does s.e.s 0->A->B->C->0 of profinite groups imply C=B/A and A<B topologically? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T15:39:19Zhttp://mathoverflow.net/feeds/question/117999http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/117999/does-s-e-s-0-a-b-c-0-of-profinite-groups-imply-cb-a-and-ab-topologicallydoes s.e.s 0->A->B->C->0 of profinite groups imply C=B/A and A<B topologically?Toink2013-01-03T21:57:15Z2013-01-03T21:57:15Z
<p>Assume $A, B, C$ are profinite groups and $0\to A\to B\to C\to 0$ is an exact sequence of continuous maps. Which of the following assertions follows?:</p>
<p>(i) the subspace-topology induced on $A$ via $A\hookrightarrow B$ agrees with the given one.</p>
<p>(ii) the quotient-topology induced on $C$ via $B\twoheadrightarrow C$ agrees with the given one.</p>