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A topological group $G$ is called

$\bullet$ minimal if it admits no strictly weaker Hausdorff group topology;

$\bullet$ completely minimal if it is Raikov-complete in each weaker Hausdorff group topology.

It is easy to see that each Raikov-complete minimal topological group is completely minimal and each completely minimal topological group is Raikov-complete.

Problem. Is each completely minimal topological group minimal? Equivalently, is an topological group completely minimal if and only if it is minimal and Raikov-complete?

Remark. The answer to this problem is affirmative for $\omega$-narrow topological groups of countable pseudocharacter and also for Abelian topological groups.

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  • $\begingroup$ check thm 3.7 in Categorically compact topological groups by Dikranjan and Uspenskij sciencedirect.com/science/article/pii/S0022404996001399 $\endgroup$ – Uri Bader May 31 '17 at 20:24
  • $\begingroup$ @UriBader Thank you for the reference to the paper of Dikranjan-Uspenski. This theorem is the proof of the statement in Remark 2. So Problems 1,2 concerns injectivity of the homomorphisms. $\endgroup$ – Taras Banakh May 31 '17 at 22:02
  • $\begingroup$ @ThomasRot The group $\mathbb Z$ dmits an injective homomorphism with non-closed image into the circle group. $\endgroup$ – Taras Banakh May 31 '17 at 22:04
  • $\begingroup$ The definition is about $G$ which doesn't appear in the actual definition. $\endgroup$ – Asaf Karagila Jun 1 '17 at 10:57
  • $\begingroup$ @Asasf-Karagila Thank you for the remark. I have corrected this misprint. $\endgroup$ – Taras Banakh Jun 1 '17 at 12:30

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