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Feb 24, 2010 at 6:16 history edited Reid Barton CC BY-SA 2.5
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Feb 24, 2010 at 6:06 comment added Reid Barton I don't know of a specific place where it is written down, but invertibility of an endofunctor is an (∞,1)-categorical (as opposed to (∞,2)-categorical) notion, so it's directly analogous to the situation in classical algebra.
Feb 24, 2010 at 6:03 history edited Reid Barton CC BY-SA 2.5
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Feb 24, 2010 at 5:42 comment added Dmitri Pavlov Is the operation of adjoining an inverse to an endofunctor explained somewhere in Lurie's papers or anywhere else?
Feb 24, 2010 at 5:38 vote accept Dmitri Pavlov
Feb 24, 2010 at 5:28 comment added Reid Barton I should have mentioned that I'm working in the world of presentable (∞,1)-categories. I'm pretty sure my new statement is correct.
Feb 24, 2010 at 5:23 comment added Tyler Lawson It's even worse than that, you only get the category of suspension spectra. Also, I had some other argument written here about inverting the loop functor which suffered from me accidentally getting an adjunction on the wrong side. Remember, kids: no Math Overflow late at night.
Feb 24, 2010 at 5:01 history edited Reid Barton CC BY-SA 2.5
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Feb 24, 2010 at 4:39 comment added Chris Schommer-Pries Minor quible. Don't you only get connective spectra by starting with Top and inverting the suspension functor? Also a question: If you start with Top and invert the loops functor do you also get the category of (connective) spectra?
Feb 24, 2010 at 4:25 history answered Reid Barton CC BY-SA 2.5