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If U and V are topological subgroups of a topological group G, such that G=UV, and V normal in G, we know that the map $t: \frac{U}{U\cap V}\to \frac{UV}{V}, x(U\cap V)\mapsto xV$ is a isomorphism of groups. But this map is not a homeomorphism in general, right? Does anybody have a counterexample? Which properties for U and V we need such that t is a an open map? I would imagine, if you consider $f:U\to \frac{UV}{V}, x\mapsto xV$ continuous homormophism of top. groups, $\pi_{U\cap V}:U\to \frac{U}{U\cap V}, x\mapsto x(U\cap V)$ surjective continuous quotientmap, s.t. $t\circ \pi_{U\cap V}=f$, in the case that f is open (that isn't true in general), then t has to be open? Best.

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    Not sure the tag algebraic-topology is valid here (which usually refers to algebraic invariants of topological spaces like homotopy or homology groups). Not sure it's invalid, either. –  Feb 21 '14 at 08:13
  • Ok, I remove the tag. Regards –  Feb 21 '14 at 08:16
  • Could you clarify "in general"? As in what is the condition you are dropping in saying so? I presume it's the normality of $V$ you wish to omit. Already, there isn't any normality condition on U. Then UV is probably not even a subgroup! This hopefully (and probably) would throw the door open to counterexamples! – Karthik C Feb 21 '14 at 10:01
  • I do not understand the question as stated: What quotient are you taking, left or right? Whatever the quotient is, without normality assumption, the quotient is not a group, what do you mean by an isomorphism then? Are you assuming that both U and V are normal? – Moishe Kohan Feb 21 '14 at 10:24
  • $t$ is open if and only if $f$ is open. $f$ is open if and only if for every neighbourhood $W$ of $e$ in $G$, the set $(W\cap U)\cdot V$ is a neighbourhood of $e$ in $G$. There must be a better characterisation, but I don't see it now. – Daniel Fischer Feb 21 '14 at 13:16
  • i have edited my post a littlebit, V has to be a normal subgroup of G ( so V is normal in UV). With "t is an isomorphism" that means t is an isomorphism of groups. I want something like an isomorphism theorem for topological groups and i'm searching for conditions (like compactness or something like this for U and V) such that i) t is open and ii) t is a homeomorphism . In this case it is G=UV, I dont know how important it is. –  Feb 21 '14 at 14:29
  • Yesterday a few hours later I found this http://math.stackexchange.com/questions/448303/a-counter-example-of-the-second-isomorphism-theorem-for-topological-groups –  Feb 22 '14 at 20:02

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