Let $A$ and $B$ be $C^{\ast}$-algebras with centers $Z$ and $Z'$ respectively. Let $\phi:A \to B$ be surjective $C^{\ast}$-morphism. Then

$\phi(Z)=Z'$ if and only if the map $\Phi: \operatorname{Prim}(Z') \to \operatorname{Prim}(Z)$ defined as $\Phi(J) = \phi^{-1}(J)$ is injective.

Can someone please give me reference for the above result?

The above result is mentioned without proof in the paper titled On the homomorphic image of Center of $C^{\ast}$-algebras by Vesterstrom.

surjective$*$-homomorphism! This takes care of Jamie Gabe's objection (without $\phi$ being surjective, indeed $\Phi$ makes no sense). I am following Blackadar's book, II.6.5.4, if $J=\ker\phi$ then $B\cong A/J$ and $\newcommand{\prim}{\operatorname{Prim}}\prim(A/J) \cong \{K\in\prim(A) : J\subseteq K \}$. I think to understand Proposition 1 you'll need to know about the Dauns-Hofman Theorem. $\endgroup$ – Matthew Daws Jul 27 '20 at 13:096more comments