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S Apr 25, 2016 at 11:25 history suggested CommunityBot
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S Apr 25, 2016 at 11:25
Apr 20, 2016 at 3:43 answer added Igor Belegradek timeline score: 8
Apr 19, 2016 at 15:04 comment added Mikhail Katz @Igor, in the end Yamaguchi is not needed for the solution.
Apr 19, 2016 at 14:42 history edited user21574
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Apr 19, 2016 at 13:09 answer added Mikhail Katz timeline score: 10
Apr 14, 2016 at 9:49 history edited asv CC BY-SA 3.0
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Apr 13, 2016 at 14:17 vote accept asv
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Apr 13, 2016 at 14:16
Apr 13, 2016 at 13:27 answer added Igor Belegradek timeline score: 10
Apr 13, 2016 at 3:54 comment added asv @IgorBelegradek: I agree that some reading would be very helpful to me and I would greatly appreciate a reference. I just wanted to make sure that the above reference does contain the answer to my question since I am wondering about a specific fact rather than about more general question "how one can think about these matters". Anyway many thanks for your comments.
Apr 13, 2016 at 1:30 comment added Igor Belegradek I suggest you look at the paper by Shioya-Yamaguchi I mentioned and get the conclusion yourself. I cannot help wondering how one can think about these matters and did not know about Yamaguchi's fibration theorem. I think some reading is in order.
Apr 13, 2016 at 1:24 comment added asv @IgorBelegradek: I am not sure, what is the conclusion of the use of the collapsing theory?
Apr 13, 2016 at 1:23 comment added asv @AlexandreEremenko: Thanks, corrected.
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Apr 12, 2016 at 21:46 comment added Igor Belegradek I wish to add that if you wish to black box the collapsing theory (and you shouldn't!) you could just multiply your collapsing sequence by a circle, so that you now have a sequence of $3$-manifolds collapsing to a cylinder. Then Shioya-Yamaguchi's paper formally applies, see projecteuclid.org/euclid.jdg/1090347524. Of course this is an overkill.
Apr 12, 2016 at 20:17 comment added Alexandre Eremenko Are your surfaces compact? You did not say this. If con-compact are permitted, take cylinders $I\times S$ where $I$ is an interval and $S$ is a small circle, and equip with the standard metric of curvature $0$. You obtain a segment in the limit.
Apr 12, 2016 at 16:46 comment added Igor Belegradek If the limit is smooth (e.g. a circle), then by the Yamaguchi's fibration theorem $M_i$ is a fiber bundle over the limit (for large $i$). The only orientable closed surface that fibers over the circle is the torus (because the fiber is also a circle for dimension reasons). Reading Yamaguchi's more recent papers on low-dimensional collapse will help you understand the case when the limit is an interval.
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