fa.functional analysis - When is A : C(X) --> C(Y) a composition operator?

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 $Ksubseteqmathbb R$ is compact, then it's bounded, hence in some $[-n,n]$. It's a k-space, for if $Ksubseteqmathbb R$ 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 $betamathbb R$, and we can apply Jonas's construction: just pick a point $winbetamathbb Rsetminus mathbb R$ and evaluate there. This gives an algebra homomorphism $C^b(mathbb R)rightarrowmathbb C$ 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...