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.