Let $A, B, C $ be abelian topological groups such that we have the following exact sequence : $$0\to A \to B \to C \to 0. $$ Assume also that A, C are compact and all the maps are open. Then it's it true that $B$ is also compact?
If this is false, I would be interested in possible ways to strengthen the hypothesis so that it is true. If it's true, I would also be interested in various ways to weaken the hypothesis. In particular I would like to get rid of the open hypothesis if possible.
First I give names to the maps by
$$ 0\to A \xrightarrow{f} B \xrightarrow{g} C \to 0. $$
If $A$ is open in $B$, then consider
$$ B = \coprod_{[x] \in B/A } x.A .$$
Now each $x.A$ is open so this is an open cover of $B$. But this means $C$ is discrete and compact, so $C$ is finite. So $B$ is covered by a finite number of open compact sets thus compact.
If $g$ is not open this is in general not true. Just consider the inclusion of $G_{\text{discrete}}$ in $G$ for any compact topological group.
By the way, if $B$ is compact, then $B/A$ is compact and so the continuous bijective homomorphism from $B/A$ to $C$ is an iso. So the $g$ is a quotient map of topological groups and thus open.
So $g$ has to be necessarily open. We also can show that $g$ is closed by this result. The interesting question is what happens when $f$ is not open?