Hm, shouldn't it be the case that, given a symmetric monoidal infinity category, and an E_n-algebra R, the "free R-module functor: C-->LMod_R is E_{n-1}-monoidal? Does anyone know where this would be in Lurie?
Hah, well, Rune knows the answer to all my questions.
Here's another question: given a space $X$, we can look at the map $Y\to X$ that takes all of $Y$ to some point in $X$. What's (simplicial) presheaf $X\to Top$ associated to this? The one that takes the target of $Y$ to $Y$ and everything else to... the empty space?
I guess, to make sure that $Y\to X$ is a fibration, we need to assume that $Y$ is a Kan complex.
The cartesianness condition over a groupoid implies that to be a grothendieck fibration if x~y in the base, they have equivalent fibres and are therefore locally constant. Also I think it means that every categorical fibration over a Kan complex is a bicartesian fibration, if I am not mistaken
@HarryGindi yeah, right. good point. I was sort of thinking about it incorrectly. and I kept thinking "but if it's anything deserving of the name fibration, the fibers should be equivalent..." but that's the way to explain that. thanks to you and @Arpon!
@JonathanBeardsley in case you didn't resolve this yet, i believe (under hypotheses on C) this is 4.8.5.16–17 (using that E_n-Alg(C) = Alg(Alg(...(Alg(C)))))
what I mean is, what is the subcategory generated by HZ under colimits and desuspensions. Certainly this category consists of HZ-local objects (and contains all bounded above spectra), but does it contain everything?
If R is a non-zero (homotopy) ring spectrum such that R \wedge HZ = 0, then F(HZ, R) = 0 and consequently R is not in the localizing subcategory generated by HZ. Apparently R=K(1) is an example.
@TomBachmann it's definitely not all spectra. everything in that localizing category has trivial homology for all the Morava K-theories, and has a bunch of other goofy properties like not having any nontrivial maps to a finite p-torsion complex
@TylerLawson indeed. I discover time and again that I do not know enough "classical" homotopy theory. Do you also happen to know the answer for spaces?
It's a perfectly natural question with a tricky answer. For spaces everything is easier because truncated spaces like K(A,0) or K(Z,1) generate everything
@TylerLawson clearly every space is a colimit of 0-truncated spaces. But is also every space a filtered colimit of truncated (clearly not n-truncated for some fixed n) spaces?
