From: Ali Caglayan <alizter@gmail.com> To: Homotopy Type Theory <HomotopyTypeTheory@googlegroups.com> Subject: [HoTT] Hurewicz theorem in HoTT Date: Sat, 27 Jul 2019 06:18:45 -0700 (PDT) [thread overview] Message-ID: <f50d39d3-0017-4820-9c76-877760449e78@googlegroups.com> (raw) [-- Attachment #1.1: Type: text/plain, Size: 1479 bytes --] Is there any progress on proving the Hurewicz theorem in HoTT? I stumbled across this mathoverflow question: https://mathoverflow.net/questions/283199/an-abstract-nonsense-proof-of-the-hurewicz-theorem I wonder if we can adapt the following argument: Define H_n(X; Z) as [S^{n+t}, X /\ K(X, t)] for some large t and pointed space X. The Hurewicz map is induced by a generator g : S^t->K(Z, t) of H_n(S^n). Given by postcomposition with (id_X /\ g). H : [S^{n+t}, X /\ S^t] ---> [S^{n+k}, X /\ K(Z, t)] Now since X is (n-1)-connected and it can be shown that g is n-connected (an (n+1)-equivalence in the answer), then it follows that (id_X /\ g)_* is an isomorphism. The only trouble I see with this argument working is the definition of homology. Instead of having a large enough t floating around we would have to use a colimit and that might get tricky. Showing that g is n-connected is possible I think using some lemmas about modalities I can't name of the top of my head. Do you think this argument will work? Let me know what you all think. Thanks, Ali Caglayan -- You received this message because you are subscribed to the Google Groups "Homotopy Type Theory" group. To unsubscribe from this group and stop receiving emails from it, send an email to HomotopyTypeTheory+unsubscribe@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/HomotopyTypeTheory/f50d39d3-0017-4820-9c76-877760449e78%40googlegroups.com. [-- Attachment #1.2: Type: text/html, Size: 2125 bytes --]

next reply other threads:[~2019-07-27 13:18 UTC|newest]Thread overview:2+ messages / expand[flat|nested] mbox.gz Atom feed top2019-07-27 13:18 Ali Caglayan [this message]2019-07-30 15:44 ` Luis Scoccola

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