Proof. The implications $(1)\Rightarrow (2)\Rightarrow (3)$ are immediate and $(2)\Rightarrow (4)$ follows from van Dantzig’s theorem.

Assume statement $(3)$ and choose $U\in \mathcal {C}_{o}(G)$ such that $\phi (U)$ is open in $H$ . The restriction

$$ \begin{align*} \phi|_U:U\to \phi(U) \end{align*} $$

is a continuous surjective homomorphism from the compact group $U$ onto the compact Hausdorff group $\phi (U)$ , and hence, the quotient map by its kernel. In particular, it is open. Let $O$ be open in $G$ and take $x\in O$ . Put

$$ \begin{align*} W=U\cap x^{-1}O. \end{align*} $$

Then, $W$ is open in $U$ with $xW\subseteq O$ and $\phi (W)$ is open in $\phi (U)$ , and hence, in $H$ because $\phi (U)$ is open there. Therefore, $\phi (xW)=\phi (x)\phi (W)$ is open in $H$ and

$$ \begin{align*} \phi(O)=\bigcup_{x\in O}\phi(x(U\cap x^{-1}O)) \end{align*} $$

is open. Hence, $\phi $ is open, giving statement (1).

For $(4)\Rightarrow (1)$ , let $O$ be open in $G$ and take $x\in O$ . Choose $U\in \mathcal B$ with $U\subseteq x^{-1}O$ . Then, $xU\subseteq O$ and $\phi (xU)=\phi (x)\phi (U)$ is open in $H$ . Hence, $\phi (O)$ is open.

If $\phi $ is open and $U\in \mathcal {C}_{o}(G)$ , then $\phi (U)$ is a nonempty compact open subgroup of $H$ , so it has positive Haar measure and is nonmeagre. This gives $(1)\Rightarrow (5)$ and $(1)\Rightarrow (6)$ .

For the two converses, recall that a subgroup with nonempty interior is open, so in each case, it suffices to find an interior point of $\phi (U)$ . Under statement $(5)$ , $\phi (U)$ is a compact, and hence measurable, subgroup of positive Haar measure. By Steinhaus’ theorem, $\phi (U)\phi (U)^{-1}$ is a neighbourhood of the identity and it equals $\phi (U)$ because $\phi (U)$ is a subgroup. Hence, $\phi (U)$ has nonempty interior. (This is the compact open subgroup case of the measure criterion of [Reference Cowling, Hofmann and Morris2, Corollary 8].) Under statement $(6)$ , $\phi (U)$ is a compact, and hence closed, subgroup of $H$ . Since locally compact groups are Baire and $\phi (U)$ is nonmeagre, it has nonempty interior. Either way, $\phi (U)$ is open in $H$ and statement $(3)$ holds.

Proof. Analogous quotient statements are known for Chabauty spaces of closed subgroups under proper quotient maps [Reference Hamrouni and Kadri5, Corollaries 2.4 and 2.7]. We give the direct verification needed here, in the relative Vietoris topology.

If $U\in \mathcal {C}_{o,N}(G)$ , then $q(U)$ is a compact open subgroup of $G/N$ . Conversely, if $V\in ~\mathcal {C}_{o}(G/N)$ , then $q^{-1}(V)$ is an open subgroup of $G$ containing $N=\ker q$ and it is compact because $q$ is proper. Thus, $q^{-1}(V)\in \mathcal {C}_{o,N}(G)$ . Since

$$ \begin{align*} q^{-1}(q(U))=UN=U \end{align*} $$

for $U\in \mathcal {C}_{o,N}(G)$ and since $q(q^{-1}(V))=V$ , the map $q_{\#}$ is bijective with inverse $V\mapsto ~q^{-1}(V)$ .

Continuity of $q_{\#}$ follows immediately from the Vietoris topology: for a basic open set ${\langle W\mid W_1,\dots ,W_n\rangle }$ in $\mathcal {K}(G/N)$ , its inverse image under $q_{\#}$ is

$$ \begin{align*} \langle q^{-1}(W)\mid q^{-1}(W_1),\dots,q^{-1}(W_n)\rangle\cap\mathcal{C}_{o,N}(G). \end{align*} $$

It remains to check continuity of the inverse. Let $V\in \mathcal {C}_{o}(G/N)$ and suppose that

$$ \begin{align*} q^{-1}(V)\in \langle O\mid O_1,\dots,O_n\rangle\cap\mathcal{C}_{o,N}(G) \end{align*} $$

for open subsets $O,O_1,\dots ,O_n$ of $G$ . Since $q^{-1}(V)\subseteq O$ ,

$$ \begin{align*} V\subseteq (G/N)\setminus q(G\setminus O). \end{align*} $$

The set $(G/N)\setminus q(G\setminus O)$ is open because $q$ is closed. Moreover, each $q(O_i)$ is open and $V\cap q(O_i)\neq \emptyset $ since $q^{-1}(V)\cap O_i\neq \emptyset $ . Hence,

$$ \begin{align*} \langle (G/N)\setminus q(G\setminus O)\mid q(O_1),\dots,q(O_n)\rangle\cap\mathcal{C}_{o}(G/N) \end{align*} $$

is a neighbourhood of $V$ and every member $L$ of this neighbourhood satisfies

$$ \begin{align*} q^{-1}(L)\in \langle O\mid O_1,\ldots,O_n\rangle\cap\mathcal{C}_{o,N}(G). \end{align*} $$

Thus, $V\mapsto q^{-1}(V)$ is continuous. For related quotient maps in Chabauty spaces of locally compact abelian groups, see [Reference Cornulier1].

