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).

  • $\begingroup$ A further thought: the map $\pi$ can be thought of as $C(\beta H)\rightarrow C(\beta G)$ and is hence associated to some continuous map $\tilde\theta:\beta G\rightarrow\beta H$. Now, it's pretty easy to see that $\pi$ is surjective if and only if $\tilde\theta$ is injective. Furthermore, $\tilde\theta$ extends $\theta$. So, is it possible for $\theta$ to inject without closed range, and for $\tilde\theta$ to inject? $\endgroup$ Jun 20, 2010 at 9:03

2 Answers 2


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. $\square$

  • $\begingroup$ If you take a point x in Q, and some compact interval [a,b] in R which contains Q, then every open cover of [a,b] cap Q is dense in an open cover of [a,b], which has a finite subcover, so the associated subcover over Q is finite, too (and still a cover). So Q with the induced topology is still locally compact. Right? $\endgroup$ Jun 20, 2010 at 9:24
  • $\begingroup$ Matthew: You are right, this simple trick doesn't work under the local compactness assumption. To get a "bad" open cover, just consider the complements of shrinking closed neighborhoods of an irrational point (this is equivalent to your Baire space argument). $\endgroup$ Jun 20, 2010 at 10:16
  • $\begingroup$ Transferred comments to the proof. $\endgroup$ Jun 20, 2010 at 15:00
  • $\begingroup$ I've deleted some comments which no longer apply. This looks promising. I was worried for a bit as to why G' being locally compact implies that it was closed in H. However, I think this is because of the following. WLOG, we may suppose that G' is dense (and it has the subspace topology). By secure.wikimedia.org/wikipedia/en/wiki/… we have that G' is thus open in H, but as G' is a subgroup, it must also be closed, and hence equal to H as required. $\endgroup$ Jun 20, 2010 at 16:47
  • $\begingroup$ Unless I get a massively inspired answer, I'll accept Victor's, as it's nice, and gave me a lot of inspiration... $\endgroup$ Jun 20, 2010 at 20:51

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???

  • 1
    $\begingroup$ I think this works and I'd also be surprised if it weren't known. Where have you looked? By the way, the fact that a locally compact subgroup of a Hausdorff top group must be closed is in Bourbaki, Top gen, 3rd ed, III-3-3, Cor 2 and the fact that a locally compact subset of a Hausdorff top space is locally closed is I-9-7, Pr 12. $\endgroup$ Jun 21, 2010 at 6:25
  • $\begingroup$ I haven't looked very far: it's been the weekend, so I'm away from my books! But I'm not really sure where to look... such results for compact spaces seem to be folklore (and are easy to prove) whereas the generalisation to the locally compact case seem trickier, but equally folklorish. If anyone has an hints for sources in the literature, I'd be grateful... $\endgroup$ Jun 21, 2010 at 8:16
  • $\begingroup$ Oh, there isn't a shortage of places to look, starting with books on abstract harmonic analysis and $C^*$-algebras. A.Weil and Bourbaki certainly were known for emphasizing local compactness, but I've scarcely ever opened the analytic volumes. In fact, as I was looking over the proof of Urysohn's lemma there, which I cannot remember reading before, it occurred to me that your lemma may appear along the way. $\endgroup$ Jun 22, 2010 at 1:34

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