$X\times\hat A$ homeomorphic to $\widehat{C(X,A)}$ for unital $C^*$-algebras $A$ Let $X$ be a compact Hausdorff space and $A$ a unital $C^*$-algebra, then
I know that
\begin{align*}
\Psi:X\times\hat A &\to \widehat{C(X,A)} \\
(x,[\pi]) &\mapsto [\pi\circ\epsilon_x]
\end{align*}
is a bijection where $\epsilon_x:C(X,A)\to A$ is the point evaluation $\epsilon_x(f):=f(x)$ (see J. Dixmier "$C^*$-Algebras", Corollary 10.4.4). I want to show that $\Psi$ is a homeomorphism with respect to the Hull-Kernel-Topology on the irreducible representations $\hat A$ and $\widehat{C(X,A)}$. I was able to show that $\Psi^{-1}$ is continuous, so I only need the continuity of $\Psi$.
$\newcommand\Hull{\operatorname{Hull}}$
$\newcommand\Ker{\operatorname{Ker}}$
So far I am starting with a net $(x_\lambda,[\pi_\lambda])$ that converges to some $(x,[\pi])$ and want to show that
$$\Psi(x,[\pi])\in\overline{\Psi(\{(x_\lambda,[\pi_\lambda]\}_\lambda)}=\Hull\Ker\{[\pi_\lambda\circ\epsilon_{x_\lambda}]\}_\lambda$$
To this end let $f\in\Ker\{[\pi_\lambda\circ\epsilon_{x_\lambda}]\}_\lambda$, then we have to show that $\pi(f(x))\equiv\pi\circ\epsilon_x(f)=0$. Since $f\in\ker(\pi_\lambda\circ\epsilon_{x_\lambda})$ for all $\lambda$, we know that $\pi_\lambda(f(x_\lambda))\equiv\pi\circ\epsilon_x(f)=0$ for all $\lambda$. So the continuity of $\Psi$ follows when we can show that
$$ [\forall \lambda:\pi_\lambda(f(x_\lambda))=0,\; \pi_\lambda\to\pi,\;x_\lambda\to x] \quad\Rightarrow\quad \pi(f(x))=0 $$
I think it's safe to reduce this to the simpler statement
$$ [\forall \lambda:\pi_\lambda(a_\lambda)=0,\; \pi_\lambda\to\pi,\;a_\lambda\to a\in A] \quad\Rightarrow\quad \pi(a)=0 $$
This certainly looks like it should be true, but I'm not able to show this.
For the interested reader: One can show the continuity of $\Psi^{-1}$ using nets: if $[\pi_\lambda\circ\epsilon_{x_\lambda}]\to[\pi\circ\epsilon_x]$, then one can proof separately that $[\pi_\lambda]\to[\pi]$ and $x_\lambda\to x$ by contradiction, which shows $(x_\lambda,[\pi_\lambda])\to(x,[\pi])$ . Alternatively one can show for all subsets $X_0\subset X$, $A_0\subset\hat A$, that
$$\Ker(\Psi(X_0\times A_0))=I_{X_0,A_0}:=\{f\in C(X,A)\,|\,f(X_0)\subset\ker A_0\}$$
This can be used to show for closed $X_0,A_0$, that $\Psi(X_0\times A_0)$ is also closed, which shows that $\Psi$ is closed, therefore $\Psi^{-1}$ must be continuous.
 A: $\newcommand\Hull{\operatorname{Hull}}\newcommand\Ker{\operatorname{Ker}}$Answering my own question, continuing where I left off: Assuming $\pi_\lambda\to\pi\in\hat A$, $a_\lambda\to a\in A$ and $\pi_\lambda(a_\lambda)=0$ for all $\lambda$, we want to show that $\pi(a)=0$. Because $\pi_\lambda(a_\lambda)=0$ we get
$$ \|\pi_\lambda(a)\| = \|\pi_\lambda(a)-\pi_\lambda(a_\lambda)\| = \|\pi_\lambda(a-a_\lambda)\| \le \|a-a_\lambda\|\to 0 $$
So we have $\|\pi_\lambda(a)\|\to 0$, and from this alone together with $\pi_\lambda\to\pi$ we will conclude $\pi(a)=0$ (so we can forget the $(a_\lambda)_\lambda$ now). The map
$$ \phi : \hat A \to [0,\infty), \; \rho\mapsto\|\rho(a)\| $$
is lower semicontinuous (see J. Dixmier "$C^*$-Algebras" Prop. 3.3.2). Therefore the claim follows from
$$ \|\pi(a)\| = \phi(\pi) \le \lim_\lambda\phi(\pi_\lambda) = \lim_\lambda\|\pi_\lambda(a)\| = 0 $$
I will copy the lower semicontinuity proof here: First we show that for all closed $L\subset\mathbb R$ and hermitian $a\in A$, that
$Z := \{\pi\in\hat A\,|\,\sigma(\pi(a))\subset L\}$ is closed, where $\sigma(\pi(a))$ denotes the spectrum. Assume there is some $\rho\in\overline Z$ where $\sigma(\rho(a))$ contains some $\alpha\notin L$. Then choose a continuous function $f:\mathbb R\to\mathbb R$ with $f|_L=0$ and $f(\alpha)\ne 0$. For all $\pi\in Z$ it follows that $\pi(f(a))=f(\pi(a))=0$, so $f(a)\in\Ker Z$. But $\rho(f(a))=f(\rho(a))\ne 0$, so $\rho\notin\Hull\Ker Z=\overline Z$, contradiction.
Now we show the lower semicontinuity of $\phi:\hat A\to[0,\infty)$, $\pi\mapsto\|\pi(a)\|$. Since $\|\pi_\lambda(a)\|^2=\|\pi_\lambda(a^*a)\|$, we can assume that $a\in A^+$ is positive. Let $L>0$, then
$$\phi^{-1}([0,L])=\{\pi\in\hat A\,|\,\|\pi(a)\|\le L\}\overset{a\ge 0}=\{\pi\in\hat A\,|\,\sigma(\pi(a))\subset[0,L]\}$$
is closed by the previous observation. Therefore $\phi$ is lower semicontinuous.
