Let $G$ be a locally compact totally disconnected group and let $\phi$ be a surjective homomorphism from $G\to H$ (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



Is the image of a locally compact totally disconnected group also locally compact and totally disconnected?
The quotient group $H$ will be totally disconnected and locally compact if it is Hausdorff (or at least [$T\_1$](https://en.wikipedia.org/wiki/T1_space)) and $H$ inherits the quotient topology. 


Let $\phi: G\to H$ be a surjective homomorphism of Hausdorff topological groups, $G$ locally compact and totally disconnected. Let $1, h \in H$, and $1\in U, h\notin U$, $U \subset H$ 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 $1\in G$ [has a basis of clopen subgroup neighborhoods](https://en.wikipedia.org/wiki/Totally_disconnected_group#Locally_compact_case), so we can find $C < G$ clopen with $C \subset \phi^{-1}(U)$. Then $C\cdot N \subset \phi^{-1}(U)$ is an open subgroup, hence $C\cdot N$ is closed (the union of its non-trivial cosets is an open complement). Moreover $\phi$ is an open map, and $1 \in \phi(C), h\in \phi(G-C\cdot N)$ 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, $\mathbb{Z}\_p/\mathbb{Z}$ has the trivial quotient topology, since $
\mathbb{Z}$ is dense. Usually [topological groups are required to be Hausdorff](https://en.wikipedia.org/wiki/Topological_group#Formal_definition), so I'm not sure if you were allowing this pathology. 


