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Jan 13, 2016 at 22:08 comment added Bruno Stonek I've found this note by Cary Malkiewich which is quite enlightening.
Jun 13, 2014 at 12:24 vote accept Bruno Stonek
Apr 8, 2014 at 12:22 comment added Bruno Stonek Thank you all for your answers, they are all very enlightening! It's hard for me to accept just one.
Apr 6, 2014 at 0:22 answer added Tom Goodwillie timeline score: 21
Apr 5, 2014 at 18:45 answer added Ben Wieland timeline score: 6
Apr 5, 2014 at 16:59 answer added Peter May timeline score: 8
Apr 5, 2014 at 13:22 answer added Denis Nardin timeline score: 6
Apr 2, 2014 at 14:48 comment added Mark Grant At the level of spaces, not every map $f:X\to Y$ is a fibre inclusion (in particular, the homotopy fibre would have to be homotopy equivalent to a loop space $\Omega Z_f$). So probably the fact that they are working with spectra matters here.
Apr 2, 2014 at 13:56 history migrated from math.stackexchange.com (revisions)
Feb 21, 2014 at 12:44 comment added Bruno Stonek @ZhenLin: I'm sorry, I'm a beginner in this subject. Would you care to upgrade your comment into an answer? Thanks
Feb 21, 2014 at 12:08 comment added Zhen Lin This has to do with questions of homotopy coherence. The homotopy fibre has a universal property with respect to maps with a specified nullhomotopy, etc.
Feb 21, 2014 at 11:25 history asked Bruno Stonek CC BY-SA 3.0