and continuous linear maps.

Any references and/or corrections would be very welcome!

But an important note: I am not looking for well-known modifications, like "try the $C^\ast$-algebra instead" or the von Neumann algebra, etc.. I have good reasons for considering the ring $C(X)$ with the compact-open topology, and I don't wish to mess with it.

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There's a book called "Rings of continuous functions" by Gillman and Jerison that I looked at ages ago. I'm not sure if it has what you want, but it might be worth a look. –  Ramsey Apr 28 '11 at 21:21
I've heard of it, from looking here on MO. I requested it from the library, but it's going to take at least a few days to get from inter-library loan. And google books doesn't have it :( –  Marty Apr 28 '11 at 21:26
Try Library Genesis : gen.lib.rus.ec/book/… –  Somnath Basu Apr 28 '11 at 21:37
Note, $C(X)$ is a locally $C^*$ algebra, which naturally has the compact-open topology. –  George Lowther Apr 28 '11 at 22:11
@Marty: It is a nice book. The recent "Super real fields" by Dales and Woodin is in a sense a direct follow up. –  Andrés Caicedo Apr 28 '11 at 22:12

Let $X=Y_1\sqcup Y_2$, with both $Y_i$ homeomorphic to $Y$. Then $C(X)=C(Y_1)\oplus C(Y_2)$. Given $a\in C(Y)$ let $\phi\colon a\mapsto a\oplus 0$, in the obvious way. This $\phi$ cannot be any $f^*$, since $f^*$ would necessarily map $1\mapsto 1\oplus 1$. I believe, this is a counter example to your putative theorem, which shows that you may want a connectedness hypothesis on your spaces.
The strongest relevant result from that book is Theorem 10.8, which states that a homomorphism $\mathfrak{s}\colon C(Y)\to C(X)$ determines a unique continuous map $\tau\colon E\to \upsilon Y$ with the properties like what you want. Here $E$ is a clopen subset of $X$ and $\upsilon Y$ is the (Hewitt) realcompactification of $Y$, which is a bigger space than $Y$. See the book for full details. The hypotheses on $X$ and $Y$ (implicitly) include complete regularity, which is weaker than local compactness. Note that the homomorphism $\mathfrak{s}$ is not assumed to be continuous in any topologies on $C(Y)$ and $C(X)$. Perhaps your continuity requirement is enough to cut $\upsilon Y$ down to $Y$ and give you the desired result.
The proof finishes: A unital continuous ring homom $\phi$ from $C(Y)$ to $C(X)$ determines a unique continuous map $f$ from $X$ to $vY$ such that for all $g \in C(Y)$, and all $x \in X$, $g^v(f(x)) = (\phi(g))(x)$ by Thm 10.8 of Gillman-Jerison (here $g^v$ is the extension of $g$ to $vY$). The map $ev_x \circ \phi$, sending $g \mapsto (\phi(g))(x)$ is continuous since $\phi$ is continuous and evaluation at $x$ is continuous. All such continuous functionals on $C(Y)$ arise from evaluation at some (unique) $y \in Y$, when $Y$ is loc.cpt. (R.E.Edwards,1957,Mathematika). So $f(x) = y \in Y$. –  Marty Apr 29 '11 at 6:40
Great to hear that you've found the answer. A tiny comment though is that Gillman and Jerison insist that the realcompactification notation $\upsilon$ is an upsilon and not a v. :-) –  Igor Khavkine Apr 29 '11 at 7:15