Is every character of the algebra of continuous functions on a locally compact space some evaluation?

Question

Given any locally compact Hausdorff space $X$, let $C(X)$ denote the complex algebra of all complex-valued continuous functions on $X$.

Question. Given an arbitrary character (i.e. a non-zero homomorphism of algebras) $\chi : C(X) \to \mathbf{C}$, is it true that there exists a (unique) $x \in X$, such that $\chi$ is exactly the evaluation at $x$?

One can show without much difficulty that this holds for paracompact $X$. After playing around with some weird spaces like the long real line, I am now quite curious whether the above question still has an affirmative answer for all locally compact spaces. The whole question boils down to whether such a $\chi$ must or need not restrict to $0$ on the (non-unital) subalgebra $C_0(X)$ of all continuous functions vanishing at infinity.

Answer 1

If $X=\omega_1$ with the order topology, then $X$ is locally compact, but whenever $(Y,d)$ is a metric space, every continuous function $f:X\rightarrow Y$ is eventually constant, so the function $f:X\rightarrow Y$ extends to a unique continuous function $\overline{f}:\omega_1+1\rightarrow Y$.

As another example, suppose that $X$ is a discrete space whose cardinality is at least the first uncountable measurable cardinal. Then let $\mathcal{U}$ be a $\sigma$-complete non-principal ultrafilter on the set $X$. Then every function in $C(X)$ will be constant almost everywhere with respect to the ultrafilter $\mathcal{U}$, but the mapping that takes a function in $C(X)$ modulo this ultrafilter is not a point evaluation map. The ultraproduct by $\mathcal{U}$ is a homomorphism that maps $C(X)$ to $\mathbb{C}$, but the ultraproduct is not a point evaluation map.

More generally, if $X$ is a completely regular space and $K\in\{\mathbb{R},\mathbb{C}\}$, then every continuous function $f:X\rightarrow K$ extends to a unique continuous function $\overline{f}:\upsilon X\rightarrow K$ where $\upsilon X$ is the Hewitt realcompactification of $X$. If $X$ is completely regular but not realcompact, then $\upsilon X\neq X$, so there will be a ring homomorphism $\chi:C(X)\rightarrow\mathbb{C}$ that does not agree with the point evaluation mapping.