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?



Is an extension of compact Hausdorff topological groups compact?
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".


