# Need reference for $\phi(Z)=Z'$ if and only if $\Phi: \operatorname{Prim}(Z')\to \operatorname{Prim}(Z)$ is injective

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.

• Which parts of this have you been able to prove yourself? What have you tried so far? Jul 27 '20 at 4:40
• Also, something looks wrong with your notation. What is $\tilde{\phi\vert_Z}$ supposed to be? Jul 27 '20 at 4:53
• @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 Jul 27 '20 at 5:06
• 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. Jul 27 '20 at 13:09
• 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 Jul 27 '20 at 19:47

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.
• Sorry where did you use the fact that $A$ and $B$ are unital $C^{\ast}-$ algebras? Jul 31 '20 at 18:06