fa.functional analysis - Group homomorphisms and maps between function spaces

Inspired by Victor's idea:

Setup: Let $theta:Grightarrow H$ 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 $Grightarrow [0,1]$ vanishing at infinity, is a base for the topology on $G$.

Proof: Let $G_infty$ be the one-point compactification, let $sin G$, and let $Usubseteq G$ be open with compact closure (which is okay, as $G$ is locally compact) with $sin U$. By Urysohn, there exists a continuous $f:G_inftyrightarrow [0,1]$ with $f(s)=1$ and $f|_{G_inftysetminus U} equiv 0$. Then $f|_G$ vanishes at infinity, and $sin f^{-1}((1/2,infty)) subseteq U$. 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 $phi:Grightarrow G'$ 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 $fin C^b_{mathbb R}(G)$, and let $g=pi^{-1}(f)in C^b(H)$, so that $g(theta(s)) = f(s)$ for $sin G$. Let $U=f^{-1}((1/2,infty))$ so $theta(U) = { tin H : exists s, phi(s)=t, f(s)>1/2 }$ $= { tin G' : f(phi^{-1}(s))>1/2}$ $= { tin 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???