Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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. –  Andres Caicedo Apr 28 '11 at 22:12
show 1 more comment

1 Answer

up vote 1 down vote accepted

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.

For more general information, I second Ramsey's recommendation of "Rings of continuous functions" by Gillman and Jerison. Though, I don't think it has the exact theorem you are looking for.

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.

share|improve this answer
When I say "rings", I mean unital rings (and unital homomorphisms). I definitely don't want any connectedness hypothesis. But thank you for the Theorem 10.8 reference. I'll check it out this evening. –  Marty Apr 29 '11 at 0:07
Theorem 10.8 (with your comments) goes most of the way towards what I want. Thank you again! I think I'll check in some earlier papers of Hewitt, and see if I can finish off the proof. –  Marty Apr 29 '11 at 3:14
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
Thank you for getting this started. Perhaps I'll use an upsilon for the first time since Greek class. –  Marty Apr 29 '11 at 14:53
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.