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A composition operator $C\_T : C(X) \to C(Y)$ with $T \in C(Y, X)$ is defined by $C\_T f := f \circ T, f \in C(X)$.

I read in the book about Composition Operators by Singh and others that a nontrivial algebra homomorphism $A : C(X) \to C(Y)$ is a composition operator (meaning there is a $T$ with $A = C\_T$) if $A(\overline{f}) = \overline{A(f)}$ holds for all $f \in C(X)$. This is true for $X$, $Y$ compact Hausdoff spaces. The proof is not difficult if one uses the isometric isomorphism $j(X) = M(C(X))$ ($j$ mapping $X$ into the space of dirac functionals, $M$ being the spectrum of the algebra $C(X)$).

Is this still true if $X, Y$ are hemicompact k-spaces?

If not can you give a counterexample?

Def.: A topological space $X$ is hemicompact if there is a sequence $(K_n)$ of compact sets in $X$ with $\bigcup_n K_n = X$ and $K_n \subset K_{n+1}$ for all natural $n$ and if for any compact $K$ in $X$ there is an $n$ with $K \subset K_n$.

Def.: A topological space $X$ is a k-space if every subset intersecting each compact subset in a closed set is itself closed.

EDIT: As was rightfully pointed out I forgot to mention that $A$ has to be an algebra homomorphism. I have corrected this now and added the definitions of hemicompact and k-space.

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You should probably define all of your terms. –  Qiaochu Yuan Dec 2 '09 at 17:32
You must have missed something out: as written, couldn't I pick some (real-valued) measure $\mu$ on X and define $A:C(X)\rightarrow C(Y)$ by $A(f) = \mu(f) 1$. Then $A(\overline{f}) = \overline{A(f)}$ but $A$ would only be induced by a composition operator if $\mu$ were a point mass at $x_0$, with $T(y)=x_0$ for all $y\in Y$. If $A$ is also a homomorphism, then it's fine (and, AFAIK, you don't actually need it to be a $*$-homomorphism...) What is a hemicompact k-space? –  Matthew Daws Dec 2 '09 at 20:05
Seeing as my general topology is, despite Kelley's injunction, pretty shoddy - what exactly is the interest in the particular hypotheses you've chosen? For instance, is it the case that these conditions (hemicompact k-space) are the weakest which allow one to detect properties of the space X from the algebra C_R(X), in some sense? Or are you just choosing some general conditions because they seem interesting to you? –  Yemon Choi Dec 3 '09 at 3:18
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3 Answers

In this other question on mathoverflow, Eric Wofsey says that for any topological space $X$, the maximal ideals of $C(X)$ correspond to the points of the Stone-Cech compactification $\beta X$. He then says that if $C(X)/I$ is isomorphic to $\mathbb{C}$, then every continuous function on $X$ extends continuously to that point in $\beta X$. My intuition is that you'll get what you want if you can construct a proper continuous function from $X$ to the real numbers; as usual proper means that the inverse image of any compact set is compact. I don't know that you would need conditions on $Y$. I also don't know whether your conditions on $X$ yield such a function, but they look similar.

(This is not meant as a complete answer, but it is something.)

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For hemicompact k-space $X$ the space of continuous homomorphisms of algebra $C(X)$ to ℂ is $X$ (up to the obvious isomorphism). The proof can be found, for example, in H. Goldmann "Uniform Frechet Algebras". Then the same construction as for compact spaces give you the map $T$.

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Ok, thanks. I looked it up. That is a very neat trick and it answers my question, but if we drop the k-space hypothesis the original approach doesn't work anymore. I would still like to see a counterexample or alternative proof in this case. –  santker heboln Dec 4 '09 at 23:39
I don't remember if there are examples of hemicompact spaces (without k-space propery) such that $M(C(X))\neq X$ (I don't have Goldmann's book at home). Here, $M(C(X))$ is the space of all continuous homomorphisms of the algebra $C(X)$ to ℂ. It seems, that $C(M(C(X)))=C(X)$ (is it?). Then try $Y=M(C(X))$ and identity maps between $C(X)$ and $C(Y)$. –  Oleg Eroshkin Dec 5 '09 at 1:29
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I'll risk making this a post, not a commment.

I think the real numbers $\mathbb R$ are a hemicompact $k$-space. Certainly $\mathbb R = \bigcup_n [-n,n]$ and if $K\subseteq\mathbb R$ is compact, then it's bounded, hence in some $[-n,n]$. It's a k-space, for if $K\subseteq\mathbb R$ has closed intersection with all compacts, then by looking at sequences, it's easy to see that $K$ is closed.

But $\mathbb R$ is not compact, so I guess you really mean to look at $C^b(\mathbb R)$, the algebra/space of all bounded continuous functions. Is that right? If not, then it's a whole new ball game (as $C(\mathbb R)$ the space of all continuous functions is not a Banach space).

But if so, then $C^b(\mathbb R)$ has character space $\beta\mathbb R$, and we can apply Jonas's construction: just pick a point $w\in\beta\mathbb R\setminus \mathbb R$ and evaluate there. This gives an algebra homomorphism $C^b(\mathbb R)\rightarrow\mathbb C$ which is not a composition operator.

Edit: Yes, the original question was about all continuous functions on X, not just the bounded ones. My mistake...

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No I don't mean C^b(X). C(X) is not a Banach space, it's a Frechet space. –  santker heboln Dec 4 '09 at 14:29
Santker's questions is much easier for $C_b(X)$. The answer would then be no if $\beta X$ has any points that are not in $X$. –  Greg Kuperberg Dec 4 '09 at 18:57
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