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Apr 28, 2021 at 9:57 vote accept CommunityBot
Apr 28, 2021 at 9:57 history edited user178109 CC BY-SA 4.0
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Apr 22, 2021 at 20:35 comment added HJRW @YCor: I omitted to mention that there is an extra hypthesis needed, namely that the quotient $H$ should be of type $F_3$.
Apr 22, 2021 at 15:51 history became hot network question
Apr 22, 2021 at 10:34 history rollback user178109
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Apr 22, 2021 at 8:48 comment added HJRW @YCor: the fact that they are finitely presented follows from the 1-2-3 theorem of Baumslag--Bridson--Miller--Short, but the proof doesn't give an explicit presentation.
Apr 22, 2021 at 8:47 answer added HJRW timeline score: 17
Apr 22, 2021 at 8:36 history edited user178109 CC BY-SA 4.0
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Apr 22, 2021 at 8:35 history undeleted user178109
Apr 22, 2021 at 8:21 history deleted user178109 via Vote
Apr 22, 2021 at 8:13 comment added YCor If you checked that the fiber product is finitely presentable, don't you get an explicit finite presentation?
Apr 22, 2021 at 8:10 comment added user178109 added generators of the kernel to the input
Apr 22, 2021 at 8:09 history edited user178109 CC BY-SA 4.0
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Apr 22, 2021 at 8:07 comment added YCor I haven't checked whether your condition now implies finite presentability. But what is the input then? Just asking whether the kernel is finitely generated is not algorithmically solvable in general.
Apr 22, 2021 at 8:04 comment added user178109 @YCor I edited is it correct now?
Apr 22, 2021 at 8:02 history edited user178109 CC BY-SA 4.0
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Apr 22, 2021 at 8:00 comment added YCor If $F$ is free of finite rank and $Q$ is an infinite quotient of $F$ with infinite kernel, then the fiber product $F\times_Q F$ is not finitely presentable.
Apr 22, 2021 at 7:55 history edited user178109 CC BY-SA 4.0
added 30 characters in body; edited title
Apr 22, 2021 at 7:48 history asked user178109 CC BY-SA 4.0