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## does s.e.s 0->A->B->C->0 of profinite groups imply C=B/A and A<B topologically?

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?:

(i) the subspace-topology induced on $A$ via $A\hookrightarrow B$ agrees with the given one.

(ii) the quotient-topology induced on $C$ via $B\twoheadrightarrow C$ agrees with the given one.

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Both assertions are true, using twice that a continuous bijection between compact spaces is a homeomorphism. – Yves Cornulier Jan 3 at 22:03
@Yves: You also need that the profinite topology is Hausdorff and the quotient topology on $B/A$ is Hausdorff for your argument to work. I don't quite see how to prove $B/A$ is Hausdorff after thinking for just a minute. – Ian Agol Jan 3 at 23:44
@Agol: I was thinking of the French sense of compact which includes Hausdorff (actually compact groups and locally compact groups are even in the English conventions often assumed Hausdorff). Now since $A$ is closed, the unit is closed in $B/A$; while a topological group is Hausdorff iff $\{1\}$is closed. – Yves Cornulier Jan 4 at 0:24
See Proposition 2.1 in this paper for Yves claim ($T_1\implies T_2$ for topological groups). math.wm.edu/~vinroot/PadicGroups/topgroups.pdf – Ian Agol Jan 4 at 6:17
Is this question just asking about profinite spaces, or does it matter that they are groups? Or should the question ask for the continuous maps to be homomorphisms? – Jonathan Kiehlmann Jan 4 at 13:51