Let $A$ be a $C^{\ast}$-algebra and $\widehat A$ be the space of all non-zero multiplicative linear functionals on $A$ equipped with the Gelfand topology. Let $f \in C \left (\widehat {A} \right ),$ the space of all complex valued continuous functions on $\widehat {A}.$ Let $\varphi_{0}$ denote the zero linear functional on $A.$ Consider the function $\overline {f} : \widehat {A} \cup \left \{\varphi_{0} \right \} \longrightarrow \mathbb C$ by $$\overline {f} (\varphi) = \begin{cases} f(\varphi) & \varphi \in \widehat {A} \\ 0 & \varphi = \varphi_{0} \end{cases}$$ If $\widehat {A}$ is compact then show that $\overline {f}$ is continuous on $\widehat {A} \cup \left \{\varphi_{0} \right \}.$
I am trying to show that inverse image of a closed set under $\overline {f}$ is closed. Although I can able to show it for closed subsets of $\mathbb C$ not containing $0$; I find it difficult to show when I consider a closed set containing $0.$ Could anyone please help me in this regard?
Thanks a bunch.
You don't say what topology you consider on $\tilde A=\widehat A\cup\{\varphi_0\}$. Assuming you are considering the weak$^*$ topology on $\tilde A$, this is a Hausdorff topology, so compact sets are closed. As $\widehat A$ is compact you have that $\varphi_0$ is a point on the complement of $A$, which is open. So you can define $\bar f$ to be whatever you want at $\varphi_0$, and it will still be continuous.