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Oct 4, 2016 at 19:45 comment added A Rock and a Hard Place Hey, I asked a follow-up question about the possibility of extending this to (commutative) ring spectra. If you have the time, I would be very grateful if you would take a look at that as well. Cheers!
Oct 3, 2016 at 13:38 comment added Denis Nardin In fact you can say more: if $E$ is any parametrized spectrum, $\mathrm{Map}(E,E'_X) = \mathrm{Map}(\mathrm{colim}_x E_x, E')$ (with exactly the same proof as above).
Oct 3, 2016 at 1:39 vote accept A Rock and a Hard Place
Oct 3, 2016 at 1:34 history answered Denis Nardin CC BY-SA 3.0