Group homomorphisms and maps between function spaces

Question

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).

Answer 1

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. $\square$

Answer 2

Inspired by Victor's idea:

Setup: Let $\theta:G\rightarrow H$ be a continuous dense range map between locally compact (Hausdorff) spaces. Let $G'=\theta(G)$ with the subspace topology from H. Let $\pi:C^b(H) \rightarrow C^b(G)$ be the pull-back of $\theta$.

Lemma: The collection of sets of the form $f^{-1}((1/2,\infty))$, where $f$ is a continuous map $G\rightarrow [0,1]$ vanishing at infinity, is a base for the topology on $G$.

Proof: Let $G_\infty$ be the one-point compactification, let $s\in G$, and let $U\subseteq G$ be open with compact closure (which is okay, as $G$ is locally compact) with $s\in U$. By Urysohn, there exists a continuous $f:G_\infty\rightarrow [0,1]$ with $f(s)=1$ and $f|_{G_\infty\setminus U} \equiv 0$. Then $f|_G$ vanishes at infinity, and $s\in f^{-1}((1/2,\infty)) \subseteq U$. Clearly every open set can now be written as a union of these special open sets. QED.

Claim: Suppose that $\pi$ is surjective (so $\pi$ is infact a bijection). Let $\phi:G\rightarrow G'$ be $\theta$, considered as having codomain $G'$. Then $\phi$ is a homeomorphism, and so $\theta(G)$ is open in $H$.

Proof: We show that $\phi$ is open. Let $f\in C^b_{\mathbb R}(G)$, and let $g=\pi^{-1}(f)\in C^b(H)$, so that $g(\theta(s)) = f(s)$ for $s\in G$. Let $U=f^{-1}((1/2,\infty))$ so $\theta(U) = \{ t\in H : \exists s, \phi(s)=t, f(s)>1/2 \}$ $= \{ t\in G' : f(\phi^{-1}(s))>1/2\}$ $= \{ t\in G' : g(t)>1/2 \} $ $= G' \cap g^{-1}((1/2,\infty))$. Thus $\phi(U)$ is open, as $G'$ has the subspace topology. By the lemma, this does show that $\phi$ is open. Then $G'$ is itself locally compact, and so as $G'$ is dense, it must be open in $H$. QED.

Claim: If additionally $G$ and $H$ are groups and $\theta$ a homomorphism, then $\theta$ is a surjection.

Proof: An open subgroup is closed. QED.

If this is all correct, then I'd be a little surprised if this wasn't known (say, the stuff not about groups). Any ideas???