When is $A : C(X) \to C(Y)$ a composition operator?

Question

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.

Answer 1

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 $T$.

Answer 2

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.)

Answer 3

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...