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

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

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

If not can you give a counterexample?

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

Def.: A topological space X $X$ 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 $A$ has to be an algebra homomorphism. I have corrected this now and added the definitions of hemicompact and k-space.

A composition operator C_T : C(X) --> C(Y) with T \in C(Y, X) is defined by C_T f := f o T, f \in C(X).

I read in the book about Composition Operators by Singh and others that an operator a nontrivial algebra homomorphism A : C(X) --> 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 a compact Hausdoff space. The proof is not difficult if one uses the isometric isomorphism j(X) = M(C(X)) (j mapping X in 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 union_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.

A composition operator C_T : C(X) --> C(Y) with T \in C(Y, X) is defined by C_T f := f o T, f \in C(X).

I read in the book about Composition Operators by Singh and others that an operator A : C(X) --> 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 a compact Hausdoff space. The proof is not difficult if one uses the isometric isomorphism j(X) = M(C(X)) (j mapping X in 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?