Relates between states corresponding measure on $C(X)$

Question

Let $X$ and $Y$ be topological spaces and let $\pi:C(X) \mapsto C(Y)$ be a homomorphism. Let $\omega$ and $\tilde{\omega}$ be states on $C(X)$ and $C(Y)$ respectively such that $\omega=\tilde{\omega}\circ \pi$. How does the corresponding probability measures $m_{\omega}$ and $m_{\tilde{\omega}}$ related. Can we express in closed form relating with $\pi$?

Answer 1

I will assume both $X,Y$ are Hausdorff compact. If they aren't, then $C(X)$ is not a C$^*$-algebra, and I wouldn't know what a state would, or what it would be like.

Since $\pi$ is a homomorphism, there exists $h:Y\to X$, continuous, such that $\pi(f)=f\circ h$ for all $f$.

The relation $\omega=\tilde\omega\circ\pi$ gives, for any $f\in C(X)$, $$ \int_Xf\,dm_\omega=\omega(f)=\tilde\omega(\pi(f))=\int_Yf\circ h\,dm_{\tilde\omega}=\int_Xf\,d\mu, $$ where $\mu=m_{\tilde\omega}\circ h^{-1}$. As the equality occurs for all $f\in C(X)$, the two measures are equal. That is

$$ m_\omega=m_{\tilde\omega}\circ h^{-1} $$