Let $\mathcal{A},\,\mathcal{B}$ be $\mathrm{C}^*$-algebras and $\pi:\mathcal{A}\rightarrow \mathcal{B}$ a surjective unital $*$-homomorphism. Suppose self-adjoint $f\in \mathcal{A}$ has connected spectrum $[a,b]\subset \mathbb{R}$ but $\pi(f)$ has finite spectrum and spectral decomposition in $\mathcal{B}$ [proving in case of $\mathcal{B}$ finite dimensional would be a good partial result]:
$$\pi(f)=\sum_{\lambda\in\sigma(\pi(f))}\lambda\,p_\lambda$$
Suppose that $\varphi_0\in\mathcal{S}(\mathcal{B})$ is a state such that for some $\lambda\in\sigma(\pi(f))$, $\varphi_0(p_\lambda)=1$, which implies that $\varphi_0(f)=\lambda$ and for a vector $\xi_0$ together with a unital $*$-homomorphism $\pi_0:\mathcal{B}\rightarrow B(\mathsf{H}_0)$ implementing $\varphi_0$ as a vector state:
$$\varphi_0(g)=\langle\xi_0,\pi_0(g)\xi_0\rangle \qquad (g\in\mathcal{B}),$$
we have that $\pi_0(p_\lambda)\xi_0=\xi_0$ and also $\widetilde{p_\lambda}\varphi_0=\varphi_0$ where:
$$\widetilde{p_\lambda}\varphi_0=\frac{\varphi_0(p_\lambda \cdot p_\lambda)}{\varphi_0(p_\lambda)}.$$
We can pullback $\varphi_0$ to a state on on $\mathcal{A}$ by precomposing with $\pi:\mathcal{A}\rightarrow \mathcal{B}$:
$$\varphi(f)=\varphi_0(\pi(f))\qquad(f\in\mathcal{A}).$$
The question is, do we still have something like $\varphi_0(p_\lambda)=1$ for $\varphi$? Note that in $\mathcal{B}$, $p_\lambda=\mathbf{1}_{\{\lambda\}}(\pi(f))$ is a spectral measure.
Unfortunately we cannot necessarily find a $p_\lambda\in \mathcal{A}$. Instead we must pass to the enveloping von Neumann algebra $\mathcal{A}^{**}$ to find $\mathbb{1}_{\{\lambda\}}(f)$. We must also work with the normal extension of $\varphi\in\mathcal{S}(\mathcal{A})$ to a state $\omega_\varphi\in\mathcal{S}(\mathcal{A}^{**})$.
If $\varphi_0(p_\lambda)=1$ does it follow that $\omega_\varphi(\mathbb{1}_{\{\lambda\}}(f))=1$?
I was thinking maybe that I could make a vector state out of $\varphi$ using $\pi_0\circ\pi:\mathcal{A}\rightarrow B(\mathsf{H}_0)$, i.e.
$$\varphi(f)=\langle\xi_0,\pi_0\circ\pi(f)\xi_0\rangle\qquad (f\in\mathcal{A}),$$
and then maybe showing $\mathbf{1}_{\{\lambda\}}(f)\xi_0=\xi_0$ which would probably do the trick… recall that $\mathcal{A}^{**}=\pi_U(\mathcal{A})^{''}$ so $\varphi:=\varphi_0\circ \pi$ shows up somewhere there.
Best Answer
A correspondent suggests that "Your question is not a noncommutative issue. It suffices to look at the restriction on the commutative $\mathrm{C}^*$-subalgebra generated by $f$ and by $\pi(f)$ and then everything is clear."
Following this line define $\mathcal{A}_0:=\mathrm{C}^*(f)\cong C([a,b])$, and, where $\Lambda=\sigma(\pi(f))=\{\lambda_1,\dots,\lambda_m\}$, define $\mathcal{B}_0:=\mathrm{C}^*(\pi(f))\cong C(\Lambda)$. In $\mathcal{A}_0$, $f(z)=z$ and $\pi(f)=\sum_{\lambda\in\Lambda}\lambda\,\delta_\lambda$. For $g\in \mathcal{A}_0$, $\pi(g)=\sum_{\lambda\in\Lambda}g(\lambda)\delta_\lambda$.
By assumption, implicitly using restrictions, for some $\lambda\in\Lambda$, $\varphi_0(\delta_\lambda)=1$ and so $$\varphi(g)=\varphi_0\left(\sum_{\lambda\in\Lambda}g(\lambda)\delta_\lambda\right)=g(\lambda)\Rightarrow \varphi=\operatorname{ev}_{\lambda}.$$
The enveloping von Neumann algebra $\mathcal{A}_0^{**}\cong \ell^{\infty}([a,b])\subset B(\ell^2([a,b]))$ and $\varphi$ extends to $\omega_\varphi=\operatorname{ev}_{\lambda}\in \ell^{\infty}([a,b])^*$. The spectral projection $\mathbf{1}_{\{\lambda\}}(f)$ is:
$$\ell^2([a,b])\ni \phi\mapsto \phi(\lambda)\delta_{\lambda}\in\ell^2([a,b]),$$
so that $\mathbf{1}_{\{\lambda\}}(f)$ may be identified with $\delta_{\lambda}\in\ell^{\infty}([a,b])$ and indeed $\omega_\varphi(\mathbf{1}_{\{\lambda\}}(f))=\operatorname{ev}_{\lambda}(\delta_{\lambda})=1$, as required.