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Timeline for Motivation for algebraic K-theory?

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Mar 31 at 20:39 history made wiki Post Made Community Wiki by Stefan Kohl
Dec 25, 2011 at 12:47 comment added Justin Noel Since there is now a paper on this, I would like to add that one can read about the universal characterization of higher K-theory that Tyler is discussing here: arxiv.org/abs/1001.2282
Jul 16, 2011 at 9:50 comment added Thomas Riepe Goncalo Tabuada's description of this talk mentions "the first conceptual characterization of Quillen's higher K-theory since Quillen's foundational work". Does anyone know where one could read about that?: mpim-bonn.mpg.de/node/3501
Mar 16, 2010 at 3:40 comment added Tyler Lawson That's the Barrat-Priddy-Quillen theorem. See Barrat-Priddy, "On the homology of non-connected monoids and their associated groups", and others reference Quillen, "The group completion of a simplicial monoid", Appendix Q in Friedlander-Mazur's "Filtrations on the homology of algebraic varieties". Also see Segal, "Configuration-spaces and iterated loop-spaces" for another take.
Mar 15, 2010 at 21:21 comment added roger123 Nice to read! "The K-theory of the category of finite sets captures stable homotopy groups of spheres." I often heard this but never seen a precise statement and a proof. Can anybody provide me a reference?
Nov 25, 2009 at 3:26 comment added Mariano Suárez-Álvarez It is universal in a specific sense. See the thesis of Gonçalo Tabuada.
Nov 18, 2009 at 13:30 comment added Tyler Lawson I wasn't talking specifically about Quillen's higher K-theory functor, but about something homotopy equivalent, which produces a universal functor from symmetric monoidal categories to spectra that converts the symmetric monoidal structure into addition.
Nov 18, 2009 at 9:10 comment added Shizhuo Zhang You mean Quillen Higher K-functor is universal? I doubt it very much, because Quillen never proved that. In fact, A.Rosenberg developed its a derived approach K theory which is universal and enjoy all the property of Quillen higher K-functor,moreover, it is universal.
Oct 15, 2009 at 1:46 vote accept S. Carnahan
Oct 15, 2009 at 1:43 history answered Tyler Lawson CC BY-SA 2.5