Let $G$ be a topological group and let $H$ be a central subgroup, i.e. $H\subset Z(G)$. Consider the inclusion $\iota:H\hookrightarrow G$.
Denote by $\widehat{G}$ the unitary dual of $G$, i.e. the set of isomorphism classes of irreducible unitary representations of $G$ equipped 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 $y_1,...,y_n\in\mathcal{V}$ satisfying $|\langle\pi(g)(x_i),x_j\rangle-\langle\rho(g)(y_i),y_j\rangle|<\epsilon$ for all $g\in K$, $i,j=1,...,n$, where $K\subset G$ is compact, $\epsilon>0$, and $x_1,...,x_n\in\mathcal{H}$.
We can get a map $\widehat{\iota}:\widehat{G}\to\widehat{H}$ by setting $\widehat{\iota}(\pi)(h)=\lambda_h^\pi1_\mathbb{C}$, where $\pi(h)=\lambda_h^\pi1_\mathcal{H}$ for some $\lambda_h^\pi\in\mathbb{C}$ by Schur's lemma, $\pi\in\widehat{G}$, $h\in H$.
1. Is $\widehat{\iota}$ continuous?
2. Is $\widehat{\iota}$ surjective?
My attempts:
Ad 1: The Fell topology on $\widehat{H}$ is just the compact-open topology if we identify $\widehat{H}$ with the Pontryagin dual of $H$. So we have to prove that for every $K\subset H$ compact, $U\subset S^1$ open, the set $\widehat{\iota}^{-1}(\{\chi\in\widehat{H}\ |\ \chi(K)\subset U\})$ is open in $\widehat{G}$. This set is just $V:=\{\pi\in\widehat{G}\ |\ \lambda_h^\pi\in U\ \forall h\in K\}$, so let $\pi_0\in V$. We have to find $K',\epsilon,x_1,...,x_n$ as above such that $\pi_0\in U_{K',\epsilon,x_1,...,x_n}\subset V$. My idea: Choose $K'=K$, $n=1$, $x_1\in\mathcal{H}$ such that $\|x_1\|=1$. Since $U$ is open, for all $h\in K$, there exist $\epsilon_h>0$ such that $B_{\epsilon_h}(\lambda_h^{\pi_0})\subset U$. Setting $y_1=x_1$, $\rho\in U_{K,\epsilon_h,x_1}$ just means $|\lambda_h^{\pi_0}-\lambda_h^\rho|<\epsilon_h$, i.e. $\lambda_h^\rho\in B_{\epsilon_h}(\lambda_h^{\pi_0})\subset U$. Now I think the problem is that this is dependent on $h$ and that I can't just take $\inf_{h\in K}\epsilon_h$ or something. Is my attempt correct so far? Is it possible to make this independent of $h$?
Edit: Since a continuous image of a compact set is compact and compact subsets of $\mathbb{R}$ admit an infimum, it would suffice to show that one can choose the $\epsilon_h$, $h\in K$, such that $K\ni h\mapsto\epsilon_h\in\mathbb{R}$ is continuous.
Ad 2: One can consider the case $H=Z(G)$ and the problem is to extend a unirrep of $Z(G)$ to $G$. By Schur's lemma, such a unirrep is just a character of $Z(G)$ but I'm not sure if that's helpful here. I'm aware of the induced representation functor, which constructs from a unirep of $Z(G)$ a unirep of $G$, but this representation is usually not irreducible, so I don't know how to extend in an irreducible way. Any ideas? Or counterexamples?
Addressing the surjectivity part of your question, the answer is yes, and the method of proof is pretty much what you indicated: induced representations!
Given a character $\varphi $ on $H$, we want to produce an irreducible representation $\pi $ of $G$, such that $$ \pi (h) = \varphi (h)I, \quad \forall h\in H. \tag 1 $$
Viewing $\varphi $ as a one-dimensional representation of $H$, let us begin by describing the representation of $G$ induced by $\varphi $, sometimes denoted $\text{Ind}_H^G(\varphi )$, but which I'd rather denote simply by $\pi $.
Consider the space $K(G)$ formed by the continuous, compactly supported functions of $G$, and let $\pi _0$ be the (algebraic) representaion of $G$ on $K(G)$ defined by $$ \pi _0(r) f |_s = f(r^{-1}s), \quad \forall r, s\in G, \ \forall f\in K(G). $$
This is sometimes called the left regular representation of $G$, but the inner product to be considered will soon change this.
We then define an inner product on $K(G)$ by $$ \langle f,g\rangle = \int_H \int_G \Delta (t^{-1}) \overline{g(t^{-1})} f(t^{-1}s) \varphi (s) \, dt\, ds, \quad \forall f,g\in K(G). \tag 2 $$
It might be interesting to notice that this may be obtained as the composition of the following processes:
Starting with $f$ and $g$, we take the convolution product $g^**f$, where $g^*(t) = \Delta (t^{-1}) \overline{g(t^{-1})}$, and $\Delta $ is the modular function of $G$,
Next we restrict $g^**f$ to $H$, obtaining a function $E(g^**f)\in K(H)$,
Finally we compute the integrated form $\tilde \varphi $ on $E(g^**f)$, leading up to $$ \int_H E(g^**f)(s) \varphi (s) \,ds, $$ which is precisely the right-hand-side of (2).
The Hilbert space of $\pi $ is then the completion of $K(G)$ relative to this inner product and the representation $\pi $ of $G$ there is obtained by extending every $\pi _0(t)$ by continuity.
The crucial point is that $\pi $ satisfies (1) because, for every $f,g\in K(G)$, and every $r\in H$, one has that $$ \langle \pi _0(r)f,g\rangle = \int_H \int_G \Delta (t^{-1}) \overline{g(t^{-1})} f(r^{-1}t^{-1}s) \varphi (s) \, dt\, ds = $$$$ = \int_H \int_G \Delta (t^{-1}) \overline{g(t^{-1})} f(t^{-1}s') \varphi (rs') \, dt\, ds' = $$$$ = \varphi (r)\int_H \int_G \Delta (t^{-1}) \overline{g(t^{-1})} f(t^{-1}s') \varphi (s') \, dt\, ds' = $$$$ = \varphi (r)\langle f,g\rangle , $$ where we performed the change of variables $s=rs'$ in the second step above.
Now, there is no guarantee that $\pi $ is irreducible, but any irreducible representation of $G$ weakly contained in $\pi $ will also satisfy (1).