Unitary dualization functor continuous? Let $G$ be a topological group and denote its unitary dual by $\hat{G}:=\{\pi:G\to\text{U}(\mathcal{H})\text{ irreducible unitary representation}\}/_\cong$. If $H$ is another topological group and $\varphi:H\to G$ is a continuous surjective group homomorphism, then we have the map $\hat{\varphi}:\hat{G}\to\hat{H}$, $\pi\mapsto\pi\circ\varphi$. This yields the contravariant dualization functor from the category of topological groups to the category of sets.
Now, I want to equip my unitary duals with the Fell topology, i.e. a neighborhood of $\pi:G\to\text{U}(\mathcal{H})$ is generated by sets of the form $U_{K,\epsilon,x_1,...,x_n}$ consisting of irreducible unitary representations $\rho:G\to\text{U}(\mathcal{V})$ such that there exist $y_1,...,y_n\in\mathcal{V}$ satsifying $|\langle\pi(g)(x_i),x_j\rangle-\langle\rho(g)(y_i),y_j\rangle|<\epsilon$ for all $g\in K$, $i,j=1,...,n$, where $K\subset G$ is compact, $\epsilon>0$ and $x_1,...,x_n\in\mathcal{H}$.
My question is: If we equip $\hat{G}$ and $\hat{H}$ with the Fell topology, is $\hat{\varphi}$ continuous? I.e. can we consider the dualization functor as a functor having the category of topological spaces as its codomain?
I tried proving this directly via the definition of continuity but, even under the assumption that my groups were abelian, I couldn't get anywhere. Any help is very much appreciated!
 A: Yes, given any continuous, surjective group homomorphism $\varphi :H\to G$, the map
$$
  \hat \varphi : \pi \in \hat G\mapsto  \pi \circ \varphi \in \hat H
  $$
is continuous for the corresponding Fell topologies.
The proof doesn't involve any big idea and the only possible difficulty is unraveling the somewhat complex description of basic
open sets in the Fell topology.
In oder to prove it, fix $\pi $ in
$\hat G$, and let us verify continuity of $\hat \varphi $ at $\pi $.  For this, let  $V$ be a neighborhood of $\hat\varphi (\pi )$.
Without loss of generality we may assume that $V$ is a basic neighborhood, namely
$$
  V=U_{\hat\varphi (\pi ),  K,\varepsilon ,x_1,...,x_n} = $$$$ = \Big \{\rho \in \hat H: \exists y_1,...,y_n\in \mathcal H_\rho , \text { such that  } $$$$
  |\langle \hat\varphi (\pi )(h)x_i,x_j\rangle -\langle \rho (h)y_i,y_j\rangle |<\varepsilon ,  $$$$
  \forall h\in K,   \ \forall i,j=1,...,n\Big\},
  $$
where $K$ is a compact subset of $H$, $\varepsilon $ is a positive real number, and $x_1,...,x_n$ are vectors in $\mathcal H_{\hat\varphi (\pi )}$,
namely the representation space of $\hat\varphi (\pi )$.
Let $L=\varphi (K)$, so that $L$ is a compact subset of $G$.  Moreover observe that $\mathcal H_\pi =\mathcal H_{\hat\varphi (\pi )}$, meaning that $\pi $ and $\hat\varphi (\pi )$ are
representations (of different groups) on the same Hilbert space.  Therefore we may view the $x_i$ as vectors in $\mathcal H_\pi $, and consequently
$$
  W:= U_{\pi , L,\varepsilon ,x_1,...,x_n}
  $$
is a neighborhood of $\pi $ in $\hat G$,
and we claim that $\hat \varphi (W)\subseteq V$.  In fact, given $\tau \in W$, choose $y_1,\ldots ,y_n$ in $\mathcal H_\tau $ such that
$$
  |\langle \pi (g)x_i,x_j\rangle -\langle \tau (g)y_i,y_j\rangle |<\varepsilon ,
  $$
for all $g$ in $L$, and all $i,j=1,...,n$.  Replacing $g$ by $\varphi (h)$, where $h$ is any element of $K$, we have that $g$
lies in $L$,  and
$$
  |\langle \pi (\varphi (h))x_i,x_j\rangle -\langle \tau (\varphi (h))y_i,y_j\rangle |<\varepsilon .
  $$
Recalling that $y_1,\ldots ,y_n\in \mathcal H_\tau =\mathcal H_{\hat\varphi (\tau )}$, we then see that $\hat \varphi (\tau ) = \tau \circ \varphi $ lies in $V$.
This proves that $\hat \varphi (W)\subseteq V$,  and hence that $\hat \varphi $ is continuous at $\pi $, as desired.
