Given a type one $C^\*$-algebra $A$, its center $Z$ acts by scalars on each irreducible representation space. Mapping a representation to its central character yields a continuous map from the structure space $\hat A$ to the structure space $\hat Z$.
Is this map open?



Is restriction to the center an open map?
This map is not always open. Take for example
$$
A=\Big\{f\in C([0,1],M\_2(\mathbb C)): f(0)=\begin{pmatrix}\lambda &0\\ 0&\mu\end{pmatrix},\, f(1)=\begin{pmatrix}\lambda &0\\ 0&\lambda\end{pmatrix},\, \lambda,\mu\in \mathbb C\Big\}.
$$
(Continuous functions from $[0,1]$ to $M\_2(\mathbb C)$ with endpoint conditions.)


Here the center is the functions that are scalar multiples of the identity for all $t$ and $f(0)=f(1)$. So we get $Z(A)\cong C([0,1]/(0\sim 1))\cong C(\mathbb T)$.


Besides the 2-dimensional representations of pointwise evaluation $\pi\_t$ for $0<t<1$, we have two 1-dimensional representations, $\pi\_\lambda$ and $\pi\_\mu$, that pick-up the corresponding scalars at the endpoints.


The set $\{\pi\_t:0<t<1/2\}\cup \{\pi\_\mu\}$ is open in the Jacobson topology. But the central characters induced by these representations give us a non-open arc in $\mathbb T$. We get $\{0,1\}\cup (0,1/2)$ in $[0,1]/(0\sim 1)$.


The map described in the question is some times called the complete regularization map. A relevant reference is "Quasi-standard C\*-algebras", by Archbold and Somerset. Their Theorem 2.1 lists a number of conditions equivalent to the complete regularization map being open.


