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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.

Asvin
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  • Just to clarify: the maps are topologically continuous as well as open and group homomorphisms, I presume? – Patrick Stevens Jul 19 '16 at 11:30
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    Yes, all those conditions hold. – Asvin Jul 19 '16 at 11:30
  • Forgive me if I'm being dense, but doesn't this follow from "continuous image of a compact space is compact"? – Patrick Stevens Jul 19 '16 at 11:31
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    @PatrickStevens $B$ is not the image of $A$ (generally) – anon Jul 19 '16 at 11:31
  • A stupid remark : if $B$ is connected then $A=B$ so in particular $B$ is compact. –  Jul 19 '16 at 11:35
  • I think the maps aren't open in your example – Asvin Jul 19 '16 at 11:37
  • I'm not familiar with topological groups, but a splitting condition could work? Also demanding that the maps are open looks like demanding that an inverse (if it exists) is continuous, so splitness seems like a natural condition. – Mathematician 42 Jul 19 '16 at 11:49
  • @N.H. In that case you'll have $A \cong B$, not $A=B$. This is a common mistake for beginners. – Nitrogen Jul 19 '16 at 12:02
  • @Nitrogen : could you explain the difference to me please ? I tried to google it but I couldn't find anything relevant in the litterature. Asvin : if the first map is open, then the image of $A$ is compact and open. Assuming $B$ Hausdorff the image of $A$ is open and closed, so equal to $B$ if we assume that $B$ is connected. Therefore, the second map is simply the projection to a point. Both maps are open. (But this example isn't really interesting I agree). –  Jul 19 '16 at 12:13
  • I was actually replying to someone else who deleted his comment, anyway I think the distinction between isomorphic and equal is simply that we don't think of A as a subset of B, just isomorphic to some subset. It doesn't really make a difference in this context. – Asvin Jul 19 '16 at 12:14
  • N.H also there is a simpler proof, open subgroups are always closed so if B is connected, A is either trivial or all of B. To see that open subgroups are closed, nor that the costets of the subgroup are all open, so it's their union which is exactly the complement of our subgroup. – Asvin Jul 19 '16 at 12:16

1 Answers1

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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?

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