Let $1 \rightarrow A \xrightarrow{a} B \xrightarrow{c} C \rightarrow 1$ be a short exact sequence of topological groups (i.e., all maps are continuous, $A = \mathrm{Ker}(c)$, and $C = \mathrm{Coker}(a)$). Suppose that $A$ and $C$ are compact Hausdorff. Does it follow that $B$ is compact?
1 Answer
You have to assume that $a$ is a homeomorphism onto its image. Indeed, if you don't then you get counterexamples with $C=1$: let $A$ be a finite group with the discrete topology and let $B$ be the same group with the indiscrete topology. You also have to assume that $c$ is a topological quotient map. Indeed, if you don't then you get counterexamples with $A=1$: let $B$ be the circle with the discrete topology and $C$ be the circle with the usual compact Hausdorff topology.
With these extra topological assumptions, everything works out fine. First note that $a(A) = \ker c$ is closed in $B$ since $C$ is Hausdorff, so the identity point of $B$ is closed as we can check it in the Hausdorff $A$ (using that $a$ is a homeomorphism onto a closed image). Hence, $B$ is Hausdorff (as for any topological group with closed identity point). This didn't use the compactness (and is also not at all interesting).
Now that $B$ is Hausdorff, we can argue with nets: if $\{x_i\}$ is a net in $B$ then we can pass to a subnet so that $\{c(x_i)\}$ converges to some $y \in C$. Writing $y = c(x)$ and replacing $x_i$ with $x_i x^{1}$ then reduces us to the case that $c(x_i) \rightarrow 1$ in $C$. But $C$ has the quotient topology from $B$ by hypothesis, so after passing to a subnet we can choose $a_i \in A$ such that $x_i a_i^{1} \rightarrow 1$ in $B$. Since $A$ is compact Hausdorff, passing to a further subnet allows us to arrange that $a_i \rightarrow a$ in $A$, so $x_i a^{1} \rightarrow 1$ in $B$; i.e., $x_i \rightarrow a$ in $B$.
QED
You may like to consider the more interesting/useful case when "compact" is relaxed to "locally compact".

1$\begingroup$ One might mention that regular monomorphisms in the category $Top$ are subspace embeddings, and regular epimorphisms in $Top$ are topological quotients, so under the categorically correct meaning of short exact sequence in the category of topological groups, one gets these desired hypotheses. $\endgroup$– Todd Trimble ♦Jul 12, 2014 at 4:38

$\begingroup$ Thanks! Could you elaborate the step where passage to a subnet allows a choice of $a_i \in A$ with $x_i a_i^{1} \rightarrow 1$, i.e., how to choose such a subnet? $\endgroup$ Jul 12, 2014 at 14:45

1$\begingroup$ @QuestionMark: Consider the poset of open neighborhoods of 1 in $B$. For each such $U$, $UA$ is open; hence, $c(U) = UA/A$ is open in $C$ and by definition of the quotient topology this gives the poset of open neighborhoods of 1 in $C$ except with possibly massive redundancies. Anyway, for each $U$ we can find $i(U)$ so that $c(x_i)\in c(U)$ for all $i\ge i(U)$. Let $J$ be the poset of pairs $(U,i)$ with $i \ge i(U)$. The natural map $J\rightarrow I$ is cofinal, and for each $j=(i,U)\in J$ we define $x_j=x_i$ and can pick $a_j \in A$ so that $x_j a_j^{1}\in U$. Now rename $J$ as $I$. $\endgroup$ Jul 12, 2014 at 15:00

$\begingroup$ Thanks. I did not recall previously that the definition of a subnet allows an indexing set that need not be a subset of the original indexing set, which caused my troubles at this step. $\endgroup$ Jul 12, 2014 at 15:17

4$\begingroup$ In case anybody is interested in this: for the locally compact case alluded to at the end of the answer one may consult Thm. 5.25 of Hewitt, Ross "Abstract harmonic analysis I". $\endgroup$ Sep 25, 2014 at 23:43