Is the image of a locally compact totally disconnected group also locally compact and totally disconnected? [duplicate]

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Let $G$ be a locally compact totally disconnected group and let $phi$ be a surjective homomorphism from $Gto H$ (added later: where $H$ has the topology coinduced by $phi$). Is H also locally compact and totally disconnected? If not is there a homomorphism from such a group to a Lie group?

More generally what happens if we remove local compactness? Is total disconnectedness still preserved

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The quotient group $H$ will be totally disconnected and locally compact if it is Hausdorff (or at least $T_1$) and $H$ inherits the quotient topology.

Let $phi: Gto H$ be a surjective homomorphism of Hausdorff topological groups, $G$ locally compact and totally disconnected. Let $1, h in H$, and $1in U, hnotin U$, $U subset H$ an open subset. Then $phi^-1(U)$ is an open subset of $G$, saturated by cosets of the closed normal subgroup $phi^-1(1)= N lhd G$. Then $1in G$ has a basis of clopen subgroup neighborhoods, so we can find $C < G$ clopen with $C subset phi^-1(U)$. Then $Ccdot N subset phi^-1(U)$ is an open subgroup, hence $Ccdot N$ is closed (the union of its non-trivial cosets is an open complement). Moreover $phi$ is an open map, and $1 in phi(C), hin phi(G-Ccdot N)$ are clopen sets separating $1$ from $h$. Hence $H$ is totally disconnected.

On the other hand, if $H$ is not Hausdorff, then it might be connected. For example, $mathbbZ_p/mathbbZ$ has the trivial quotient topology, since $
mathbbZ$ is dense. Usually topological groups are required to be Hausdorff, so I'm not sure if you were allowing this pathology.

No [Update: well, the answer was no to the original question, but the OP later without comment edited the question in a way that invalidates this answer.] . Let $mu_p^infty$ be the group of $p$th-power roots of unity in $mathbf C^times$, with the topology it gets as a subset of $mathbf C^times$. This group is not locally compact, since there is no compact neighborhood of 1 in $mu_p^infty$: a nonempty compact Hausdorff space without isolated points is uncountable, while the full group $mu_p^infty$ is countable. We'll show $mu_p^infty$ with its topology inside $mathbf C^times$ is the image of $mathbf Q_p$ under a continuous homomorphism.

Set $chi colon mathbf Q_p rightarrow mu_p^infty$ by
$chi(x) = e^2pi ix_p$, where $x_p$ is the $p$-adic fractional part of the $p$-adic number $x$. This $chi$ is a homomorphism with kernel $mathbf Z_p$, which is open in $mathbf Q_p$, so $chi$ is locally constant on $mathbf Q_p$ and thus is continuous. It is surjective since for $0 leq a < p^n$ we have $e^2pi i a/p^n = chi(a/p^n)$.

Remark: as abstract groups $mu_p^infty cong mathbf Q_p/mathbf Z_p$, but as topological groups they are not the same since $mathbf Q_p/mathbf Z_p$ (with its quotient topology) has the discrete topology. The group $mathbf Q_p/mathbf Z_p$ with its discrete topology is locally compact and totally disconnected, so that would not provide a counterexample for your question.

A rather trivial counterexample: Take $H$ to be your favourite topological group that fails to be locally compact and/or totally disconnected; take $G$ to be $H$ with the discrete topology. Then $G$ is locally compact and totally disconnected (any discrete space is both), and the identity function $G to H$ is a continuous surjective homomorphism.

So in sum: Neither local compactness nor total disconnectedness is preserved under images, either separately or together.

Edit: I notice after posting that Yemon Choi already posted this in comments; but since it answers the question, I’ll leave this up unless Yemon would prefer to make it an answer himself.