Prove a unique point satisfies $\theta f(p)=f(q)$ if $\theta: C(X)\rightarrow C(Y)$ denotes an isomorphism of complex algebra

Question

Let $X$ and $Y$ be compact Hausdorff spaces, and $\theta: C(X)\rightarrow C(Y)$ denotes an isomorphism (just in algebraic sense) of complex algebra. Let $p\in Y$, I want to show that there is a unique point $q\in X$ s.t. $$\theta f(p)=f(q), f\in C(X)$$

Answer 1

Since you're speaking about complex algebras, I assume you mean by $C(X)$ the space of continuous functions $X\to\mathbb{C}$, and that $\theta$ is a *-isomorphism, i.e., a bijective linear map that is multiplicative and satisfies $\theta(f^*)=\theta(f)^*$, where $f^*(x)=\overline{f(x)}$ for each $x\in X$. By Gelfand duality, there is a homeomorphism $h:Y\to X$ such that $\theta(f)=f\circ h$. Hence your question translates to the question whether there is a unique point $q\in X$ such that $f\circ h(p)=f(q)$. Since $h$ is a homeomorphism, it is clear that there is such a point, namely $q=h(p)$. It is also unique, since if there is another point $q'\in X$ such that $f\circ h(p)=f(q')$ for each $f\in C(X)$, we have $f(q)=f(q')$ for each $f\in C(X)$. The Stone-Weierstrass Theorem assures that functions in $C(X)$ separate points of $X$, and since $f(q)=f(q')$ for each $f\in C(X)$, it implies $q=q'$.

Answer 2

Given a compact Hausdorff space $Z$ and a point $r \in Z$, we have a natural non-zero homomorphism of complex algebras $\operatorname{ev}_r \colon C(Z) \rightarrow \mathbb{C}$ given by $\operatorname{ev}_r(f) = f(r)$ (it is non-zero by Urysohn's lemma). Start by showing that any non-zero homomorphism $\varphi \colon C(Z) \rightarrow \mathbb{C}$ is of the form $\operatorname{ev}_r$ for some unique $r \in Z$. This can be done directly using Urysohn's lemma and a partition of unity argument or by applying to the theory of $C^{*}$-algebras.

Returning to your question, given $p \in Y$, consider the non-zero homomorphism $\operatorname{ev}_p \circ \theta \colon C(X) \rightarrow \mathbb{C}$. By the above, there exists a unique $q \in X$ such that $\operatorname{ev}_p \circ \theta = \operatorname{ev}_q$ giving you the required result.

Answer 3

This question can be found on Arveson's 'A short course on spectral theory', with the same unicity assumption. I don't see why it should be unique :

Note first that $\theta$ sends non-invertible elements to non-invertible elements and invertibles to invertibles (being an isomorphism), thus $$range(f)= \sigma(f)= \sigma(\theta f)= range(\theta f)$$ If $p\in Y$ is any point, $\theta f(p)\in range(f)$ so $\exists \ q\in X: \theta f(p)= f(q)$ .

This point need not be unique, take for example the unit element, the constant function $f:x\to 1$, you have $\theta f= 1_{C(Y)}$ and for any $p\in Y$ all of the points in $X$ satisfy $1_{C(Y)}(p)= f(q)$, so it suffice to take your $X$ to have more than one element.