Let G and H be locally compact groups, and let $\theta:G\rightarrow H$ be a continuous group homomorphism. This induces a \*-homomorphism $\pi:C^b(H) \rightarrow C^b(G)$ between the spaces of bounded continuous functions on H and G.


If $\theta$ is an injection with closed range, then as locally compact groups are normal, you can use the Tietze extension theorem to show that $\pi$ is a surjection.


Conversely, if $\pi$ surjects, then $\theta$ must be an injection. Need $\theta(G)$ be closed in H??


(If G and H are just locally compact spaces, and $\theta$ just a continuous map, then no: you could let G be non-compact and $H=\beta G$ the Stone-Cech compactification, with $\theta$ being the canonical inclusion. The resulting map $\pi$ is just $C^b(H) = C(\beta G) \rightarrow C^b(G) = C(\beta G)$, which is the identity, once suitably interpreted. Of course, here $\theta$ has open range, and in a topological group, an open subgroup is closed, so maybe there's hope... hence my question).


**More thoughts:** As in my comment, we can extend $\theta$ to a map $\tilde\theta:\beta G\rightarrow\beta H$ between the Stone-Cech compactifications: this induces the map $\pi:C(\beta H)\rightarrow C(\beta G)$. As these are compact, it follows that $\pi$ is surjective if and only if $\tilde\theta$ is injective. By replacing $H$ with the closure of $\theta(G)$, we may suppose that $\theta$ has dense range: this forces $\tilde\theta$ to be a bijection, and hence a homeomorphism. So is it possible for $\theta$ to be an injection with dense range, and $\tilde\theta$ a homeomorphism, but without $\theta$ being onto? For example, certainly H cannot be compact, as then $\beta G$ would be a topological group, which is possible only if $G$ is compact (I think).



Group homomorphisms and maps between function spaces
The answer is yes, at least if the group $G$ is metrizable $\iff$ $G$ is Hausdorff and has countable basis of neighborhoods of the identity element $e$. This follows from the following general statement.



> 
> **Proposition.** Let $G$ and $G'$ be topological groups, with $G$ locally compact and metrizable and $f:G\to G'$ be a continuous homomorphism such that   
> 
> (a) $f$ is a bijection; and   
> 
> (b) the induced map $f^\*:C^b(G')\to C^b(G)$ of the spaces of bounded continuous functions is surjective.   
> 
> Then $f$ is a homeomorphism.
> 


Let $G'=\theta(G)\subset H$ with the subspace topology, $f$ be the same map as $\theta$, but with codomain $G'$. Then $G'$ is also locally compact, therefore, it is closed in $H.$


**Proof.** Let $d:G\to\mathbb{R}$ be the distance to $e$. Without loss of generality, $d$ may be assumed to be bounded. Consider the function $d':G'\to \mathbb{R}, d'(y)=d(f^{-1}(y)).$ Then   



(1) $f^{\*}d'=d$;  

(2) by (a), $f^{\*}$ is injective, so $d'$ is the only pre-image of $d$ under $f^\*$; and   
(3) by (b), $d'$ is continuous.   



The open ball in $B(e,r)\subset G$ consists of all $x\in G$ such that $d(x)<r$, so $$f(B(e,r))=\{y\in G':d(f^{-1}(y))<r\}=d'^{-1}((-\infty,r))$$
is open in $G'$ by (3). Since open balls form a neighborhood basis of $e$, the map $f$ is a homeomorphism. $\square$