Proof. Since $\phi $ is open, $\phi (U)\in \mathcal {C}_{o}(H)$ whenever $U\in \mathcal {C}_{o}(G)$ . Being a restriction of the image map on compact subsets, $\phi _{\#}$ is continuous: for a basic Vietoris set $\langle O\mid ~\!\!O_1,\dots ,O_n\rangle $ in $\mathcal {K}(H)$ , its inverse image under $\phi _{\#}$ is

$$ \begin{align*} \langle \phi^{-1}(O)\mid \phi^{-1}(O_1),\dots,\phi^{-1}(O_n)\rangle\cap\mathcal{C}_{o}(G). \end{align*} $$

This direct verification is the standard continuity of the induced hyperspace map in the Vietoris topology, as in [Reference Donjuán, Jonard-Pérez and López-Poo3, Reference Michael9].

Put $N=\ker \phi $ . Since $\phi $ is open and surjective, the induced bijective morphism $\bar \phi :G/N\to H$ , with $\phi =\bar \phi \circ q$ and $q:G\to G/N$ the quotient map, is an isomorphism of topological groups and such an isomorphism carries spaces of compact open subgroups homeomorphically onto one another, since the image map of a homeomorphism is a homeomorphism for the Vietoris topology. Write ${\bar \phi _{\#}:\mathcal {C}_{o}(G/N)\to \mathcal {C}_{o}(H)}$ , $\bar \phi _{\#}(W)=\bar \phi (W)$ , for the map it induces. Since $N$ is compact, Proposition 3.2 gives the homeomorphism $q_{\#}:\mathcal {C}_{o,N}(G)\to \mathcal {C}_{o}(G/N)$ and composing inverses gives a continuous map

$$ \begin{align*} s:=q_{\#}^{-1}\circ\bar\phi_{\#}^{-1}:\mathcal{C}_{o}(H)\to \mathcal{C}_{o}(G),\quad s(V)=q^{-1}(\bar\phi^{-1}(V))=\phi^{-1}(V), \end{align*} $$

with image $\mathcal C_{o,N}(G)$ . Here, $\phi ^{-1}(V)$ is compact because $\phi =\bar \phi \circ q$ is proper. Since $\phi _{\#}(s(V))=V$ for every $V\in \mathcal {C}_{o}(H)$ , the map $s$ is a continuous section and $\phi _{\#}$ is surjective. Its image $\mathcal C_{o,N}(G)$ is closed in $\mathcal {C}_{o}(G)$ . Indeed, if $L\in \mathcal {C}_{o}(G)$ does not contain $N$ , choose $n\in N\setminus L$ . Then, $O=G\setminus \{n\}$ is an open set containing $L$ and $\langle O\rangle \cap \mathcal {C}_{o}(G)$ is a neighbourhood of $L$ that misses $\mathcal C_{o,N}(G)$ .

Since $\phi _{\#}\circ s$ is the identity on $\mathcal {C}_{o}(H)$ , the restriction of $\phi _{\#}$ to $\mathcal C_{o,N}(G)$ and the map $s$ are inverse homeomorphisms. In addition, the map

$$ \begin{align*} r=s\circ\phi_{\#}:\mathcal{C}_{o}(G)\to\mathcal{C}_{o}(G) \end{align*} $$

is continuous, maps $\mathcal {C}_{o}(G)$ into $\mathcal C_{o,N}(G)$ and is the identity on that subspace: if $L=s(V)$ , then $r(L)=s(\phi _{\#}(s(V)))=s(V)=L$ . Hence, $r$ is a retraction onto $\mathcal C_{o,N}(G)$ . This parallels the closed-subgroup retraction for proper quotient morphisms in [Reference Hamrouni and Kadri5, Proposition 2.8].

Finally, a continuous surjection that admits a continuous section is a quotient map: if $A\subseteq \mathcal {C}_{o}(H)$ has $\phi _{\#}^{-1}(A)$ open in $\mathcal {C}_{o}(G)$ , then

$$ \begin{align*} A=s^{-1}(\phi_{\#}^{-1}(A)) \end{align*} $$

is open in $\mathcal {C}_{o}(H)$ .

Proof. The implication $(1)\Rightarrow (2)$ is contained in Theorem 3.3, and $(2)\Rightarrow (3)$ follows from $\mathcal {C}_{o}(H)\ne \emptyset $ . If statement $(3)$ holds for some $V\in \mathcal {C}_{o}(H)$ , then $e_H\in V$ , so ${\ker \phi \subseteq \phi ^{-1}(V)}$ . The kernel is closed in $G$ and, hence, it is a closed subset of the compact set $\phi ^{-1}(V)$ . Thus, $\ker \phi $ is compact.