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Suppose $H,G,K$ are abelian Hausdorff topological groups and $0\to H\overset{\alpha}\to G\overset{\beta}\to K\to 0$ an exact sequence of continuous homomorphisms.
If $H$ and $K$ are compact, can we conclude that $G$ is compact too?
(I know that the answer is yes if $\beta$ is open or closed, but can we drop this hypothesis?)

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Not sure, but if you can somehow prove that $G$ is $\sigma$-compact, then $\beta$ would be open, and you'd be done. –  Prahlad Vaidyanathan Sep 27 '13 at 14:05
    
I havn't thought about it completely, but what about $H=0$,$G=S^1$ with the discrete topology,K=$S^1$ with the regular topology and $\beta$ the identity? –  archipelago Sep 27 '13 at 19:51
    
Of course this is a counterexample! Please post it as an answer, I will accept it. –  Mizar Sep 27 '13 at 20:03

1 Answer 1

up vote 2 down vote accepted

What about $H=0$, $G=S^1$ with the discrete topology, $K=S^1$ with the regular topology and $\beta$ the identity?

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