Let $G$ be a topological group with $K \le G$ a compact subgroup such that $G/K$ is compact. Is $G$ compact, too?
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Cannot we write $G$ as the continuous image of $K \times G/K$? – Henno Brandsma Mar 21 '17 at 14:35
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@HennoBrandsma I tried that first. But you have to choose continuously representatives for the cosets of $G/K$. I don't know how to do that. – principal-ideal-domain Mar 21 '17 at 14:37
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Is the quotient a closed map? If so, use a standard result on perfect maps I proved today on this site. – Henno Brandsma Mar 21 '17 at 14:40
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@HennoBrandsma Yeah, it is a closed map. Can you link your result? – principal-ideal-domain Mar 21 '17 at 14:43
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Can you show closedness? – Henno Brandsma Mar 21 '17 at 14:50
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$q: G \rightarrow G/K$ is a perfect map: continuous, all fibres are homeomorphic to $K$, hence compact, $q$ is closed (I take this on faith from the OP). We then apply this standard fact on perfect maps (and my answer for a proof sketch) to conclude compactness of $G$.
Henno Brandsma
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The closedness is equivalent to $AK$ being closed for $A\subseteq G$ closed. For that, if $x\notin AK$, the compact set $xK$ is disjoint from $A$, so $UK\cap A=\emptyset$ for some open $U\ni x$ by the tube lemma. Hence $U\cap AK=\emptyset$, too. – Hanno Mar 21 '17 at 15:04
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@Hanno thx :) but tube lemma? that is for Cartesian products? But I think I could show it, using nets. – Henno Brandsma Mar 21 '17 at 15:13
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The multiplication map $G\times K\to G$, $(x,k)\mapsto xk$, is homeomorphic to the projection $G\times K\to G$ by means of $(x,k)\leftrightarrow (xk,k)$. The tube lemma gives the closedness of the projection, so the multiplication is closed, too. Hmm, ok, this renders the rest of the proof above unnecessary. – Hanno Mar 21 '17 at 15:17