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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.

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  • $\begingroup$ Which parts of this have you been able to prove yourself? What have you tried so far? $\endgroup$
    – Yemon Choi
    Commented Jul 27, 2020 at 4:40
  • $\begingroup$ Also, something looks wrong with your notation. What is $\tilde{\phi\vert_Z}$ supposed to be? $\endgroup$
    – Yemon Choi
    Commented Jul 27, 2020 at 4:53
  • $\begingroup$ @YemonChoi: I got confused with the proof. Could not get much idea of the proof. I'm trying it. Regarding your second comment: fixed notation $\endgroup$
    – Math Lover
    Commented Jul 27, 2020 at 5:06
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    $\begingroup$ The linked paper has an important condition which is missing from the question: $\phi$ is supposed to be a 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
    Commented Jul 27, 2020 at 13:09
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    $\begingroup$ The restated version of the question finally makes sense (it would have saved everyone time if a link to the relevant paper had been posted in the original question). Without claiming to immediately see a full proof, it should be noted that fo a commutative ${\rm C}^*$-algebra primitive ideals are the same as maximal ideals, and indeed the primitive ideal space corresponds naturally to the Gelfand spectrum. So the map $\Phi$ admits a very concrete description $\endgroup$
    – Yemon Choi
    Commented Jul 27, 2020 at 19:47

1 Answer 1

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Basically it is a theorem about commutative unital C$^*$-algebras (Vesterstrom also has a blanket assumption that $A$ and $B$ are unital).

We have a map $\phi: Z\to Z'$. So $\phi(Z)$ is a C$^*$-subalgebra of $Z'$, and $\phi(Z)=Z'$ if and only if $\phi(Z)$ separates the points of ${\rm Prim}(Z')$. For $J\in{\rm Prim}(Z')$, $\Phi(J)=\phi^{-1}(J)=\{ z\in Z: \phi(z)\in J\}$.

Hence for $J_1, J_2\in{\rm Prim}(Z')$, $\Phi(J_1)=\Phi(J_2)$ if and only if for all $z\in Z$, $\phi(z)\in J_1\Leftrightarrow \phi(z)\in J_2$, and this condition holds if and only if $\phi(Z)$ fails to separate $J_1$ and $J_2$. Thus $\phi$ is surjective if and only if $\Phi$ is injective.

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  • $\begingroup$ Sorry where did you use the fact that $A$ and $B$ are unital $C^{\ast}-$ algebras? $\endgroup$
    – Math Lover
    Commented Jul 31, 2020 at 18:06
  • $\begingroup$ I am not sure that I did, but Vesterstrom has it, so I felt safer with that hypothesis. In the non-unital case, the Stone-Weierstrass theorem requires "separates the points and vanishes nowhere", if I remember, so perhaps I did use it in line 2 of the second para. $\endgroup$
    – Douglas Somerset
    Commented Jul 31, 2020 at 19:07

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